Solution

Let $\Delta= \begin{vmatrix} x^{2}-x+1 & x-1 \\ x+1 & x+1\end{vmatrix} $

$ \begin{aligned} C_1 \to C_1 & -C_2 \\ & = \begin{vmatrix} x^{2}-2 x+2 & x-1 \\ 0 & x+1 \end{vmatrix} \\ & =(x+1)(x^{2}-2 x+2)-0 \\ & =x^{3}-2 x^{2}+2 x+x^{2}-2 x+2=x^{3}-x^{2}+2 \end{aligned} $

Solution

Let $\Delta= \begin{vmatrix} a+x & y & z \\ x & a+y & z \\ x & y & a+z\end{vmatrix} $

$ C_1 \to C_1+C_2+C_3 $

$ = \begin{vmatrix} a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z \end{vmatrix} =(a+x+y+z) \begin{vmatrix} 1 & y & z \\ 1 & a+y & z \\ 1 & y & a+z \end{vmatrix} $

(Taking $a+x+y+z$ common)

$R_1 \to R_1-R_2, R_2 \to R_2-R_3$

$ =(a+x+y+z) \begin{vmatrix} 0 & -a & 0 \\ 0 & a & -a \\ 1 & y & a+z \end{vmatrix} $

Expanding along $C_1=(a+x+y+z)|1(a^{2}-0)|=a^{2}(a+x+y+z)$

Solution

Let $\Delta= \begin{vmatrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0\end{vmatrix} $

Taking $x^{2}, y^{2}$ and $z^{2}$ common from $C_1, C_2$ and $C_3$ respectively $=x^{2} y^{2} z^{2} \begin{vmatrix} 0 & x & x \\ y & 0 & y \\ z & z & 0\end{vmatrix} $

Expanding along $R_1$

$ \begin{aligned} & =x^{2} y^{2} z^{2}[0 \begin{vmatrix} 0 & y \\ z & 0 \end{vmatrix} -x \begin{vmatrix} y & y \\ z & 0 \end{vmatrix} +x \begin{vmatrix} y & 0 \\ z & z \end{vmatrix} ] \\ & =x^{2} y^{2} z^{2}[-x(0-y z)+x(y z-0)] \\ & =x^{2} y^{2} z^{2}(x y z+x y z)=x^{2} y^{2} z^{2}(2 x y z)=2 x^{3} y^{3} z^{3} \end{aligned} $

Solution

Let $\Delta= \begin{vmatrix} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z\end{vmatrix} $

$C_1 \to C_1+C_2+C_3$

$=\lvert, \begin{matrix} x+y+z & -x+y & -x+z \\ x+y+z & 3 y & z-y \\ x+y+z & y-z & 3 z\end{matrix} .$

Taking $(x+y+z)$ common from $C_1$

$ =(x+y+z) \begin{vmatrix} 1 & -x+y & -x+z \\ 1 & 3 y & z-y \\ 1 & y-z & 3 z \end{vmatrix} $

$R_1 \to R_1-R_2, R_2 \to R_2-R_3$

$=(x+y+z) \begin{vmatrix} 0 & -x-2 y & -x+y \\ 0 & 2 y+z & -y-2 z \\ 1 & y-z & 3 z\end{vmatrix} $

Expanding along $C_1$

$ \begin{aligned} & =(x+y+z)[1 \lvert, \begin{matrix} -x-2 y & -x+y \\ 2 y+z & -y-2 z \end{matrix} .] \\ & =(x+y+z)[(-x-2 y)(-y-2 z)-(2 y+z)(-x+y)] \\ & =(x+y+z)(x y+2 z x+2 y^{2}+4 y z+2 x y-2 y^{2}+z x-z y) \\ & =(x+y+z)(3 x y+3 z x+3 y z)=3(x+y+z)(x y+y z+z x) \end{aligned} $

Solution

Let $\Delta= \begin{vmatrix} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4\end{vmatrix} $

$ \begin{aligned} C_1 \to C_1 & +C_2+C_3 \\ & = \begin{vmatrix} 3 x+4 & x & x \\ 3 x+4 & x+4 & x \\ 3 x+4 & x & x+4 \end{vmatrix} \end{aligned} $

Taking $(3 x+4)$ common from $C_1$

$ =(3 x+4) \begin{vmatrix} 1 & x & x \\ 1 & x+4 & x \\ 1 & x & x+4 \end{vmatrix} $

$ R_1 \to R_1-R_2, R_2 \to R_2-R_3 $

$ =(3 x+4) \begin{vmatrix} 0 & -4 & 0 \\ 0 & 4 & -4 \\ 1 & x & x+4 \end{vmatrix} $

Expanding along $C_1$

$ =(3 x+4)[1 \begin{vmatrix} -4 & 0 \\ 4 & -4 \end{vmatrix} ]=(3 x+4)(16-0)=16(3 x+4) $

Solution

Let $\Delta= \begin{vmatrix} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{vmatrix} $

$ \begin{aligned} & R_1 \to R_1+R_2+R_3 \\ &= \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{vmatrix} \end{aligned} $

Taking $(a+b+c)$ common from $R_1$

$ \begin{aligned} & =(a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{vmatrix} \\ C_1 \to C_1-C_2, C_2 & \to C_2-C_3 \\ & =(a+b+c) \begin{vmatrix} 0 & 0 & 1 \\ b+c+a & -(b+c+a) & 2 b \\ 0 & a+b+c & c-a-b \end{vmatrix} \end{aligned} $

Taking $(b+c+a)$ from $C_1$ and $C_2$

$ =(a+b+c)^{3} \begin{vmatrix} 0 & 0 & 1 \\ 1 & -1 & 2 b \\ 0 & 1 & c-a-b \end{vmatrix} $

Expanding along $R_1$

$ =(a+b+c)^{3}[1 \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} ]=(a+b+c)^{3} \text{. } $

Using the properties of determinants in Exercises 7 to 9, prove that:

Solution

$\quad$ L.H.S. $= \begin{vmatrix} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y\end{vmatrix} $

$R_1 \to x R_1, R_2 \to y R_2, R_3 \to z R_3$ and dividing the determinant by $x y z$.

$ =\frac{1}{x y z} \begin{vmatrix} x y^{2} z^{2} & x y z & x y+z x \\ y z^{2} x^{2} & y z x & y z+x y \\ z x^{2} y^{2} & z x y & z x+z y \end{vmatrix} $

Taking $x y z$ common from $C_1$ and $C_2$

$ =\frac{x y z \cdot x y z}{x y z} \begin{vmatrix} y z & 1 & x y+z x \\ z x & 1 & y z+x y \\ x y & 1 & z x+z y \end{vmatrix} $

$C_3 \to C_3+C_1$

$ =x y z \begin{vmatrix} y z & 1 & x y+y z+z x \\ z x & 1 & x y+y z+z x \\ x y & 1 & x y+y z+z x \end{vmatrix} $

Taking $(x y+y z+z x)$ common from $C_3$

$ \begin{aligned} & =(x y z)(x y+y z+z x) \begin{vmatrix} y z & 1 & 1 \\ z x & 1 & 1 \\ x y & 1 & 1 \end{vmatrix} \\ & =(x y z)(x y+y z+z x) \begin{vmatrix} y z & 1 & 1 \\ z x & 1 & 1 \\ x y & 1 & 1 \end{vmatrix} =0 \\ & {[\because \quad C_2 \text{ and } C_3 \text{ are identical }]} \end{aligned} $

L.H.S. $=$ R.H.S. Hence proved.

Solution

$\quad$ L.H.S. $= \begin{vmatrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y\end{vmatrix} $

$C_1 \to C_1-(C_2+C_3)$

$ = \begin{vmatrix} 0 & z & y \\ -2 x & z+x & x \\ -2 x & x & x+y \end{vmatrix} $

Taking -2 common from $C_1$

$ \begin{aligned} & =-2 \begin{vmatrix} 0 & z & y \\ x & z+x & x \\ x & x & x+y \end{vmatrix} \\ R_2 \to R_2 & -R_3 \\ & =-2 \begin{vmatrix} 0 & z & y \\ 0 & z & -y \\ x & x & x+y \end{vmatrix} \end{aligned} $

Expanding along $C_1$

L.H.S. $=$ R.H.S.

$=-2[x|-z y-z y|]=-2(-2 x y z)=4 x y z$ R.H.S.

Hence, proved.

Solution

L.H.S. $= \begin{vmatrix} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{vmatrix} $

$ \begin{aligned} & R_1 \to R_1-R_2, R_2 \to R_2-R_3 \\ &= \begin{vmatrix} a^{2}-1 & a-1 & 0 \\ 2 a-2 & a-1 & 0 \\ 3 & 3 & 1 \end{vmatrix} = \begin{vmatrix} (a+1)(a-1) & a-1 & 0 \\ 2(a-1) & a-1 & 0 \\ 3 & 3 & 1 \end{vmatrix} \end{aligned} $

Taking $(a-1)$ common from $C_1$ and $C_2$

Expanding along $C_3$

$ =(a-1)(a-1) \begin{vmatrix} a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{vmatrix} $

$ \begin{aligned} & .=(a-1)^{2}[1 \lvert, \begin{matrix} a+1 & 1 \\ 2 & 1 \end{matrix} .]] \\ & =(a-1)^{2}(a+1-2)=(a-1)^{2}(a-1)=(a-1)^{3} \text{ R.H.S. } \end{aligned} $

L.H.S. $=$ R.H.S.

Hence, proved.

