Solution

Let $A= \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} $.

We have,

$A _{11}=4, A _{12}=-3, A _{21}=-2, A _{22}=1$

$\therefore adj A= \begin{vmatrix} A _{11} & A _{21} \\ A _{12} & A _{22}\end{vmatrix} = \begin{vmatrix} 4 & -2 \\ -3 & 1 \end{vmatrix} $

Solution

Let $A= \begin{cases} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{cases} $.

We have,

$A _{11}= \begin{vmatrix} 3 & 5 \\ 0 & 1\end{vmatrix} =3-0=3$

$A _{12}=- \begin{vmatrix} 2 & 5 \\ -2 & 1\end{vmatrix} =-(2+10)=-12$

$A _{13}= \begin{vmatrix} 2 & 3 \\ -2 & 0\end{vmatrix} =0+6=6$ $A _{21}=- \begin{vmatrix} -1 & 2 \\ 0 & 1\end{vmatrix} =-(-1-0)=1$

$A _{22}= \begin{vmatrix} 1 & 2 \\ -2 & 1\end{vmatrix} =1+4=5$

$A _{23}=- \begin{vmatrix} 1 & -1 \\ -2 & 0\end{vmatrix} =-(0-2)=2$

$A _{31}= \begin{vmatrix} -1 & 2 \\ 3 & 5\end{vmatrix} =-5-6=-11$

$A _{32}=- \begin{vmatrix} 1 & 2 \\ 2 & 5\end{vmatrix} =-(5-4)=-1$

$A _{33}= \begin{vmatrix} 1 & -1 \\ 2 & 3\end{vmatrix} =3+2=5$

Hence, adj $A= \begin{vmatrix} A _{11} & A _{21} & A _{31} \\ A _{12} & A _{22} & A _{32} \\ A _{13} & A _{23} & A _{33}\end{vmatrix} = \begin{vmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{vmatrix} $.

Solution

$A= \begin{vmatrix} 2 & 3 \\ -4 & -6 \end{vmatrix} $

we have,

$|A|=-12-(-12)=-12+12=0$

$\therefore|A| I=0 \begin{vmatrix} 1 & 0 \\ 0 & 1\end{vmatrix} = \begin{vmatrix} 0 & 0 \\ 0 & 0 \end{vmatrix} $

Now,

$A _{11}=-6, A _{12}=4, A _{21}=-3, A _{22}=2$

$\therefore adj A= \begin{vmatrix} -6 & -3 \\ 4 & 2 \end{vmatrix} $

Now,

$ \begin{aligned} A(adj A) & = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \\ & = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \end{aligned} $

Also, $(adj A) A= \begin{vmatrix} -6 & -3 \\ 4 & 2\end{vmatrix} \begin{vmatrix} 2 & 3 \\ -4 & -6 \end{vmatrix} $

$ = \begin{bmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} $

Hence, $A(adj A)=(adj A) A=|A| I$.

Solution

$ \begin{aligned} & A= \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \\ & |A|=1(0-0)+1(9+2)+2(0-0)=11 \\ & \therefore|A| I=11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \end{aligned} $

Now,

$ \begin{aligned} & A _{11}=0, A _{12}=-(9+2)=-11, A _{13}=0 \\ & A _{21}=-(-3-0)=3, A _{22}=3-2=1, A _{23}=-(0+1)=-1 \\ & A _{31}=2-0=2, A _{32}=-(-2-6)=8, A _{33}=0+3=3 \end{aligned} $

$\therefore adj A= \begin{cases} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{cases} $

Now,

$ \begin{aligned} A(adj A) & = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 0+11+0 & 3-1-2 & 2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9 \end{bmatrix} \\ & = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \end{aligned} $

Also,

$ \begin{aligned} (adj A) \cdot A & = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0+0+0 & 0+2+9 \end{bmatrix} \\ & = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \end{aligned} $

Hence, $A(adj A)=(adj A) A=|A| I$.

