Q.If $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$, verify that $A(\text{adj }A) = (\text{adj }A)A = |A|\,I$

Step 1: $A=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$.

Step 2: The nine cofactors work out to $A_{11}=0,\ A_{12}=-11,\ A_{13}=0,\ A_{21}=3,\ A_{22}=1,\ A_{23}=-1,\ A_{31}=2,\ A_{32}=8,\ A_{33}=3$, so $\text{adj}\,A=\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}$.

Step 3: $|A|$, expanding along row 1: $1(0\cdot3-(-2)\cdot0)-(-1)(3\cdot3-(-2)\cdot1)+2(3\cdot0-0\cdot1)=1(0)+1(9+2)+2(0)=11$.

Step 4: $A(\text{adj}\,A)=\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}=\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}$ (multiplying row by column and simplifying every entry). …

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