Find the determinant of

$A=\left(\begin{array}{ccc}1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1\end{array}\right)$

Soln. $R_1 \leftarrow R_1+R_2$

$\therefore|A|= \left(\begin{array}{ccc} 1+x^2 & x+1 & x^2+x \\ x^2 & 1 & x \\ x & x^2 & 1 \end{array}\right)$

$R_1 \leftarrow R_1+R_3$

Then $|A|=\left|\begin{array}{ccc}1+x+x^2 & 1+x+x^2 & 1+x+1 x^2 \\ x^2 & 1 & x \\ x & x^2 & 1\end{array}\right|$

$\begin{array}{l}\therefore|A|=\left(1+x+x^2\right)\left|\begin{array}{ccc}1 & 1 & 1 \\ x^2 & 1 & x \\ x & x^2 & 1\end{array}\right|\end{array}$

Now $C_1 \leftarrow C_1-C_2$

$\therefore |A|=\left(1+x+x^2\right)\left| \begin{array}{ccc}0 & 1 & 1 \\ x^2-1 & 1 & x \\ x-x^2 & x^2 & 1 \end{array}\right|$

$C_2 \leftarrow C_2-C_3$

$\therefore|A|=\left(1+x+x^2\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ x^2-1 & 1-x & x \\ x(1-x) & x^2-1 & 1\end{array}\right|$

$\therefore|A|=\left(1+x+x^2\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ x^2-1 & 1-x & x \\ x(1-x) & x^2-1 & 1\end{array}\right|$

$= \left(1+x+x^2\right)\left(\left(x^2-1\right)^2-x(1-x)^2\right)$

$= \left(1+x+x^2\right)(x-1)^2\left((x+1)^2-x\right)$

$= \left(1+x+x^2\right)(x-1)^2\left(x^2+x+1\right)$

$= (x-1)\left(x^2+x+1\right) (x-1)\left(x^2+x+1\right)$

$= \left(1-x^3\right)^2$

Find the determinant of

$A=\left(\begin{array}{ccc}\sqrt{11}+\sqrt{3} & \sqrt{20} & \sqrt{5} \\ \sqrt{15}+\sqrt{22} & \sqrt{25} & \sqrt{10} \\ 3+\sqrt{55} & \sqrt{15} & \sqrt{25}\end{array}\right)$

Soln. $|A|=\left(\begin{array}{lll}\sqrt{11} & \sqrt{20} & \sqrt{5} \\ \sqrt{22} & \sqrt{25} & \sqrt{10} \\ \sqrt{55} & \sqrt{15} & \sqrt{25} \end{array}\right)+\left|\begin{array}{lll}\sqrt{3} & \sqrt{20} & \sqrt{5} \\ \sqrt{15} & \sqrt{25} & \sqrt{10} \\ \sqrt{3} \cdot \sqrt{3} & \sqrt{15} & \sqrt{25}\end{array}\right|$

$= \left|A_1\right|+\left|A_2\right|$

$\left|A_1\right|=\sqrt{11} \cdot \sqrt{5} \cdot \sqrt{5} \left|\begin{array}{ccc}1 & 2 & 1 \\ \sqrt{2} & \sqrt{5} & \sqrt{2} \\ \sqrt{5} & \sqrt{3} & \sqrt{5}\end{array}\right|$

Since two columns are identical the determinant $=0$

$\therefore\left|A_1\right|=0$

$\therefore|A|=\left|A_2\right|=\left|\begin{array}{lll}\sqrt{3} & \sqrt{20} & \sqrt{5} \\ \sqrt{15} & \sqrt{25} & \sqrt{10} \\ \sqrt{3} \cdot \sqrt{3} & \sqrt{15} & \sqrt{25}\end{array}\right|$

$\therefore|A|=\sqrt{3} \cdot \sqrt{5} \cdot \sqrt{5}\left|\begin{array}{ccc}1 & 2 & 1 \\ \sqrt{5} & \sqrt{5} & \sqrt{2} \\ \sqrt{3} & \sqrt{3} & \sqrt{5}\end{array}\right|$

Its determinants

$=1 \cdot(5-\sqrt{6})-2(5-\sqrt{6})$

$+1(\sqrt{15}-\sqrt{15})$

$=-1(5-\sqrt{6})=(\sqrt{6}-5)$

$\therefore|A|=5 \sqrt{3} \times (\sqrt{6}-5)$

Find the determinant of

$A=\left(\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right)$

Soln. $R_1 \leftarrow R_1+R_2+R_3$

$|A|= \left|\begin{array}{ccc}a+b+c & b+c+a & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

$\therefore |A|=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$

$|A|=(a+b+c)\left|\begin{array}{ccc}0 & 1 & 1 \\ 0 & b-c-a & 2b \\ a+b+c & 2c & c-a-b\end{array}\right|$

$\begin{array}{l}\therefore|A|=(a+b+c)(-1)^{3+1}(a+b+c)(2 b-(b-c-a))\end{array}$

$=(a+b+c)^3 Ans$

$A=\left|\begin{array}{ccc}x & x^2 & y z \\ y & y^2 & z x \\ z & z^2 & x y\end{array}\right|$

Soln. $R_1=R_1-R_2$

$|A|=\left|\begin{array}{ccc}x-y & (x-y)(x+y) & z(y-x) \\ y & y^2 & z x \\ z & z^2 & x y\end{array}\right|$