Solution

L.H.S. $= \begin{vmatrix} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1\end{vmatrix} $

Expanding along $C_1$

$ \begin{aligned} =1 \begin{vmatrix} 1 & \cos A \\ \cos A & 1 \end{vmatrix} -\cos C \begin{vmatrix} \cos C & \cos B \\ \cos A & 1 \end{vmatrix} \\ +\cos B \begin{vmatrix} \cos C & \cos B \\ 1 & \cos A \end{vmatrix} \end{aligned} $

$ \begin{aligned} = & 1(1-\cos ^{2} A)-\cos C(\cos C-\cos A \cos B) \\ & +\cos B(\cos A \cos C-\cos B) \\ = & \sin ^{2} A-\cos ^{2} C+\cos A \cos B \cos C \\ & +\cos A \cos B \cos C-\cos ^{2} B \\ = & \sin ^{2} A-\cos ^{2} B-\cos ^{2} C+2 \cos A \cos B \cos C \\ = & -\cos (A+B) \cdot \cos (A-B)-\cos ^{2} C+2 \cos A \cos B \cos C \\ & \quad[\because \quad \sin ^{2} A-\cos ^{2} B=-\cos (A+B) \cdot \cos (A-B)] \\ = & -\cos (-C) \cdot \cos (A-B)+\cos C(2 \cos A \cos B-\cos C) \end{aligned} $

$ \begin{aligned} = & -\cos C(\cos A \cos B+\sin A \sin B) \\ & +\cos C(2 \cos A \cos B-\cos C) \\ = & -\cos C(\cos A \cos B+\sin A \sin B-2 \cos A \cos B+\cos C) \\ = & -\cos C(-\cos A \cos B+\sin A \sin B+\cos C) \\ = & \cos C(\cos A \cos B-\sin A \sin B-\cos C) \\ = & \cos C[\cos (A+B)-\cos C] \\ = & \cos C[\cos (-C)-\cos C] \\ = & \cos C[\cos C-\cos C]=\cos C \cdot 0=0 \text{ R.H.S. } \end{aligned} $

L.H.S. $=$ R.H.S.

Hence, proved.

Solution

Area of triangle whose vertices are $(x_1, y_1),(x_2, y_2)$ and $(x_3, y_3)$

$=\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix} $

Let $\Delta=\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix} \Rightarrow \Delta^{2}=\frac{1}{4} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix} ^{2}$

But area of equilateral triangle whose side is ’ $a$ ’ $=\frac{\sqrt{3}}{4} a^{2}$

$ \therefore \quad(\frac{\sqrt{3}}{4} a^{2})^{2}=\frac{1}{4} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} ^{2} $

$ \Rightarrow \frac{3}{16} a^{4}=\frac{1}{4} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} ^{2} \Rightarrow \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} ^{2}=\frac{3}{16} a^{4} \times 4=\frac{3}{4} a^{4} $

Hence, proved.

Solution

Let $\quad A= \begin{bmatrix} 1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2 \end{bmatrix} =0$

$C_1 \to C_1-C_2$

$ |A|= \begin{vmatrix} 1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2 \end{vmatrix} =0 $

$\Rightarrow \quad \begin{vmatrix} 0 & 1 & \sin 3 \theta \\ -7 & 3 & \cos 2 \theta \\ 14 & -7 & -2\end{vmatrix} =0$

Taking 7 common from $C_1$

$ \begin{matrix} \Rightarrow & 7 \begin{vmatrix} 0 & 1 & \sin 3 \theta \\ -1 & 3 & \cos 2 \theta \\ 2 & -7 & -2 \end{vmatrix} =0 \\ \Rightarrow & \begin{vmatrix} 0 & 1 & \sin 3 \theta \\ -1 & 3 & \cos 2 \theta \\ 2 & -7 & -2 \end{vmatrix} =0 \end{matrix} $

Expanding along $C_1$

$ \begin{matrix} \Rightarrow & 1 \begin{vmatrix} 1 & \sin 3 \theta \\ -7 & -2 \end{vmatrix} +2 \begin{vmatrix} 1 & \sin 3 \theta \\ 3 & \cos 2 \theta \end{vmatrix} =0 \\ \Rightarrow & -2+7 \sin 3 \theta+2(\cos 2 \theta-3 \sin 3 \theta)=0 \\ \Rightarrow & -2+7 \sin 3 \theta+2 \cos 2 \theta-6 \sin 3 \theta=0 \\ \Rightarrow & -2+2 \cos 2 \theta+\sin 3 \theta=0 \\ \Rightarrow & -2+2(1-2 \sin ^{2} \theta)+3 \sin \theta-4 \sin ^{3} \theta=0 \\ \Rightarrow & -2+2-4 \sin ^{2} \theta+3 \sin \theta-4 \sin ^{3} \theta=0 \end{matrix} $

$ \begin{aligned} & \Rightarrow \quad-4 \sin ^{3} \theta-4 \sin ^{2} \theta+3 \sin \theta=0 \\ & \Rightarrow \quad-\sin \theta(4 \sin ^{2} \theta+4 \sin \theta-3)=0 \\ & \sin \theta=0 \quad \text{ or } \quad 4 \sin ^{2} \theta+4 \sin \theta-3=0 \\ & \therefore \quad \theta=n \pi \quad \text{ or } 4 \sin ^{2} \theta+6 \sin \theta-2 \sin \theta-3=0 \\ & \Rightarrow \quad 2 \sin \theta(2 \sin \theta+3)-1(2 \sin \theta+3)=0 \\ & \Rightarrow \quad(2 \sin \theta+3)(2 \sin \theta-1)=0 \\ & \Rightarrow \quad 2 \sin \theta+3=0 \quad \text{ or } \quad 2 \sin \theta-1=0 \\ & \sin \theta=\frac{-3}{2} \quad \text{ or } \quad \sin \theta=\frac{1}{2} \end{aligned} $

$\sin \theta=\frac{-3}{2}$ is not possible as $-1 \leq x \leq 1$

$\therefore \quad \sin \theta=\frac{1}{2} \quad \Rightarrow \quad \sin \theta=\sin \frac{\pi}{6} \quad \Rightarrow \quad \theta=n \pi+(-1)^{n} \cdot \frac{\pi}{6}$

Hence, $\theta=n \pi \quad$ or $\quad n \pi+(-1)^{n} \frac{\pi}{6}$

Solution

Let $\quad A= \begin{bmatrix} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{bmatrix} =0$

$|A|= \begin{vmatrix} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{vmatrix} =0$

$R_1 \to R_1+R_2+R_3$

$\Rightarrow \quad \begin{vmatrix} 12+x & 12+x & 12+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{vmatrix} =0$

Taking $(12+x)$ common from $R_1$,

$ \begin{aligned} & \Rightarrow \quad(12+x) \begin{vmatrix} 1 & 1 & 1 \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{vmatrix} =0 \\ & C_1 \to C_1-C_2, C_2 \to C_2-C_3 \end{aligned} $

$ (12+x) \begin{vmatrix} 0 & 0 & 1 \\ 2 x & -2 x & 4+x \\ 0 & 2 x & 4-x \end{vmatrix} =0 $

Expanding along $R_1$

$ \begin{aligned} & \Rightarrow \quad(12+x)[1 \cdot \begin{vmatrix} 2 x & -2 x \\ 0 & 2 x \end{vmatrix} ]=0 \\ & \Rightarrow \quad x=-12 \text{ or } x=0 \end{aligned} $

Solution

If $a_1, a_2, a_3, \ldots a_r$ be the terms of G.P., then

$ a_n=AR^{n-1} $

(where $A$ is the first term and $R$ is the common ratio of the G.P. )

$ \begin{aligned} a _{r+1} & =AR^{r+1-1}=AR^{r} ; a _{r+5}=AR^{r+5-1}=AR^{r+4} \\ a _{r+9} & =AR^{r+9-1}=AR^{r+8} ; a _{r+7}=AR^{r+7-1}=AR^{r+6} \\ a _{r+11} & =AR^{r+11-1}=AR^{r+10} ; a _{r+15}=AR^{r+15-1}=AR^{r+14} \\ a _{r+17} & =AR^{r+17-1}=AR^{r+16} ; a _{r+21}=AR^{r+21-1}=AR^{r+20} \end{aligned} $

$\therefore \quad$ The determinant becomes

$ \begin{vmatrix} AR^{r} & AR^{r+4} & AR^{r+8} \\ AR^{r+6} & AR^{r+10} & AR^{r+14} \\ AR^{r+10} & AR^{r+16} & AR^{r+20} \end{vmatrix} $

Taking $AR^{r}, AR^{r+6}$ and $AR^{r+10}$ common from $R_1, R_2$ and $R_3$ respectively.

$ \begin{aligned} & AR^{r} \cdot AR^{r+6} \cdot AR^{r+10} \begin{vmatrix} 1 & R^{4} & R^{8} \\ 1 & R^{4} & R^{8} \\ 1 & R^{6} & R^{10} \end{vmatrix} \\ = & AR^{r} \cdot AR^{r+6} \cdot AR^{r+10}|0| \\ = & 0 \quad[\because \quad R_1 \text{ and } R_2 \text{ are identical rows }] \end{aligned} $

Hence, the given determinant is independent of $r$.

Solution

If the given points lie on a straight line, then the area of the triangle formed by joining the points pairwise is zero.

So, $ \begin{vmatrix} a+5 & a-4 & 1 \\ a-2 & a+3 & 1 \\ a & a & 1\end{vmatrix} $

$R_1 \to R_1-R_2, R_2 \to R_2-R_3$

$\Rightarrow \quad \begin{vmatrix} 7 & -7 & 0 \\ -2 & 3 & 0 \\ a & a & 1\end{vmatrix} $

Expanding along $C_3$

$ 1 \cdot \begin{vmatrix} 7 & -7 \\ -2 & 3 \end{vmatrix} =21-14=7 \text{ units } $

As $7 \neq 0$. Hence, the three points do not lie on a straight line for any value of $a$.

