Question
If A = $\left[\begin{array}{lll} {1} & {3} & {3} \\ {1} & {4} & {3} \\ {1} & {3} & {4} \end{array}\right]$then verify that $A$ adj $A = | A| I.$ Also find $A^{–1}.$

Answer

We have $A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 \neq 0$
Now $A_{11} = 7, A_{12} = –1, A_{13} = –1, A_{21} = –3, A_{22} = 1,A_{23} = 0, A_{31} = –3, A_{32} = 0, A_{33} = 1$
Therefore $\text { adj } \mathrm{A}=\left[\begin{array}{rrr} {7} & {-3} & {-3} \\ {-1} & {1} & {0} \\ {-1} & {0} & {1} \end{array}\right]$
Now $A ($adj $A) = \left[\begin{array}{ccc} {1} & {3} & {3} \\ {1} & {4} & {3} \\ {1} & {3} & {4} \end{array}\right]\left[\begin{array}{ccc} {7} & {-3} & {-3} \\ {-1} & {1} & {0} \\ {-1} & {0} & {1} \end{array}\right]$
= $\left[\begin{array}{ccc} {7-3-3} & {-3+3+0} & {-3+0+3} \\ {7-4-3} & {-3+4+0} & {-3+0+3} \\ {7-3-4} & {-3+3+0} & {-3+0+4} \end{array}\right]$
$= \left[\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]=(1) \cdot\left[\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]=|\mathrm{A}| . \mathrm{I}$
Also $\mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|} \text { adj } \mathrm{A}$ = $\frac{1}{1}\left[\begin{array}{rrr} {7} & {-3} & {-3} \\ {-1} & {1} & {0} \\ {-1} & {0} & {1} \end{array}\right]=\left[\begin{array}{rrr} {7} & {-3} & {-3} \\ {-1} & {1} & {0} \\ {-1} & {0} & {1} \end{array}\right]$

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