$\therefore|A|=(x-y) \cdot\left|\begin{array}{ccc}1 & x+y & -z \\ y & y^2 & z x \\ z & z^2 & x y\end{array}\right|$

$\quad R_2 \leftarrow R_2-R_3$

$|A|=(x-y)\left|\begin{array}{ccc}1 & x+y & -z \\ y-z & (y-z)(y+z) & x(z-y) \\ z & z^2 & x y\end{array}\right|$

$\therefore|A|=(x-y)(y-z)\left|\begin{array}{ccc}1 & x+y & -z \\ 1 & y+z & -x \\ z & z^2 & x y\end{array}\right|$

$\therefore R_2=R_2-R_1$ We have

$\therefore|A|=(x-y)(y-z)\left|\begin{array}{ccc}1 & x+y & -z \\ 0 & z-x & z-x \\ z & z^2 & x y\end{array}\right|$

$= (x-y)(y-z)(z-x)\\ =(x-y)(y-z)(z-x) |B|\left|\begin{array}{ccc}1 & x+y & -z \\ 0 & 1 & 1 \\ z & z^2 & x y\end{array}\right|$

$|B|=1 \cdot\left(x y-z^2\right)+z(x+y-(-z))$

$=\left(x y-z^2\right)+(x+y+z) z$

$=x y-z^2+x z+y z+z^2$

$=x y+y z+z x$

$\therefore|A| =(x-y)(y-z)(z-x)(x y+y z+z x)$

$A \cdot\left(\begin{array}{ccc}a^2+1 & a b & a c \\ a b & b^2+1 & b c \\ c a & c b & c^2+1\end{array}\right)$

Soln. $\therefore|A|=\frac{1}{a b c}\left|\begin{array}{ccc}a\left(a^2+1\right) & a^2 b & a^2 c \\ a b^2 & b\left(b^2+1\right) & b^2 c \\ c^2 a & c^2 b & c\left(c^2+1\right)\end{array}\right|$

$\therefore|A|=\frac{a b c}{a b c}\left|\begin{array}{lll}a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1\end{array}\right|$

$\therefore |A|=\left|\begin{array}{ccc}a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1\end{array}\right|$

$\therefore R_1 \leftarrow R_2+R_3+R_1$

$|A|=\left|\begin{array}{ccc}a^2+b^2+c^2+1 & a^2+b^2+c^2+1 & a^2+b^2+1 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1\end{array}\right|$

$|A|=(a^2+b^2+c^2+1)\left|\begin{array}{ccc}1 & 1 & 1 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1\end{array}\right|$

$C_1 \leftarrow C_1-C_3$

$C_2 \leftarrow C_2-C_3$

$|A|=(a^2+b^2+c^2+1)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & b^2 \\ -1 & -1 & c^2+1\end{array}\right|$

$\therefore |A| = (1+a^2+b^2+c^2)$

Find the determinant of

$A= \left(\begin{array}{ccc} n & ^np_n & n_{C_n} \\ n+1 & ^{n+1}P_{n+1} & ^{n+1}C_{n+1} \\ n+2 & ^{n+2}P_{n+2} & ^{n+2}C_{n+2}\end{array}\right)$

Soln. $\therefore A=\left(\begin{array}{ccc}n & n ! & 1 \\ n+1 & (n+1) ! & 1 \\ n+2 & (n+2) ! & 1\end{array}\right)$

$|A|=n !\left|\begin{array}{ccc}n & 1 & 1 \\ n+1 & (n+1) & 1 \\ n+2 & (n+1)(n+2) & 1\end{array}\right|$

$R_3=R_3-R_2$

$|A|=n !\left|\begin{array}{ccc}n & 1 & 1 \\ n+1 & n+1 & 1 \\ 1 & (n+1)^2 & 0\end{array}\right|$

$R_2=R_2-R_1$

$|A|=n !\left|\begin{array}{ccc}n & 1 & 1 \\ 1 & n & 0 \\ 1 & (n+1)^2 & 0\end{array}\right|$

$\therefore|A|=n !(-1)^{1+3}\left((n+1)^2-n\right)$

$=n !\left(n^2+n+1\right)$

Verification: $\quad n=2$

$A=\left|\begin{array}{lll}2 & 2 ! & 1 \\ 3 & 3 ! & 1 \\ 4 & 4 ! & 1\end{array}\right|$

$\therefore|A|=\left|\begin{array}{lll}2 & 2 & 1 \\ 3 & 6 & 1 \\ 4 & 24 & 1\end{array}\right|$

$|A|=\left|\begin{array}{ccc}2 & 2 & 1 \\ 3 & 6 & 1 \\ 1 & 18 & 0\end{array}\right|$

$R_3=R_3-R_2$

$=\left|\begin{array}{ccc}2 & 2 & 1 \\ 1 & 4 & 0 \\ 1 & 18 & 0\end{array}\right|$

$R_2=R_2-R_1$

$=|A|=1\cdot(18-4)=14$

We got for general $n$

$|A|=n !\left(n^2+n+1\right) \text {. }$

Putting $x=2$

We have $n!\left(n^2+n+1\right)$

$=2 !\left(2^2+2+1\right)$

$=2(4+2+1)$

$=2 \cdot 7=14$