Solution

$ \begin{vmatrix} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B & \cos ^{2} C+\cos C \end{vmatrix} =0 . $

$C_1 \rarr C_1-C_2,C_2 \rarr C_2-C_3$

$\Rightarrow \left|\begin{array}{ccc} 0 & 0 & 1 \\ \cos A-\cos B & \cos B-\cos C & 1+\cos C \\ \cos ^2 A+\cos A & \cos ^2 B+\cos B & \cos ^2 C+\cos C \\ -\cos ^2 B-\cos B & -\cos ^2 C-\cos C & \end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc} 0 & 0 & 1 \\ \cos A-\cos B & \cos B-\cos C & 1+\cos C \\ \cos ^2 A-\cos ^2 B & \cos ^2 B-\cos ^2 C & \cos ^2 C+\cos C \\ +\cos A-\cos B & +\cos B-\cos C & \end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc} 0 & 0 & 1 \\ \cos A-\cos B & \cos B-\cos C & 1+\cos C \\ (\cos A+\cos B) \times & (\cos B+\cos C) \times & \\ (\cos A-\cos B) & (\cos B-\cos C) & \cos ^2 C+\cos C \\ +(\cos A-\cos B) & +(\cos B+\cos C) & \end{array}\right|=0$

Taking $(\cos A-\cos B)$ and $(\cos B-\cos C)$ common from $C_1$ and $C_2$ respectively.

$\Rightarrow \quad(\cos A-\cos B)(\cos B-\cos C) \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & 1+\cos C \\ \cos A+ & \cos B+ & \cos ^{2} C+ \\ \cos B+1 & \cos C+1 & \cos C\end{vmatrix} =0$

Expanding along $R_1$

$\Rightarrow(\cos A-\cos B)(\cos B-\cos C)\left[1\left|\begin{array}{cc} 1 & 1 \\ \cos \mathrm{A}+ & \cos \mathrm{B}+ \\ \cos \mathrm{B}+1 & \cos \mathrm{C}+1 \end{array}\right|\right]=0$

$\Rightarrow(\cos A-\cos B)(\cos B-\cos C) \begin{vmatrix} (\cos B+\cos C+1)- \\ (\cos A+\cos B+1) \end{vmatrix} =0$

$\Rightarrow(\cos A-\cos B)(\cos B-\cos C)[\cos B+\cos C+1-\cos A-\cos B-1]=0$

$\Rightarrow(\cos A-\cos B)(\cos B-\cos C)(\cos C-\cos A)=0$

$\cos A-\cos B=0$ or $\cos B-\cos C=0$

or $\cos C-\cos A=0$

$\Rightarrow \cos A=\cos B$ or $\cos B=\cos C$ or $\cos C=\cos A$

$\Rightarrow \angle A=\angle C$ or $\angle B=\angle C \Rightarrow \angle A=\angle B$

Hence, $\triangle ABC$ is an isosceles triangle.

Solution

Here, $\quad A= \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} $

$ \begin{aligned} |A| & =0 \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} -1 \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} +1 \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} \\ & =0-1(0-1)+1(1-0) \\ & =1+1=2 \neq 0 \text{ (non-singular matrix.) } \end{aligned} $

Now, co-factors,

$ \begin{aligned} & a _{11}=+ \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} =-1, \quad a _{12}=- \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} =1, \quad a _{13}=+ \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} =1 \\ & a _{21}=- \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} =1, \quad a _{22}=+ \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} =-1, \quad a _{23}=- \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} =1 \end{aligned} $

$ \begin{aligned} & a _{31}=+ \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} =1, \quad a _{32}=- \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} =1, \quad a _{33}=+ \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} =-1 \\ & Adj(A)= \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} ^{\prime}= \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \\ & \therefore \quad A^{-1}=\frac{1}{|A|} Adj(A)=\frac{1}{2} \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \\ & \text{ Now, } \quad A^{2}=A \cdot A= \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0+1+1 & 0+0+1 & 0+1+0 \\ 0+0+1 & 1+0+1 & 1+0+0 \\ 0+1+0 & 1+0+0 & 1+1+0 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} \\ & \text{ Hence, } A^{2}= \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} \end{aligned} $

Now, we have to prove that $A^{-1}=\frac{A^{2}-3 I}{2}$

$ \begin{aligned} \text{ R.H.S. } & =\frac{ \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} -3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} }{2} \\ & =\frac{ \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} }{2}=\frac{1}{2} \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \\ & =A^{-1}=\text{ L.H.S. } \end{aligned} $

Hence, proved.

Solution

Given that

$ \begin{aligned} A & = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \\ |A| & =1 \begin{vmatrix} -1 & -2 \\ -1 & 1 \end{vmatrix} -2 \begin{vmatrix} -2 & -2 \\ 0 & 1 \end{vmatrix} +0 \begin{bmatrix} -2 & -1 \\ 0 & -1 \end{bmatrix} \\ & =1(-1-2)-2(-2-0)+0 \\ & =-3+4=1 \neq 0 \text{ (non-singular matrix.) } \end{aligned} $

Now co-factors,

$ \begin{aligned} & a _{11}=+ \begin{vmatrix} -1 & -2 \\ -1 & 1 \end{vmatrix} =-3, a _{12}=- \begin{vmatrix} -2 & -2 \\ 0 & 1 \end{vmatrix} =2, a _{13}=+ \begin{vmatrix} -2 & -1 \\ 0 & -1 \end{vmatrix} =2 \\ & a _{21}=- \begin{vmatrix} 2 & 0 \\ -1 & 1 \end{vmatrix} =-2, a _{22}=+ \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} =1, a _{23}=- \begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix} =1 \\ & a _{31}=+ \begin{vmatrix} 2 & 0 \\ -1 & -2 \end{vmatrix} =-4, a _{32}=- \begin{vmatrix} 1 & 0 \\ -2 & -2 \end{vmatrix} =2, a _{33}=+ \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} =3 \\ & Adj(A)= \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix} ^{\prime}= \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix} \\ & \therefore \quad A^{-1}=\frac{1}{|A|} Adj(A)=\frac{1}{1} \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix} \\ & \Rightarrow \quad A^{-1}= \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix} \end{aligned} $

Now, the system of linear equations is given by $x-2 y=10$, $2 x-y-z=8$ and $-2 y+z=7$, which is in the form of $CX=D$.

$ \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} $

where $C= \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1\end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z\end{bmatrix} $ and $D= \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} $

$ \begin{aligned} & \because \quad(A^{T})^{-1}=(A^{-1})^{T} \\ & \therefore \quad C^{T}= \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} =A \\ & \therefore \begin{bmatrix} x \\ y \\ z \end{bmatrix} =C^{-1} D \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix} \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -30+16+14 \\ -20+8+7 \\ -40+16+21 \end{bmatrix} \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ -5 \\ -3 \end{bmatrix} \end{aligned} $

Hence, $x=0, y=-5$ and $z=-3$

Solution

Given that

$ \begin{aligned} & 3 x+2 y-2 z=3 \\ & x+2 y+3 z=6 \\ & 2 x-y+z=2 \\ & A= \begin{bmatrix} 3 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{bmatrix} \text{ and } B= \begin{bmatrix} 3 \\ 6 \\ 2 \end{bmatrix} \\ & |A|=3 \begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix} -2 \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} -2 \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \\ & =3(2+3)-2(1-6)-2(-1-4) \\ & =15+10+10=35 \neq 0 \text{ non-singular matrix } \end{aligned} $

Now, co-factors,

$ \begin{aligned} & a _{11}=+ \begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix} =5, a _{12}=- \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} =5, a _{13}=+ \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} =-5 \\ & a _{21}=- \begin{vmatrix} 2 & -2 \\ -1 & 1 \end{vmatrix} =0, a _{22}=+ \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} =7, a _{23}=- \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} =7 \\ & a _{31}=+ \begin{vmatrix} 2 & -2 \\ 2 & 3 \end{vmatrix} =10, a _{32}=- \begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} =-11, a _{33}=+ \begin{vmatrix} 3 & 2 \\ 1 & 2 \end{vmatrix} =4 \\ & Adj(A)= \begin{bmatrix} 5 & 5 & -5 \\ 0 & 7 & 7 \\ 10 & -11 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{bmatrix} \end{aligned} $

$ \begin{aligned} & \therefore \quad A^{-1}=\frac{1}{|A|} Adj(A)=\frac{1}{35} \begin{bmatrix} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{bmatrix} \\ & \text{ Now, } X=A^{-1} B \\ & \therefore \begin{bmatrix} x \\ y \\ z \end{bmatrix} =\frac{1}{35} \begin{bmatrix} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{bmatrix} \begin{bmatrix} 3 \\ 6 \\ 2 \end{bmatrix} =\frac{1}{35} \begin{bmatrix} 15+0+20 \\ 15+42-22 \\ -15+42+8 \end{bmatrix} =\frac{1}{35} \begin{bmatrix} 35 \\ 35 \\ 35 \end{bmatrix} \\ & { \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} } \end{aligned} $

Hence, $x=1, y=1$ and $z=1$.

Solution

We have, $A= \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{bmatrix} $ and $B= \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} $

$ \begin{aligned} BA & = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \\ & = \begin{bmatrix} 2+4+0 & 2-2+0 & -4+4+0 \\ 4-12+8 & 4+6-4 & -8-12+20 \\ 0-4+4 & 0+2-2 & 0-4+10 \end{bmatrix} \\ & = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} =6 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =6 I \\ \therefore \quad B^{-1} & =\frac{1}{6} A=\frac{1}{6} \begin{bmatrix} 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \end{aligned} $

The given equations can be re-write as,

$ \begin{aligned} & x-y=3,2 x+3 y+4 z=17 \text{ and } y+2 z=7 \\ & \therefore \quad \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} \\ & \Rightarrow \quad \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} ^{-1} \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} \\ & =\frac{1}{6} \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} \\ & =\frac{1}{6} \begin{bmatrix} 6+34-28 \\ -12+34-28 \\ 6-17+35 \end{bmatrix} =\frac{1}{6} \begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix} \end{aligned} $

Hence, $x=2, y=-1$ and $z=4$

Solution

Given that: $a+b+c \neq 0$ and $ \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} =0$

$C_1 \to C_1+C_2+C_3$

$ \begin{aligned} & \Rightarrow \quad \begin{bmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{bmatrix} =0 \\ & \Rightarrow \quad(a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} =0 \quad \begin{matrix} \text{ (Taking } a+b+c \\ \text{ common from } C_1 \text{ ) } \end{matrix} \\ & \Rightarrow \quad a+b+c \neq 0 \quad \therefore \quad \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} =0 \end{aligned} $

$\Rightarrow \quad \begin{vmatrix} 0 & b-c & c-a \\ 0 & c-a & a-b \\ 1 & a & b\end{vmatrix} =0$

Expanding along $C_1$

$ \begin{aligned} & 1 \begin{vmatrix} b-c & c-a \\ c-a & a-b \end{vmatrix} =0 \\ & \Rightarrow \quad(b-c)(a-b)-(c-a)^{2}=0 \\ & \Rightarrow \quad a b-b^{2}-a c+b c-c^{2}-a^{2}+2 a c=0 \\ & \Rightarrow \quad-a^{2}-b^{2}-c^{2}+a b+b c+a c=0 \\ & \Rightarrow \quad a^{2}+b^{2}+c^{2}-a b-b c-a c=0 \\ & \Rightarrow \quad 2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 a c=0 \end{aligned} $

(Multiplying both sides by 2 )

$\Rightarrow(a^{2}+b^{2}-2 a b)+(b^{2}+c^{2}-2 b c)+(a^{2}+c^{2}-2 a c)=0$

$\Rightarrow \quad(a-b)^{2}+(b-c)^{2}+(a-c)^{2}=0$

It is only possible when $(a-b)^{2}=(b-c)^{2}=(a-c)^{2}=0$

$\therefore a=b=c \quad$ Hence, proved.