Solution

$|A|=2 \times 3-4 \times-2=14$

Minors are $M_{11}=3 M_{12}=4 M_{21}=-2 M_{22}=2$

Cofactors are $C_{11}=3 C_{12}=-4 C_{21}=2 C_{22}=2$

Adj. $A=\left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right]$

$A^{-1}=\frac{1}{|A|} A d j . A=\frac{1}{14}\left[\begin{array}{cc} 3 & 2 \\ -4 & 2 \end{array}\right]$

Solution

Let $A= \begin{vmatrix} -1 & 5 \\ -3 & 2 \end{vmatrix} $.

we have,

$|A|=-2+15=13$

Now,

$A _{11}=2, A _{12}=3, A _{21}=-5, A _{22}=-1$

$\therefore adj A= \begin{vmatrix} 2 & -5 \\ 3 & -1 \end{vmatrix} $

$\therefore A^{-1}=\frac{1}{|A|} adj A=\frac{1}{13} \begin{vmatrix} 2 & -5 \\ 3 & -1 \end{vmatrix} $

Solution

Let $A= \begin{vmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{vmatrix} $.

We have,

$|A|=1(10-0)-2(0-0)+3(0-0)=10$

Now,

$A _{11}=10-0=10, A _{12}=-(0-0)=0, A _{13}=0-0=0$

$A _{21}=-(10-0)=-10, A _{22}=5-0=5, A _{23}=-(0-0)=0$

$A _{31}=8-6=2, A _{32}=-(4-0)=-4, A _{33}=2-0=2$

$\therefore adj A= \begin{vmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{vmatrix} $

$\therefore A^{-1}=\frac{1}{|A|} adj A=\frac{1}{10} \begin{vmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{vmatrix} $

Solution

Let $A= \begin{vmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{vmatrix} $.

We have,

$|A|=1(-3-0)-0+0=-3$

Now,

$A _{11}=-3-0=-3, A _{12}=-(-3-0)=3, A _{13}=6-15=-9$

$A _{21}=-(0-0)=0, A _{22}=-1-0=-1, A _{23}=-(2-0)=-2$

$A _{31}=0-0=0, A _{32}=-(0-0)=0, A _{33}=3-0=3$

$\therefore adj A= \begin{vmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{vmatrix} $

$\therefore A^{-1}=\frac{1}{|A|}$ adj $A=-\frac{1}{3} \begin{vmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{vmatrix} $

Solution

Let $A= \begin{vmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{vmatrix} $.

We have,

$ \begin{aligned} |A| & =2(-1-0)-1(4-0)+3(8-7) \\ & =2(-1)-1(4)+3(1) \\ & =-2-4+3 \\ & =-3 \end{aligned} $

Now,

$ \begin{aligned} & A _{11}=-1-0=-1, A _{12}=-(4-0)=-4, A _{13}=8-7=1 \\ & A _{21}=-(1-6)=5, A _{22}=2+21=23, A _{23}=-(4+7)=-11 \\ & A _{31}=0+3=3, A _{32}=-(0-12)=12, A _{33}=-2-4=-6 \\ & \therefore \text{ adj } A= \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix} \\ & \therefore A^{-1}=\frac{1}{|A|} \text{ adjA }=-\frac{1}{3} \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix} \end{aligned} $

Solution

Let $A= \begin{vmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{vmatrix} $.

By expanding along $C_1$, we have:

$|A|=1(8-6)-0+3(3-4)=2-3=-1$

Now,

$A _{11}=8-6=2, A _{12}=-(0+9)=-9, A _{13}=0-6=-6$

$A _{21}=-(-4+4)=0, A _{22}=4-6=-2, A _{23}=-(-2+3)=-1$

$A _{31}=3-4=-1, A _{32}=-(-3-0)=3, A _{33}=2-0=2$

$\therefore adj A= \begin{vmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{vmatrix} $

$\therefore A^{-1}=\frac{1}{|A|}$ adj $A=- \begin{vmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2\end{vmatrix} = \begin{vmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{vmatrix} $

Solution

Let $A= \begin{vmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{vmatrix} $.

