Q.27 If \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), prove that \(A\cdot(\operatorname{adj} A) = (\operatorname{adj} A)\cdot A = |A|\cdot I\)
Best answer
Answer: Each product equals \(\begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} = |A|\cdot I\); proved.
\(|A| = 4 - 6 = -2\).
Cofactors: \(A_{11} = 4,\ A_{12} = -3,\ A_{21} = -2,\ A_{22} = 1\). So \(\operatorname{adj} A = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\).
\(A\cdot(\operatorname{adj} A) = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} 4 - 6 & -2 + 2 \\ 12 - 12 & -6 + 4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}\)
\((\operatorname{adj} A)\cdot A = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 - 6 & 8 - 8 \\ -3 + 3 & -6 + 4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}\)
\(|A|\cdot I = -2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}\)
Hence \(A\cdot(\operatorname{adj} A) = (\operatorname{adj} A)\cdot A = |A|\cdot I\).
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