Solution

Let $\Delta= \begin{vmatrix} b c-a^{2} & c a-b^{2} & a b-c^{2} \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & a c-b^{2}\end{vmatrix} $

$C_1 \to C_1+C_2+C_3$

$\Rightarrow \quad \begin{vmatrix} a b+b c+a c-a^{2}-b^{2}-c^{2} & c a-b^{2} & a b-c^{2} \\ a b+b c+a c-a^{2}-b^{2}-c^{2} & a b-c^{2} & b c-a^{2} \\ a b+b c+a c-a^{2}-b^{2}-c^{2} & b c-a^{2} & a c-b^{2}\end{vmatrix} $

Taking $a b+b c+a c-a^{2}-b^{2}-c^{2}$ common from $C_1$

$(a b+b c+a c-a^{2}-b^{2}-c^{2}) \begin{vmatrix} 1 & c a-b^{2} & a b-c^{2} \\ 1 & a b-c^{2} & b c-a^{2} \\ 1 & b c-a^{2} & a c-b^{2}\end{vmatrix} $

$R_1 \to R_1-R_2$ and $R_2 \to R_2-R_3$

$ \begin{aligned} & \Rightarrow \quad(a b+b c+a c-a^{2}-b^{2}-c^{2}) \\ & \begin{vmatrix} 0 & c a-b^{2}-a b+c^{2} & a b-c^{2}-b c+a^{2} \\ 0 & a b-c^{2}-b c+a^{2} & b c-a^{2}-a c+b^{2} \\ 1 & b c-a^{2} & a c-b^{2} \end{vmatrix} \\ & \Rightarrow \quad(a b+b c+a c-a^{2}-b^{2}-c^{2}) \\ & \begin{vmatrix} 0 & a(c-b)+(c+b)(c-b) & b(a-c)+(a+c)(a-c) \\ 0 & b(a-c)+(a+c)(a-c) & c(b-a)+(b+a)(b-a) \\ 1 & b c-a^{2} & a c-b^{2} \end{vmatrix} \\ & \Rightarrow \quad(a b+b c+a c-a^{2}-b^{2}-c^{2}) \\ & \begin{vmatrix} 0 & (c-b)(a+b+c) & (a-c)(a+b+c) \\ 0 & (a-c)(a+b+c) & (b-a)(a+b+c) \\ 1 & b c-a^{2} & a c-b^{2} \end{vmatrix} \\ & \Rightarrow \quad(a b+b c+a c-a^{2}-b^{2}-c^{2})(a+b+c)(a+b+c) \\ & \begin{vmatrix} 0 & c-b & a-c \\ 0 & a-c & b-a \\ 1 & b c-a^{2} & a c-b^{2} \end{vmatrix} \\ & \Rightarrow \quad(a+b+c)^{2}(a b+b c+a c-a^{2}-b^{2}-c^{2}) \\ & \begin{vmatrix} 0 & c-b & a-c \\ 0 & a-c & b-a \\ 1 & b c-a^{2} & a c-b^{2} \end{vmatrix} \end{aligned} $

Expanding along $C_1$

$\Rightarrow(a+b+c)^{2}(a b+b c+a c-a^{2}-b^{2}-c^{2})[1 \begin{vmatrix} c-b & a-c \\ a-c & b-a\end{vmatrix} ]$

$\Rightarrow(a+b+c)^{2}(a b+b c+a c-a^{2}-b^{2}-c^{2})[(c-b)(b-a)-(a-c)^{2}]$

$\Rightarrow(a+b+c)^{2}(a b+b c+a c-a^{2}-b^{2}-c^{2})(b c-c a-b^{2}+a b-a^{2}-c^{2}+2 a c)$

$\Rightarrow(a+b+c)^{2}(a b+b c+a c-a^{2}-b^{2}-c^{2})(a b+b c+c a-a^{2}-b^{2}-c^{2})$

$\Rightarrow(a+b+c)^{2}(a b+b c+a c-a^{2}-b^{2}-c^{2})^{2}$

$\Rightarrow(a+b+c)(a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-a c)^{2}$

Hence, the given determinant is divisible by $a+b+c$ and the quotient is

$ \begin{aligned} & (a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-a c)^{2} \\ \Rightarrow \quad & (a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-a c)(a^{2}+b^{2}+c^{2}-a b-b c-a c) \\ \Rightarrow \quad & (a^{3}+b^{3}+c^{3}-3 a b c)(a^{2}+b^{2}+c^{2}-a b-b c-a c) \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad-(a^{3}+b^{3}+c^{3}-3 a b c)(2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 a c) \\ & \Rightarrow \quad \frac{1}{2}(a^{3}+b^{3}+c^{3}-3 a b c)[(a-b)^{2}+(b-c)^{2}+(a-c)^{2}] \end{aligned} $

Solution

L.H.S.

$ \text{ Let } \Delta= \begin{vmatrix} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{vmatrix} $

Expanding along $R_1$

$ \begin{aligned} & \Rightarrow \quad x a \begin{vmatrix} z a & x b \\ x c & y a \end{vmatrix} -y b \begin{vmatrix} y c & x b \\ z b & y a \end{vmatrix} +z c \begin{vmatrix} y c & z a \\ z b & x c \end{vmatrix} \\ & \Rightarrow \quad x a(y z a^{2}-x^{2} b c)-y b(y^{2} a c-x z b^{2})+z c(x y c^{2}-z^{2} a b) \\ & \Rightarrow \quad x y z a^{3}-x^{3} a b c-y^{3} a b c+x y z b^{3}+x y z c^{3}-z^{3} a b c \\ & \Rightarrow \quad x y z(a^{3}+b^{3}+c^{3})-a b c(x^{3}+y^{3}+z^{3}) \\ & \Rightarrow \quad x y z(a^{3}+b^{3}+c^{3})-a b c(3 x y z) \\ & \quad[(\because \quad x+y+z=0)(\therefore \quad x^{3}+y^{3}+z^{3}=3 x y z)] \\ & \Rightarrow x y z(a^{3}+b^{3}+c^{3}-3 a b c) \end{aligned} $

R.H.S. $x y z \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a\end{vmatrix} $

$ R_1 \to R_1+R_2+R_3 $

$ \begin{aligned} & \Rightarrow \quad x y z \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ c & a & b \\ b & c & a \end{vmatrix} \\ & \Rightarrow \quad x y z(a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ c & a & b \\ b & c & a \end{vmatrix} \end{aligned} $

$C_1 \to C_1-C_2, C_2 \to C_2-C_3$

$\Rightarrow \quad x y z(a+b+c) \begin{vmatrix} 0 & 0 & 1 \\ c-a & a-b & b \\ b-c & c-a & a\end{vmatrix} $

Expanding along $R_1$ $$ \begin{aligned} & \Rightarrow \quad x y z(a+b+c)\left[1\left|\begin{array}{cc} c-a & a-b \\ b-c & c-a \end{array}\right|\right] \\ & \Rightarrow \quad x y z(a+b+c)\left[(c-a)^2-(b-c)(a-b)\right] \\ & \Rightarrow \quad x y z(a+b+c)\left(c^2+a^2-2 c a-a b+b^2+a c-b c\right) \\ & \Rightarrow \quad x y z(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) \\ & \Rightarrow \quad x y z\left(a^3+b^3+c^3-3 a b c\right) \\ & \quad\left[a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right] \end{aligned} $$ L.H.S. $=$ R.H.S.

Hence, proved.

Solution

Given that

$ \begin{aligned} & \Rightarrow \quad \begin{vmatrix} 2 x & 5 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix} \\ & \Rightarrow \quad 2 x^{2}-40=18+14 \Rightarrow 2 x^{2}=32+40 \\ & \Rightarrow \quad 2 x^{2}=72 \Rightarrow x^{2}=36 \\ & \therefore \quad x= \pm 6 \end{aligned} $

Hence, the correct option is (c).

• Option (a) 3 is incorrect because solving the determinant equation results in ( x^2 = 36 ), which gives ( x = \pm 6 ). The value 3 does not satisfy this equation.

• Option (b) (\pm 3) is incorrect because solving the determinant equation results in ( x^2 = 36 ), which gives ( x = \pm 6 ). The values (\pm 3) do not satisfy this equation.

• Option (d) 6 is incorrect because while 6 is one of the solutions, the equation ( x^2 = 36 ) also has another solution, which is -6. Therefore, the correct answer must include both (\pm 6).