We have,

$|A|=1(-\cos ^{2} \alpha-\sin ^{2} \alpha)=-(\cos ^{2} \alpha+\sin ^{2} \alpha)=-1$

Now,

$A _{11}=-\cos ^{2} \alpha-\sin ^{2} \alpha=-1, A _{12}=0, A _{13}=0$

$A _{21}=0, A _{22}=-\cos \alpha, A _{23}=-\sin \alpha$

$A _{31}=0, A _{32}=-\sin \alpha, A _{33}=\cos \alpha$

$\therefore adj A= \begin{vmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{vmatrix} $

$\therefore A^{-1}=\frac{1}{|A|} \cdot adj A=- \begin{vmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha\end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{vmatrix} $

Solution

Let $A= \begin{vmatrix} 3 & 7 \\ 2 & 5 \end{vmatrix} $.

We have,

$|A|=15-14=1$

Now,

$A _{11}=5, A _{12}=-2, A _{21}=-7, A _{22}=3$

$\therefore adj A= \begin{vmatrix} 5 & -7 \\ -2 & 3 \end{vmatrix} $

$\therefore A^{-1}=\frac{1}{|A|} \cdot adj A= \begin{vmatrix} 5 & -7 \\ -2 & 3 \end{vmatrix} $

Now, let $B= \begin{vmatrix} 6 & 8 \\ 7 & 9 \end{vmatrix} $.

We have,

$|B|=54-56=-2$

$\therefore adj B= \begin{vmatrix} 9 & -8 \\ -7 & 6 \end{vmatrix} $

$\therefore B^{-1}=\frac{1}{|B|} adj B=-\frac{1}{2} \begin{vmatrix} 9 & -8 \\ -7 & 6\end{vmatrix} = \begin{vmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{vmatrix} $

Now,

$$ \begin{align*} B^{-1} A^{-1} & = \begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \\ & = \begin{bmatrix} -\frac{45}{2}-8 & \frac{63}{2}+12 \\ \frac{35}{2}+6 & -\frac{49}{2}-9 \end{bmatrix} = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \tag{1} \end{align*} $$

Then,

$ \begin{aligned} A B & = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} \\ & = \begin{bmatrix} 18+49 & 24+63 \\ 12+35 & 16+45 \end{bmatrix} \\ & = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix} \end{aligned} $

Therefore, we have $|A B|=67 \times 61-87 \times 47=4087-4089=-2$.

Also,

$$ \begin{align*} & adj(A B)= \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} \\ & \begin{align*} \therefore(A B)^{-1}=\frac{1}{|A B|} adj(A B) & =-\frac{1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} \\ & = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} . \end{align*} \end{align*} $$

From (1) and (2), we have:

$(A B)^{-1}=B^{-1} A^{-1}$

Hence, the given result is proved.

Solution

$A= \begin{vmatrix} 3 & 1 \\ -1 & 2 \end{vmatrix} $

$A^{2}=A \cdot A= \begin{vmatrix} 3 & 1 \\ -1 & 2\end{vmatrix} \begin{vmatrix} 3 & 1 \\ -1 & 2\end{vmatrix} = \begin{vmatrix} 9-1 & 3+2 \\ -3-2 & -1+4\end{vmatrix} = \begin{vmatrix} 8 & 5 \\ -5 & 3 \end{vmatrix} $

$\therefore A^{2}-5 A+7 I$

$= \begin{vmatrix} 8 & 5 \\ -5 & 3\end{vmatrix} -5 \begin{vmatrix} 3 & 1 \\ -1 & 2\end{vmatrix} +7 \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} $

$= \begin{vmatrix} 8 & 5 \\ -5 & 3\end{vmatrix} - \begin{vmatrix} 15 & 5 \\ -5 & 10\end{vmatrix} + \begin{vmatrix} 7 & 0 \\ 0 & 7 \end{vmatrix} $

$= \begin{vmatrix} -7 & 0 \\ 0 & -7\end{vmatrix} + \begin{vmatrix} 7 & 0 \\ 0 & 7\end{vmatrix} = \begin{vmatrix} 0 & 0 \\ 0 & 0 \end{vmatrix} $

Hence, $A^{2}-5 A+7 I=O$.

