Subjective Type

Verify $$A (adj A) = (adj A) A = |A| I$$ $$A=\begin{bmatrix} 1 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 3 \end{bmatrix}$$

Solution

Let $$A=\begin{bmatrix} 1 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 3 \end{bmatrix}$$

Here, $$|A|=-1(9-2)=-7$$ (expanding along the second column)

$$|A|I=-7I$$ .....(1)

We know that, $$adj A =C^{T}$$

So, we will find out co-factors of each element of A.
$$C_{ 11 }=(-1)^{ 1+1 }\begin{vmatrix} 0 & 2 \\ 0 & 3 \end{vmatrix}$$
$$\Rightarrow C_{11}=0$$

$$C_{ 12 }=(-1)^{ 1+2 }\begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix}$$
$$\Rightarrow C_{12}=-(9-2)=-7$$

$$C_{ 13 }=(-1)^{ 1+3 }\begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix}$$
$$\Rightarrow C_{13}=0$$

$$C_{ 21 }=(-1)^{ 2+1 }\begin{vmatrix} 1 & 2 \\ 0 & 3 \end{vmatrix}$$
$$\Rightarrow C_{21}=-(3-0)=-3$$

$$C_{ 22 }=(-1)^{ 2+2 }\begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix}$$
$$\Rightarrow C_{22}=3-2=1$$

$$C_{ 23 }=(-1)^{ 2+3 }\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}$$
$$\Rightarrow C_{23}=-(0-1)=1$$

$$C_{ 31 }=(-1)^{ 3+1 }\begin{vmatrix} 1 & 2 \\ 0 & 2 \end{vmatrix}$$
$$\Rightarrow C_{31}=2-0=2$$

$$C_{ 32 }=(-1)^{ 3+2 }\begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix}$$
$$\Rightarrow C_{32}=-(2-6)=4$$

$$C_{ 33 }=(-1)^{ 3+3 }\begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix}$$
$$\Rightarrow C_{33}=0-3=-3$$

So, the cofactor matrix $$C=\begin{bmatrix} 0 & -7 & 0 \\ -3 & 1 & 1 \\ 2 & 4 & -3 \end{bmatrix}$$

$$\Rightarrow adj A= { C }^{ T }=\begin{bmatrix} 0 & -3 & 2 \\ -7 & 1 & 4 \\ 0 & 1 & -3 \end{bmatrix}$$

Consider $$ (adj A)A=\begin{bmatrix} 0 & -3 & 2 \\ -7 & 1 & 4 \\ 0 & 1 & -3 \end{bmatrix}\begin{bmatrix} 1 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 3 \end{bmatrix}$$

$$=\begin{bmatrix} 0-9+2 & 0+0+0 & 0-6+6 \\ -7+3+4 & -7+0+0 & -14+2+12 \\ 0+3-3 & 0+0+0 & 0+2-9 \end{bmatrix}$$

$$=\begin{bmatrix} -7 & 0 & 0 \\ 0 & -7 & 0 \\ 0 & 0 & -7 \end{bmatrix}$$

$$=-7\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\Rightarrow (adj A)A=-7I$$ .....(2)

Now, $$ A(adjA)=\begin{bmatrix} 1 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 0 & -3 & 2 \\ -7 & 1 & 4 \\ 0 & 1 & -3 \end{bmatrix}$$

$$=\begin{bmatrix} 0-7+0 & -3+1+2 & 2+4-6 \\ 0+0+0 & -9+0+2 & 6+0-6 \\ 0+0+0 & -3+0+3 & 2+0-9 \end{bmatrix}$$

$$=\begin{bmatrix} -7 & 0 & 0 \\ 0 & -7 & 0 \\ 0 & 0 & -7 \end{bmatrix}$$

$$=-7\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\Rightarrow A(adj A)=-7I$$ ....(3)

From (1), (2) and (3), we get
$$A(adj A)=(adj A)A=|A|I$$

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