Solution

Here, we have $ \begin{vmatrix} a-b & b+c & a \\ b-a & c+a & b \\ c-a & a+b & c\end{vmatrix} $

$C_2 \to C_2+C_3$

$ \begin{aligned} & \Rightarrow \begin{vmatrix} a-b & a+b+c & a \\ b-a & a+b+c & b \\ c-a & a+b+c & c \end{vmatrix} \\ & \Rightarrow \quad(a+b+c) \begin{vmatrix} a-b & 1 & a \\ b-a & 1 & b \\ c-a & 1 & c \end{vmatrix} \quad \text{ (Taking } a+b+c \text{ common } \\ & R_1 \to R_1-R_2, R_2 \to R_2-R_3 \\ & \Rightarrow \quad(a+b+c) \begin{vmatrix} 2(a-b) & 0 & a-b \\ b-c & 0 & b-c \\ c-a & 1 & c \end{vmatrix} \end{aligned} $

Taking $(a-b)$ and $(b-c)$ common from $R_1$ and $R_2$ respectively

$ \Rightarrow \quad(a+b+c)(a-b)(b-c) \begin{vmatrix} 2 & 0 & 1 \\ 1 & 0 & 1 \\ c-a & 1 & c \end{vmatrix} $

Expanding along $C_2$

$ \begin{aligned} & \Rightarrow \quad(a+b+c)(a-b)(b-c)[-1 \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} ] \\ & \Rightarrow \quad(a+b+c)(a-b)(b-c)(-1) \\ & \Rightarrow \quad(a+b+c)(a-b)(c-b) \end{aligned} $

Hence, the correct option is $(d)$.

• Option (a) $a^{3}+b^{3}+c^{3}$ is incorrect because the determinant simplifies to a product of linear factors involving $(a+b+c)$, $(a-b)$, and $(c-b)$, not a sum of cubes.

• Option (b) $3 b c$ is incorrect because the determinant does not simplify to a constant multiple of $bc$. The correct expression involves the product of three linear terms.

• Option (c) $a^{3}+b^{3}+c^{3}-3 a b c$ is incorrect because the determinant does not simplify to a polynomial expression involving cubes and a product of $abc$. The correct expression is a product of linear terms.

Solution

Area of triangle with vertices $(x_1 y_1),(x_2 y_2)$ and $(x_3, y_3)$ will be:

$ \begin{matrix} \Delta & =\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \Rightarrow \Delta=\frac{1}{2} \begin{vmatrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{vmatrix} \\ \Rightarrow \quad & =\frac{1}{2}[-3 \begin{vmatrix} 0 & 1 \\ k & 1 \end{vmatrix} -0 \begin{vmatrix} 3 & 1 \\ 0 & 1 \end{vmatrix} +1 \begin{vmatrix} 3 & 0 \\ 0 & k \end{vmatrix} ] \\ \Rightarrow \quad & =\frac{1}{2}[-3(-k)-0+1(3 k)] \\ \Rightarrow \quad & =\frac{1}{2}(3 k+3 k) \quad \Rightarrow \quad \frac{1}{2}(6 k)=3 k \end{matrix} $

$\qquad 3k=9 \Rightarrow k=3$

Hence, the correct option is (b).

• Option (a) 9: This option is incorrect because substituting ( k = 9 ) into the area formula would result in an area of ( 27 ) square units, not ( 9 ) square units.
• Option (c) -9: This option is incorrect because substituting ( k = -9 ) into the area formula would result in an area of ( 27 ) square units, not ( 9 ) square units.
• Option (d) 6: This option is incorrect because substituting ( k = 6 ) into the area formula would result in an area of ( 18 ) square units, not ( 9 ) square units.

Solution

Let

$ \begin{aligned} \Delta & = \begin{vmatrix} b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2} \end{vmatrix} \\ & = \begin{vmatrix} b(b-a) & b-c & c(b-a) \\ a(b-a) & a-b & b(b-a) \\ c(b-a) & c-a & a(b-a) \end{vmatrix} \text{ (Taking }(b-a) \text{ common } \\ \text{ from } C_1 \text{ and } C_3) \\ & =(b-a)^{2} \begin{vmatrix} b & b-c & c \\ a & a-b & b \\ c & c-a & a \end{vmatrix} \\ -C_3 & \\ & =(a-b)^{2} \begin{vmatrix} b-c & b-c & c \\ a-b & a-b & b \\ c-a & c-a & a \end{vmatrix} \quad C_1 \text{ and } C_2 \text{ identical columns.) } \\ & =(a-b)^{2} \cdot 0 \\ & =0 \end{aligned} $

$ C_1 \to C_1-C_3 $

Hence, the correct option is $(d)$.

• Option (a) $a b c(b-c)(c-a)(a-b)$ is incorrect because the determinant simplifies to zero, not a product involving $a$, $b$, and $c$.
• Option (b) $(b-c)(c-a)(a-b)$ is incorrect because the determinant simplifies to zero, not a non-zero product of differences.
• Option (c) $(a+b+c)(b-c)(c-a)(a-b)$ is incorrect because the determinant simplifies to zero, not a product involving the sum $a+b+c$ and the differences.

Solution

Given that

$ \begin{aligned} & C_1 \to C_1+C_2+C_3 \\ & \Rightarrow \end{aligned} $

Taking $2 \cos x+\sin x$ common from $C_1$

$ \begin{aligned} & \Rightarrow \quad(2 \cos x+\sin x) \begin{vmatrix} 1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x \end{vmatrix} =0 \\ & R_1 \to R_1-R_2, R_2 \to R_2-R_3 \\ & \Rightarrow(2 \cos x+\sin x) \begin{vmatrix} 0 & \cos x-\sin x & 0 \\ 0 & \sin x-\cos x & \cos x-\sin x \\ 1 & \cos x & \sin x \end{vmatrix} =0 \\ & \Rightarrow(2 \cos x+\sin x)[1 \begin{vmatrix} \cos x-\sin x & 0 \\ \sin x-\cos x & \cos x-\sin x \end{vmatrix} ] \\ & \Rightarrow \quad(2 \cos x+\sin x)(\cos x-\sin x)^{2}=0 \\ & 2 \cos x+\sin x=0 \\ & 2+\tan x=0 \\ & \therefore \quad \tan x=-2 \\ & \Rightarrow \quad \tan x=1 \\ & -\frac{\pi}{4} \leq x \leq \frac{\pi}{4} \\ & (\cos x-\sin x)^{2}=0 \\ & \cos x-\sin x=0 \\ & \Rightarrow \quad \tan x=\tan \frac{\pi}{4} \\ & \therefore \quad x=\frac{\pi}{4} \in[\frac{-\pi}{4}, \frac{\pi}{4}] \end{aligned} $

So, $x$ has no solution. So, it will have only one real root.

Hence, the correct option is (c).

• Option (a) 0: This option is incorrect because the equation does have a real root. The solution process shows that there is at least one value of ( x ) that satisfies the equation, specifically ( x = \frac{\pi}{4} ).

• Option (b) 2: This option is incorrect because the equation does not have two distinct real roots. The solution process reveals only one distinct real root, ( x = \frac{\pi}{4} ).

• Option (d) 3: This option is incorrect because the equation does not have three distinct real roots. The solution process identifies only one distinct real root, ( x = \frac{\pi}{4} ).

Solution

Let $\Delta= \begin{vmatrix} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1\end{vmatrix} $

$C_1 \to a C_1+b C_2+c C_3$

$\Rightarrow \begin{vmatrix} -a+b \cos C+c \cos B & \cos C & \cos B \\ a \cos C-b+c \cos A & -1 & \cos A \\ a \cos B+b \cos A-C & \cos A & -1\end{vmatrix} $

$ \begin{aligned} & \Rightarrow \begin{vmatrix} -a+a & \cos C & \cos B \\ -b+b & -1 & \cos A \\ -c+c & \cos A & -1 \end{vmatrix} \begin{bmatrix} \because \quad \text{ From projection formula } \\ a=b \cos C+c \cos B \\ b=a \cos C+c \cos A \\ c=b \cos A+a \cos B \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 0 & \cos C & \cos B \\ 0 & -1 & \cos A \\ 0 & \cos A & -1 \end{bmatrix} =0 \end{aligned} $

Hence, the correct option is (a).

• Option (b) -1 is incorrect because the determinant of the given matrix simplifies to 0, not -1. The matrix transformation and simplification show that the first column becomes all zeros, leading to a determinant of 0.

• Option (c) 1 is incorrect because, as shown in the solution, the determinant of the matrix is 0. The matrix transformation and simplification clearly indicate that the determinant cannot be 1.

• Option (d) None of these is incorrect because the correct answer is indeed one of the provided options, specifically option (a) 0. The determinant calculation confirms that the value is 0.

Solution

We have $f(t)= \begin{bmatrix} \cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t \end{bmatrix} $

Expanding along $R_1$

$ \begin{aligned} & =\cos t \begin{vmatrix} t & 2 t \\ t & t \end{vmatrix} -t \begin{vmatrix} 2 \sin t & 2 t \\ \sin t & t \end{vmatrix} +1 \begin{vmatrix} 2 \sin t & t \\ \sin t & t \end{vmatrix} \\ & =\cos t(t^{2}-2 t^{2})-t(2 t \sin t-2 t \sin t)+(2 t \sin t-t \sin t) \\ & =-t^{2} \cos t+t \sin t \\ & \therefore \quad \frac{f(t)}{t^{2}}=\frac{-t^{2} \cos t+t \sin t}{t^{2}} \\ & \Rightarrow \quad \frac{f(t)}{t^{2}}=-\cos t+\frac{\sin t}{t} \\ & \Rightarrow \quad \lim _{t \to 0} \frac{f(t)}{t^{2}}=\lim _{t \to 0}(-\cos t)+\lim _{t \to 0} \frac{\sin t}{t}=-1+1=0 \end{aligned} $

Hence, the correct option is (a).

• Option (b) -1:

  • The limit calculation shows that the term involving (\cos t) approaches (-1) and the term involving (\frac{\sin t}{t}) approaches (1). Therefore, their sum is (0), not (-1).