$\therefore A \cdot A-5 A=-7 I$

$\Rightarrow A \cdot A(A^{-1})-5 A A^{-1}=-7 I A^{-1} \quad[.$ Post-multiplying by $A^{-1}$ as $.|A| \neq 0]$

$\Rightarrow A(A A^{-1})-5 I=-7 A^{-1}$

$\Rightarrow A I-5 I=-7 A^{-1}$

$\Rightarrow A^{-1}=-\frac{1}{7}(A-5 I)$

$\Rightarrow A^{-1}=\frac{1}{7}(5 I-A)$

$=\frac{1}{7}( \begin{vmatrix} 5 & 0 \\ 0 & 5\end{vmatrix} - \begin{vmatrix} 3 & 1 \\ -1 & 2\end{vmatrix} )=\frac{1}{7} \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} $

$\therefore A^{-1}=\frac{1}{7} \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} $

Solution

$A= \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} $

$\therefore A^{2}= \begin{vmatrix} 3 & 2 \\ 1 & 1\end{vmatrix} \begin{vmatrix} 3 & 2 \\ 1 & 1\end{vmatrix} = \begin{vmatrix} 9+2 & 6+2 \\ 3+1 & 2+1\end{vmatrix} = \begin{vmatrix} 11 & 8 \\ 4 & 3 \end{vmatrix} $

Now,

$A^{2}+a A+b I=O$

$\Rightarrow(A A) A^{-1}+a A A^{-1}+b I A^{-1}=O \quad \quad[$ Post-multiplying by $A^{-1}$ as $.|A| \neq 0]$

$\Rightarrow A(A A^{-1})+a I+b(I A^{-1})=O$

$\Rightarrow A I+a I+b A^{-1}=O$

$\Rightarrow A+a I=-b A^{-1}$

$\Rightarrow A^{-1}=-\frac{1}{b}(A+a I)$

Now,

$A^{-1}=\frac{1}{|A|}$ adj $A=\frac{1}{1} \begin{vmatrix} 1 & -2 \\ -1 & 3\end{vmatrix} = \begin{vmatrix} 1 & -2 \\ -1 & 3 \end{vmatrix} $

We have:

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

Comparing the corresponding elements of the two matrices, we have:

$-\frac{1}{b}=-1 \Rightarrow b=1$

$\frac{-3-a}{b}=1 \Rightarrow-3-a=1 \Rightarrow a=-4$

Hence, -4 and 1 are the required values of $a$ and $b$ respectively.

Solution

$ \begin{aligned} & A= \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \\ & A^{2}= \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \end{aligned} $

$= \begin{bmatrix} 1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{bmatrix}$

$= \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}$

$ \begin{aligned} & A^{3}=A^{2} \cdot A= \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+16+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42 \end{bmatrix} \\ & = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix} \end{aligned} $

$\therefore A^{3}-6 A^{2}+5 A+11 I$

$= \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{bmatrix} -6 \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{bmatrix} +5 \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{bmatrix} +11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $

$= \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{bmatrix} - \begin{bmatrix} 24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84\end{bmatrix} + \begin{bmatrix} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15\end{bmatrix} + \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} $

$= \begin{bmatrix} 24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84\end{bmatrix} - \begin{bmatrix} 24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84 \end{bmatrix} $

$= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} =O$

Thus, $A^{3}-6 A^{2}+5 A+11 I=O$.

Now,

$A^{3}-6 A^{2}+5 A+11 I=O$

$\Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+5 A A^{-1}+11 I A^{-1}=0 \quad[.$ Post-multiplying by $A^{-1}$ as $|A| \neq 0$ ]

$\Rightarrow A A(A A^{-1})-6 A(A A^{-1})+5(A A^{-1})=-11(I A^{-1})$

$\Rightarrow A^{2}-6 A+5 I=-11 A^{-1}$

$\Rightarrow A^{-1}=-\frac{1}{11}(A^{2}-6 A+5 I)$

Now,

$ \begin{aligned} & A^{2}-6 A+5 I \\ & = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} -6 \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} +5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} - \begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix} + \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \\ & = \begin{bmatrix} 9 & 2 & 1 \\ -3 & 13 & -14 \\ 7 & -3 & 19 \end{bmatrix} - \begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix} \\ & = \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix} \end{aligned} $

From equation (1), we have:

$ A^{-1}=-\frac{1}{11} \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix} =\frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix} $

