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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.
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MathsClass 12

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 A11 = 7, A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0, A33 = 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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