• Option (c) 2:

  • The limit calculation does not yield a value of (2). The sum of (-\cos t) and (\frac{\sin t}{t}) as (t) approaches (0) is (0), not (2).

• Option (d) 3:

  • The limit calculation does not yield a value of (3). The sum of (-\cos t) and (\frac{\sin t}{t}) as (t) approaches (0) is (0), not (3).

Solution

Given that: $\Delta= \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1+\cos \theta & 1 & 1\end{vmatrix} $ $C_1 \to C_1-C_2, C_2 \to C_2-C_3$

Expanding along $R_1$

$ = \begin{vmatrix} 0 & 0 & 1 \\ -\sin \theta & \sin \theta & 1 \\ \cos \theta & 0 & 1 \end{vmatrix} $

$ \begin{aligned} & =1 \begin{vmatrix} -\sin \theta & \sin \theta \\ \cos \theta & 0 \end{vmatrix} =-\sin \theta \cos \theta \\ \Rightarrow \quad & =-\frac{1}{2} \cdot 2 \sin \theta \cos \theta=-\frac{1}{2} \sin 2 \theta \end{aligned} $

but maximum value of $\sin 2 \theta=1 \Rightarrow|-\frac{1}{2} \cdot 1|=\frac{1}{2}$

Hence, the correct option is $(a)$.

• Option (b) $\frac{\sqrt{3}}{2}$ is incorrect because the maximum value of $\Delta$ is determined by the maximum value of $\sin 2\theta$, which is 1. Therefore, the maximum value of $\Delta$ is $-\frac{1}{2} \cdot 1 = -\frac{1}{2}$, and the absolute value is $\frac{1}{2}$, not $\frac{\sqrt{3}}{2}$.

• Option (c) $\sqrt{2}$ is incorrect because the maximum value of $\Delta$ is $-\frac{1}{2} \sin 2\theta$. Since the maximum value of $\sin 2\theta$ is 1, the maximum value of $\Delta$ is $-\frac{1}{2} \cdot 1 = -\frac{1}{2}$, and the absolute value is $\frac{1}{2}$, not $\sqrt{2}$.

• Option (d) $\frac{2 \sqrt{3}}{4}$ is incorrect because the maximum value of $\Delta$ is $-\frac{1}{2} \sin 2\theta$. Since the maximum value of $\sin 2\theta$ is 1, the maximum value of $\Delta$ is $-\frac{1}{2} \cdot 1 = -\frac{1}{2}$, and the absolute value is $\frac{1}{2}$, not $\frac{2 \sqrt{3}}{4}$, which simplifies to $\frac{\sqrt{3}}{2}$.

Solution

Given that: $f(x)= \begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{vmatrix} $

$ f(a)= \begin{vmatrix} 0 & 0 & a-b \\ 2 a & 0 & a-c \\ a+b & a+c & 0 \end{vmatrix} $

Expanding along $R_1=(a-b) \begin{vmatrix} 2 a & 0 \\ a+b & a+c\end{vmatrix} $

$ \begin{aligned} & =(a-b)[2 a(a+c)]=(a-b) \cdot 2 a \cdot(a+c) \neq 0 \\ f(b) & = \begin{vmatrix} 0 & b-a & 0 \\ b+a & 0 & b-c \\ 2 b & b+c & 0 \end{vmatrix} \end{aligned} $

Expanding along $R_1$

$ \begin{aligned} & -(b-a) \begin{vmatrix} b+a & b-c \\ 2 b & 0 \end{vmatrix} \\ & =-(b-a)[(-2 b)(b-c)]=2 b(b-a)(b-c) \neq 0 \\ f(0) & = \begin{vmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{vmatrix} \end{aligned} $

$ \begin{matrix} \text{ Expanding along } R_1=a \begin{vmatrix} a & -c \\ b & 0 \end{vmatrix} -b \begin{vmatrix} a & 0 \\ b & c \end{vmatrix} \\ =a(b c)-b(a c)=a b c-a b c=0 \end{matrix} $

Hence, the correct option is (c).

• Option (a) $f(a)=0$ is incorrect: The determinant calculation for ( f(a) ) results in a non-zero value. Specifically, it simplifies to ( (a-b) \cdot 2a \cdot (a+c) ), which is not zero unless ( a = b ) or ( a = -c ), but these conditions are not given in the problem.

• Option (b) $f(b)=0$ is incorrect: The determinant calculation for ( f(b) ) also results in a non-zero value. It simplifies to ( 2b(b-a)(b-c) ), which is not zero unless ( b = a ) or ( b = c ), but these conditions are not given in the problem.

• Option (d) $f(1)=0$ is incorrect: The problem does not provide a specific calculation for ( f(1) ), but based on the pattern observed in the calculations for ( f(a) ) and ( f(b) ), it is unlikely that ( f(1) ) would simplify to zero without specific conditions on ( a ), ( b ), and ( c ). Therefore, without additional information, we cannot assume ( f(1) = 0 ).

Solution

We have,

$ A= \begin{bmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{bmatrix} \Rightarrow|A|= \begin{vmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{vmatrix} $

Expanding along $R_1=2 \begin{vmatrix} 2 & 5 \\ 1 & 3\end{vmatrix} -\lambda \begin{vmatrix} 0 & 5 \\ 1 & 3\end{vmatrix} -3 \begin{vmatrix} 0 & 2 \\ 1 & 1\end{vmatrix} $

$ \begin{aligned} & =2(6-5)-\lambda(0-5)-3(0-2) \\ & =2+5 \lambda+6=8+5 \lambda \end{aligned} $

If $A^{-1}$ exists then $|A| \neq 0$

$\therefore \quad 8+5 \lambda \neq 0$ so $\lambda \neq \frac{-8}{5}$

Hence, the correct option is $(d)$.

• Option (a) $\lambda=2$: This option is incorrect because it does not account for the general condition under which the inverse of matrix ( A ) exists. The determinant of ( A ) must be non-zero for ( A^{-1} ) to exist. The condition ( \lambda=2 ) does not guarantee that the determinant is non-zero. Specifically, substituting ( \lambda=2 ) into the determinant expression ( 8 + 5\lambda ) gives ( 8 + 10 = 18 ), which is non-zero, but this is just one specific case. The correct condition is that ( \lambda \neq -\frac{8}{5} ).

• Option (b) $\lambda \neq 2$: This option is incorrect because it suggests that the inverse of ( A ) exists for all values of ( \lambda ) except ( \lambda=2 ). However, the determinant ( 8 + 5\lambda ) is non-zero for any ( \lambda ) except ( \lambda = -\frac{8}{5} ). Therefore, the condition ( \lambda \neq 2 ) is not sufficient to ensure that ( A^{-1} ) exists.

• Option (c) $\lambda \neq -2$: This option is incorrect because it suggests that the inverse of ( A ) exists for all values of ( \lambda ) except ( \lambda=-2 ). However, the determinant ( 8 + 5\lambda ) is non-zero for any ( \lambda ) except ( \lambda = -\frac{8}{5} ). Therefore, the condition ( \lambda \neq -2 ) is not sufficient to ensure that ( A^{-1} ) exists.

Solution

If $A$ and $B$ are two invertible matrices then

(a) $adj A=|A| \cdot A^{-1}$ is correct

(b) $det(A)^{-1}=[det(A)]^{-1}=\frac{1}{det(A)}$ is correct

(c) Also, $(AB)^{-1}=B^{-1} A^{-1}$ is correct

(d) $(A+B)^{-1}=\frac{1}{|A+B|} \cdot adj(A+B)$

$\therefore \quad(A+B)^{-1} \neq B^{-1}+A^{-1}$

Hence, the correct option is $(d)$.

• Option (a) is correct because the adjugate (adjoint) of a matrix ( A ) is given by ( \text{adj}(A) = |A| \cdot A^{-1} ) for an invertible matrix ( A ).

• Option (b) is correct because ( \det(A)^{-1} = [\det(A)]^{-1} = \frac{1}{\det(A)} ).

• Option (c) is correct because the inverse of the product of two invertible matrices ( A ) and ( B ) is given by ( (AB)^{-1} = B^{-1}A^{-1} ).

• Option (d) is incorrect because the inverse of the sum of two matrices ( A ) and ( B ) is not given by ( (A+B)^{-1} = B^{-1} + A^{-1} ). The correct expression involves the adjugate and determinant of ( A+B ), specifically ( (A+B)^{-1} = \frac{1}{|A+B|} \cdot \text{adj}(A+B) ).

Solution

Given that

$ \begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix} =0 $

Taking $x, y$ and $z$ common from $R_1, R_2$ and $R_3$ respectively.

$\Rightarrow \quad x y z \begin{vmatrix} \frac{1}{x}+1 & \frac{1}{x} & \frac{1}{x} \\ \frac{1}{y} & \frac{1}{y}+1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{vmatrix} =0$

$R_1 \to R_1+R_2+R_3$

$\Rightarrow x y z \begin{vmatrix} \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1 & \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1 & \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1 \\ \frac{1}{y} & \frac{1}{y}+1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{vmatrix} =0$

Taking $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1$ common from $R_1$

$\Rightarrow \quad x y z(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1) \begin{vmatrix} 1 & 1 & 1 \\ \frac{1}{y} & \frac{1}{y}+1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{vmatrix} =0$

$C_1 \to C_1-C_2, C_2 \to C_2-C_3$

$\Rightarrow \quad x y z(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1) \begin{vmatrix} 0 & 0 & 1 \\ -1 & 1 & \frac{1}{y} \\ 0 & -1 & \frac{1}{z}+1\end{vmatrix} =0$

Expanding along $R_1$

$ \begin{aligned} & \Rightarrow \quad x y z(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1)[1 \begin{vmatrix} -1 & 1 \\ 0 & -1 \end{vmatrix} ]=0 \\ & \Rightarrow \quad x y z(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1)(1)=0 \\ & \Rightarrow \quad \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1=0 \text{ and } x y z \neq 0 \quad(x \neq y \neq z \neq 0) \\ & \therefore \quad x^{-1}+y^{-1}+z^{-1}=-1 \end{aligned} $

Hence, the correct option is $(d)$.