Solution

$$ \begin{aligned} & A= \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \\ & .A^{2}=\begin{bmatrix} { \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} .} & \begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix} & .\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \end{bmatrix} \begin{bmatrix} 2 & -1 & 1 \\ -1 \\ 1 & 2 & -1 \\ -1 & 2 \end{bmatrix} \\ & = \begin{bmatrix} 4+1+1 & -2-2-1 & 2+1+2 \\ -2-2-1 & 1+4+1 & -1-2-2 \\ 2+1+2 & -1-2-2 & 1+1+4 \end{bmatrix} \\ & = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} \\ & A^{3}=A^{2} A= \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \\ & = \begin{bmatrix} 12+5+5 & -6-10-5 & 6+5+10 \\ -10-6-5 & 5+12+5 & -5-6-10 \\ 10+5+6 & -5-10-6 & 5+5+12 \end{bmatrix} \\ & = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} \end{aligned} $$

Now,

$ \begin{aligned} & A^{3}-6 A^{2}+9 A-4 I \\ & = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} -6 \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} +9 \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} -4 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} - \begin{bmatrix} 36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36 \end{bmatrix} + \begin{bmatrix} 18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} \\ & = \begin{bmatrix} 40 & -30 & 30 \\ -30 & 40 & -30 \\ 30 & -30 & 40 \end{bmatrix} - \begin{bmatrix} 40 & -30 & 30 \\ -30 & 40 & -30 \\ 30 & -30 & 40 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{aligned} $

$\therefore A^{3}-6 A^{2}+9 A-4 I=O$

Now,

$$ \begin{align*} & A^{3}-6 A^{2}+9 A-4 I=O \\ & \Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+9 A A^{-1}-4 I A^{-1}=O \\ & \Rightarrow A A(A A^{-1})-6 A(A A^{-1})+9(A A^{-1})=4(I A^{-1}) \\ & \Rightarrow A A I-6 A I+9 I=4 A^{-1} \\ & \Rightarrow A^{2}-6 A+9 I=4 A^{-1} \tag{1}\\ & \Rightarrow A^{-1}=\frac{1}{4}(A^{2}-6 A+9 I) \end{align*} $$

$ \Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+9 A A^{-1}-4 I A^{-1}=O \quad \quad[\text{ Post-multiplying by } A^{-1} \text{ as }|A| \neq 0] $

$A^{2}-6 A+9 I$

$= \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{bmatrix} -6 \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{bmatrix} +9 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $

$= \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{bmatrix} - \begin{bmatrix} 12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12\end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} $

$= \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix} $

From equation (1), we have:

$ A^{-1}=\frac{1}{4} \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix} $

Solution

We know that,

$(adj A) A=|A| I= \begin{bmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{bmatrix} $

$\Rightarrow|(adj A) A|= \begin{vmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A|\end{vmatrix} $

$\Rightarrow|adj A||A|=|A|^{3} \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{vmatrix} =|A|^{3}(I)$

$\therefore|adj A|=|A|^{2}$

Hence, the correct answer is B.

Solution

Since $A$ is an invertible matrix, $A^{-1}$ exists and $A^{-1}=\frac{1}{|A|}$ adj $A$.

As matrix $A$ is of order 2, let $A= \begin{vmatrix} a & b \\ c & d \end{vmatrix} $.

Then, $|A|=a d-b c$ and $a d j A= \begin{vmatrix} d & -b \\ -c & a \end{vmatrix}$.

Now,

$ \begin{aligned} & A^{-1}=\frac{1}{|A|} \text{ adj } A= \begin{vmatrix} \frac{d}{|A|} & \frac{-b}{|A|} \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix} \\ & \therefore|A^{-1}|= \begin{vmatrix} \frac{d}{|A|} & \frac{-b}{|A|} \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix} =\frac{1}{|A|^{2}} \begin{vmatrix} d & -b \\ -c & a \end{vmatrix} =\frac{1}{|A|^{2}}(a d-b c)=\frac{1}{|A|^{2}} \cdot|A|=\frac{1}{|A|} \\ & \therefore det(A^{-1})=\frac{1}{det(A)} \end{aligned} $

Hence, the correct answer is B.