• Option (a) $x y z$: This option is incorrect because the determinant condition given in the problem leads to the equation $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 = 0$. Solving this equation gives $x^{-1} + y^{-1} + z^{-1} = -1$, not $x y z$.

• Option (b) $x^{-1} y^{-1} z^{-1}$: This option is incorrect because $x^{-1} y^{-1} z^{-1}$ represents the product of the reciprocals of $x, y, z$, which is not related to the sum of the reciprocals. The correct result from the determinant condition is $x^{-1} + y^{-1} + z^{-1} = -1$.

• Option (c) $-x - y - z$: This option is incorrect because $-x - y - z$ represents the negative sum of the variables $x, y, z$ themselves, not their reciprocals. The correct result from the determinant condition is $x^{-1} + y^{-1} + z^{-1} = -1$.

Solution

Let $\Delta= \begin{vmatrix} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x\end{vmatrix} $

$C_1 \to C_1+C_2+C_3$

$= \begin{vmatrix} 3 x+3 y & x+y & x+2 y \\ 3 x+3 y & x & x+y \\ 3 x+3 y & x+2 y & x\end{vmatrix} $

$=(3 x+3 y) \begin{vmatrix} 1 & x+y & x+2 y \\ 1 & x & x+y \\ 1 & x+2 y & x\end{vmatrix} $

[Taking $(3 x+3 y)$ common from $C_1$ ]

$R_1 \to R_1-R_2, R_2 \to R_2-R_3$

$\Rightarrow 3(x+y) \begin{vmatrix} 0 & y & y \\ 0 & -2 y & y \\ 1 & x+2 y & x\end{vmatrix} $

Expanding along $C_1$

$\Rightarrow \quad 3(x+y)[1 \begin{vmatrix} y & y \\ -2 y & y\end{vmatrix} ]$

$ \Rightarrow 3(x+y)(y^{2}+2 y^{2}) \Rightarrow 3(x+y)(3 y^{2}) \Rightarrow 9 y^{2}(x+y) $

Hence, the correct option is $(b)$.

• Option (a) $9 x^{2}(x+y)$ is incorrect because the determinant calculation does not yield a term involving $x^2$. The correct calculation shows that the determinant is proportional to $y^2$ rather than $x^2$.

• Option (c) $3 y^{2}(x+y)$ is incorrect because the correct factor from the determinant calculation is $9 y^2(x+y)$, not $3 y^2(x+y)$. The factor of 3 is missing in this option.

• Option (d) $7 x^{2}(x+y)$ is incorrect because the determinant calculation does not yield a term involving $x^2$, and the factor 7 does not appear in the correct calculation. The correct result involves $9 y^2(x+y)$.

Solution

Given that, $\Delta= \begin{vmatrix} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a\end{vmatrix} =86$

Expanding along $C_1$

$ \begin{matrix} \Rightarrow & 1 \begin{vmatrix} a & -1 \\ 4 & 2 a \end{vmatrix} -2 \begin{vmatrix} -2 & 5 \\ 4 & 2 a \end{vmatrix} +0 \begin{vmatrix} -2 & 5 \\ a & -1 \end{vmatrix} =86 \\ \Rightarrow & (2 a^{2}+4)-2(-4 a-20) =86 \\ \Rightarrow & 2 a^{2}+4+8 a+40 =86 \\ \Rightarrow & 2 a^{2}+8 a+4+40-86 =0 \\ \Rightarrow & 2 a^{2}+8 a-42 =0 \\ \Rightarrow & a^{2}+4 a-21 =0 \\ \Rightarrow & a^{2}+7 a-3 a-21 =0 \\ \Rightarrow & (a-7)-3(a+7) =0 \\ \Rightarrow & (a-3)(a+7) =0 \end{matrix} $

$\therefore \quad a=3,-7$

Required sum of the two numbers $=3-7=-4$.

Hence, the correct option is (c).

• Option (a) 4: This option is incorrect because the sum of the two values of ( a ) that satisfy the determinant equation is ( 3 + (-7) = -4 ), not 4.

• Option (b) 5: This option is incorrect because the sum of the two values of ( a ) that satisfy the determinant equation is ( 3 + (-7) = -4 ), not 5.

• Option (d) 9: This option is incorrect because the sum of the two values of ( a ) that satisfy the determinant equation is ( 3 + (-7) = -4 ), not 9.

Solution

We know that for a matrix of order $3 \times 3$,

$ |KA|=K^{3}|A| $

$ \therefore \quad|3 A|=3^{3}|A|=27|\mathbf{A}| $

Solution

We know that for an invertible matrix $A$ of any order, $|A^{-1}|=\frac{1}{|\mathbf{A}|}$.

Solution

We have, $ \begin{vmatrix} (2^{x}+2^{-x})^{2} & (2^{x}-2^{-x})^{2} & 1 \\ (3^{x}+3^{-x})^{2} & (3^{x}-3^{-x})^{2} & 1 \\ (4^{x}+4^{-x})^{2} & (4^{x}-4^{-x})^{2} & 1\end{vmatrix} $

$ C_1 \to C_1-C_2 $

$ \begin{aligned} & \Rightarrow \begin{vmatrix} (2^{x}+2^{-x})^{2}-(2^{x}-2^{-x})^{2} & (2^{x}-2^{-x})^{2} & 1 \\ (3^{x}+3^{-x})^{2}-(3^{x}-3^{-x})^{2} & (3^{x}-3^{-x})^{2} & 1 \\ (4^{x}+4^{-x})^{2}-(4^{x}-4^{-x})^{2} & (4^{x}-4^{-x})^{2} & 1 \end{vmatrix} \\ & \Rightarrow \begin{vmatrix} 4 \cdot 2^{x} \cdot 2^{-x} & (2^{x}-2^{-x})^{2} & 1 \\ 4 \cdot 3^{x} \cdot 3^{-x} & (3^{x}-3^{-x})^{2} & 1 \\ 4 \cdot 4^{x} \cdot 4^{-x} & (4^{x}-4^{-x})^{2} & 1 \end{vmatrix} \quad(a+b)^{2}-(a-b)^{2}=4 a b \text{ ] } \\ & \Rightarrow \begin{vmatrix} 4 & (2^{x}-2^{-x})^{2} & 1 \\ 4 & (3^{x}-3^{-x})^{2} & 1 \\ 4 & (4^{x}-4^{-x})^{2} & 1 \end{vmatrix} \\ & \Rightarrow 4 \begin{vmatrix} 1 & (2^{x}-2^{-x})^{2} & 1 \\ 1 & (3^{x}-3^{-x})^{2} & 1 \\ 1 & (4^{x}-4^{-x})^{2} & 1 \end{vmatrix} \end{aligned} $

(Taking 4 common from $C_1$ )

$\Rightarrow \quad 4 \cdot 0=0$

( $\because C_1$ and $C_3$ are identical columns)

Solution

Given that: $\quad \cos 2 \theta=0$

$ \begin{aligned} \Rightarrow & & \cos 2 \theta & =\cos \frac{\pi}{2} \Rightarrow 2 \theta=\frac{\pi}{2} \\ & \therefore & \theta & =\frac{\pi}{4} \end{aligned} $

The determinant can be written as

$$ \begin{aligned} & \Rightarrow \begin{vmatrix} 0 & \cos \frac{\pi}{4} & \sin \frac{\pi}{4} \\ \cos \frac{\pi}{4} & \sin \frac{\pi}{4} & 0 \\ \sin \frac{\pi}{4} & 0 & \cos \frac{\pi}{4} \end{vmatrix} ^{2} \Rightarrow \begin{vmatrix} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{vmatrix} ^{2} \\ & \Rightarrow[\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \begin{vmatrix} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix} ]^{2} \quad(\begin{vmatrix} \text{ Taking } \frac{1}{\sqrt{2}} \text{ common from } \\ C_1, C_2 \text{ and } C_3 \end{vmatrix} \\ & \text{ Expanding along } C_1, \\ & \Rightarrow[.\frac{1}{2 \sqrt{2}}|-1 \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} +1 \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \rvert,]^{2} \Rightarrow[\frac{1}{2 \sqrt{2}}|-1(1)+1(0-1)|]^{2} \\ & \Rightarrow[\frac{1}{2 \sqrt{2}}|-1-1|]^{2} \Rightarrow \frac{1}{8} \cdot(4)=\frac{\mathbf{1}}{\mathbf{2}} \end{aligned} $$

Solution

For any square matrix $A,(A^{2})^{-1}=(A^{-1})^{2}$.

Solution

The order of a matrix is $3 \times 3$

$\therefore$ Total number of elements $=3 \times 3=9$

Hence, the number of minors in the determinant is 9 .

Solution

The sum of the products of elements of any row with the co-factors of corresponding elements is equal to the value of the determinant of the given matrix.

Let $\Delta= \begin{vmatrix} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33}\end{vmatrix} $

Expanding along $R_1$

$ \begin{aligned} & a _{11} \begin{vmatrix} a _{22} & a _{23} \\ a _{32} & a _{33} \end{vmatrix} -a _{12} \begin{vmatrix} a _{21} & a _{23} \\ a _{31} & a _{33} \end{vmatrix} +a _{13} \begin{vmatrix} a _{21} & a _{22} \\ a _{31} & a _{32} \end{vmatrix} \\ & \Rightarrow a _{11} M _{11}+a _{12} M _{12}+a _{13} M _{13} \end{aligned} $

$ \begin{matrix} \text{ (where } M _{11}, M _{12} \text{ and } M _{13} \text{ are the minors of the } \\ \text{ corresponding elements) } \end{matrix} $

Solution

We have, $ \begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{vmatrix} =0$

Expanding along $R_1$

$ \begin{aligned} & \Rightarrow \quad x \begin{vmatrix} x & 2 \\ 6 & x \end{vmatrix} -3 \begin{vmatrix} 2 & 2 \\ 7 & x \end{vmatrix} +7 \begin{vmatrix} 2 & x \\ 7 & 6 \end{vmatrix} =0 \\ & \Rightarrow \quad x(x^{2}-12)-3(2 x-14)+7(12-7 x)=0 \\ & \Rightarrow \quad x^{3}-12 x-6 x+42+84-49 x=0 \\ & \Rightarrow \quad x^{3}-67 x+126=0 \end{aligned} $

The roots of the equation may be the factors of 126 i.e., $2 \times 7 \times 9$ 9 is given the root of the determinant put $x=2$ in eq. (1)

$ (2)^{3}-67 \times 2+126 \Rightarrow 8-134+126=0 $

Hence, $x=\mathbf{2}$ is the other root.

Now, put $x=7$ in eq. (1)

$ (7)^{3}-67(7)+126 \Rightarrow 343-469+126=0 $

Hence, $x=7$ is also the other root of the determinant.

Solution

Let $\Delta= \begin{vmatrix} 0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0\end{vmatrix} $

$C_1 \to C_1-C_3$

Taking $(z-x)$ common from $C_1$

$ = \begin{vmatrix} z-x & x y z & x-z \\ z-x & 0 & y-z \\ z-x & z-y & 0 \end{vmatrix} $

$ =(z-x) \begin{vmatrix} 1 & x y z & x-z \\ 1 & 0 & y-z \\ 1 & z-y & 0 \end{vmatrix} $

$ R_1 \to R_1-R_2, R_2 \to R_2-R_3 $

$ =(z-x) \begin{vmatrix} 0 & x y z & x-y \\ 0 & y-z & y-z \\ 1 & z-y & 0 \end{vmatrix} $

Taking $(y-z)$ common from $R_2$

$ =(z-x)(y-z) \begin{vmatrix} 0 & x y z & x-y \\ 0 & 1 & 1 \\ 1 & z-y & 0 \end{vmatrix} $

Expanding along $C_1$

$ =(z-x)(y-z)[1 \begin{vmatrix} x y z & x-y \\ 1 & 1 \end{vmatrix} ] $

$ =(z-x)(y-z)(x y z-x+y)=(y-z)(z-x)(y-x+x y z) $

Solution

Given that

$ \begin{vmatrix} (1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \end{vmatrix} =A+B x+C x^{2}+\cdots $

Taking $(1+x)^{17},(1+x)^{23}$ and $(1+x)^{41}$ common from $R_1, R_2$ and $R_3$ respectively

$ (1+x)^{17} \cdot(1+x)^{23} \cdot(1+x)^{41} \begin{vmatrix} 1 & (1+x)^{2} & (1+x)^{6} \\ 1 & (1+x)^{6} & (1+x)^{11} \\ 1 & (1+x)^{2} & (1+x)^{6} \end{vmatrix} $

$\Rightarrow \quad(1+x)^{17} \cdot(1+x)^{23} \cdot(1+x)^{41} \cdot 0 \quad(R_1.$ and $R_3$ are identical $)$

$\therefore \quad 0=A+B x+C x^{2}+\ldots$

By comparing the like terms, we get $A=\mathbf{0}$.

Solution

Since $(A^{K})^{-1}=(A^{-1})^{K}$ where $K \in N$

So,

$ (A^{3})^{-1}=(A^{-1})^{3} \text{ is true } $

Solution

If $A$ is a non-singular square matrix, then for any non-zero scalar ’ $a$ ‘, $a A$ is invertible.

$\therefore \quad(a A) \cdot(\frac{1}{a} A^{-1})=a \cdot \frac{1}{a} \cdot A \cdot A^{-1}=I$

So, $(a A)$ is inverse of $(\frac{1}{a} A^{-1})$

$\Rightarrow \quad(a A)^{-1}=\frac{1}{a} A^{-1}$ is true.

Solution

False.

Since $|A^{-1}|=|A|^{-1}$ for a non-singular matrix.

Solution

True.

$ |3 AB|=3^{3}|AB|=27|A||B|=27 \times 5 \times 3 \quad[\because|KA|=K^{n}|A|] $

Solution

True.

Since $|A|=12$

If $A$ is a square matrix of order $n$

then $\quad|Adj A|=|A|^{n-1}$

$\therefore \quad \mid$ Adj A $.|=| A|^{3-1}=|A|^{2}=(12)^{2}=144 \quad[n=3]$

Solution

True.

Let

$ \Delta= \begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix} $

$a, b, c$ are in A.P.

$ = \begin{vmatrix} x+1 & x+2 & x+a \\ 0 & 0 & 2 b-(a+c) \\ x+3 & x+4 & x+c \end{vmatrix} $

$ \begin{aligned} \therefore b-a=c-b \Rightarrow 2 b & =a+c \\ & = \begin{vmatrix} x+1 & x+2 & x+a \\ 0 & 0 & 0 \\ x+3 & x+4 & x+c \end{vmatrix} =0 \end{aligned} $

Solution

False.

Since $|adj A|=|A|^{n-1}$ where $n$ is the order of the square matrix.

Solution

True.

Let $\Delta= \begin{vmatrix} \sin A & \cos A & \sin A+\cos B \\ \sin B & \cos A & \sin B+\cos B \\ \sin C & \cos A & \sin C+\cos B\end{vmatrix} $

Splitting up $C_3$

$= \begin{vmatrix} \sin A & \cos A & \sin A \\ \sin B & \cos A & \sin B \\ \sin C & \cos A & \sin C\end{vmatrix} + \begin{vmatrix} \sin A & \cos A & \cos B \\ \sin B & \cos A & \cos B \\ \sin C & \cos A & \cos B\end{vmatrix} $

$=0+ \begin{vmatrix} \sin A & \cos A & \cos B \\ \sin B & \cos A & \cos B \\ \sin C & \cos A & \cos B\end{vmatrix} \quad[\because C_1.$ and $C_3$ are identical $]$

$=\cos A \cos B \begin{vmatrix} \sin A & 1 & 1 \\ \sin B & 1 & 1 \\ \sin C & 1 & 1\end{vmatrix} $

[Taking $\cos A$ and $\cos B$ common from $C_2$ and $C_3$ respectively] $=\cos A \cos B(0) \quad[\because C_2.$ and $C_3$ are identical $]$ $=0$

Solution

True.

Let

$ \Delta= \begin{vmatrix} x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h \end{vmatrix} $

Splitting up $C_1$

$ \Rightarrow \begin{vmatrix} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{vmatrix} + \begin{vmatrix} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{vmatrix} $

Splitting up $C_2$ in both determinants

$ \begin{aligned} \Rightarrow & \begin{vmatrix} x & p & l+f \\ y & q & m+g \\ z & r & n+h \end{vmatrix} + \begin{vmatrix} x & u & l+f \\ y & v & m+g \\ z & w & n+h \end{vmatrix} + \begin{vmatrix} a & p & l+f \\ b & q & m+g \\ c & r & n+h \end{vmatrix} \\ & + \begin{vmatrix} a & u & l+f \\ b & v & m+g \\ c & w & n+h \end{vmatrix} \end{aligned} $

Similarly by splitting $C_3$ in each determinant, we will get 8 determinants.

Solution

True.

$ \begin{aligned} & \text{ Given that: } \\ & \Delta= \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} =16 \\ & \text{ L.H.S. } \Delta_1= \begin{vmatrix} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{vmatrix} \\ & C_1 \to C_1+C_2+C_3 \\ & = \begin{vmatrix} 2 p+2 x+2 a & a+x & a+p \\ 2 q+2 y+2 b & b+y & b+q \\ 2 r+2 z+2 c & c+z & c+r \end{vmatrix} \\ & =2 \begin{vmatrix} p+x+a & a+x & a+p \\ q+y+b & b+y & b+q \\ r+z+c & c+z & c+r \end{vmatrix} \quad \text{ [Taking } 2 \text{ common from }C_1 \end{aligned} $

$ \begin{aligned} & C_1 \to C_1-C_2=2 \begin{vmatrix} p & a+x & a+p \\ q & b+y & b+q \\ r & c+z & c+r \end{vmatrix} \\ & .C_3 \to C_3-C_1{ }^{d}=2 \begin{vmatrix} p & a+x & a \\ \text{ Splitting up } C_2 \end{vmatrix} \begin{vmatrix} r & b+y & b \\ r & c+z & c \end{vmatrix} \rvert, \\ & =2 \begin{vmatrix} p & a & a \\ q & b & b \\ r & c & c \end{vmatrix} +2 \begin{vmatrix} p & x & a \\ q & y & b \\ r & z & c \end{vmatrix} =2(0)+2 \begin{vmatrix} p & x & a \\ q & y & b \\ r & z & c \end{vmatrix} \\ & =2 \begin{vmatrix} p & x & a \\ q & y & b \\ r & z & c \end{vmatrix} \Rightarrow 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} \quad(C_1 \leftrightarrow C_3 \text{ and } C_2 \leftrightarrow C_3 \\ & =2 \times 16=32 \end{aligned} $

Solution

True.

Let $\Delta= \begin{vmatrix} 1 & 1 & 1 \\ 1 & (1+\sin \theta) & 1 \\ 1 & 1 & 1+\cos \theta\end{vmatrix} $

$C_1 \to C_1-C_2, C_2 \to C_2-C_3$

$ = \begin{vmatrix} 0 & 0 & 1 \\ -\sin \theta & \sin \theta & 1 \\ 0 & -\cos \theta & 1+\cos \theta \end{vmatrix} $

Expanding along $C_3$

$ \begin{aligned} & =1 \begin{vmatrix} -\sin \theta & \sin \theta \\ 0 & -\cos \theta \end{vmatrix} =\sin \theta \cos \theta-0=\sin \theta \cos \theta \\ & =\frac{1}{2} \cdot 2 \sin \theta \cos \theta=\frac{1}{2} \sin 2 \theta \\ & =\frac{1}{2} \times 1 \quad \text{ [Maximum value of } \sin 2 \theta=1 \text{ ] } \\ & =\frac{1}{2} \end{aligned} $