Find the adjoint of the matrixA=−1−2−221−22−21 and hence show that A(adj A) = |A|I3

Given, A=−1−2−221−22−21Let Aij be the cofactor of an element aij of A. Then, cofactors of elements of |A| areA11=(−1)1+11−2−21=(1−4)=−3A12=(−1)1+22−221=−(2+4)=−6A13=(−1)1+3212−2=(−4−2)=−6A21=(−1)2+1−2−2−21=−(−2−4)=6A22=(−1)2+2−1−221=(−1+4)=3A23=(−1)2+3−1−22−2=−(2+4)=−6A31=(−1)3+1−2−21−2=(4+2)=6A32=(−1)3+2−1−22−2=−(2+4)=−6A33=(−1)3+3−1−221=(−1+4)=3Clearly, the adjoint of the matrix A is given byadj ⁡A=A11 A21 A31A12 A22 A32A13 A23 A33=−366−63−6−6−63Now, |A|=−1−2−221−22−21=−1(1−4)+2(2+4)−2(−4−2)=−1(−3)+2(6)−2(−6)=3+12+12=27and A⋅(adj⁡A)=−1−2−221−22−21−366−63−6−6−63=3+12+12−6−6+12−6+12−6−6−6+1212+3+1212−6−6−6+12−612−6−612+12+3=270002700027=27100010001=27I3=|A|I3 Hence proved.

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Find the adjoint of the matrix
$A = [\begin{array}{ccc} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{array}]$ and hence show that A(adj A) = |A| $I_{3}$

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Detailed Solution

Given, $A = [\begin{array}{ccc} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{array}]$
$Let A_{ij} be the cofactor of an element a_{i j} of A . Then, cofactors of elements of | A | are$ $\begin{array}{l} A_{11} = (- 1)^{1 + 1} |\begin{array}{cc} 1 & - 2 \\ - 2 & 1 \end{array}| = (1 - 4) = - 3 \\ A_{12} = (- 1)^{1 + 2} |\begin{array}{cc} 2 & - 2 \\ 2 & 1 \end{array}| = - (2 + 4) = - 6 \\ A_{13} = (- 1)^{1 + 3} |\begin{array}{cc} 2 & 1 \\ 2 & - 2 \end{array}| = (- 4 - 2) = - 6 \\ A_{21} = (- 1)^{2 + 1} |\begin{array}{cc} - 2 & - 2 \\ - 2 & 1 \end{array}| = - (- 2 - 4) = 6 \\ A_{22} = (- 1)^{2 + 2} |\begin{array}{cc} - 1 & - 2 \\ 2 & 1 \end{array}| = (- 1 + 4) = 3 \\ A_{23} = (- 1)^{2 + 3} |\begin{array}{cc} - 1 & - 2 \\ 2 & - 2 \end{array}| = - (2 + 4) = - 6 \\ A_{31} = (- 1)^{3 + 1} |\begin{array}{cc} - 2 & - 2 \\ 1 & - 2 \end{array}| = (4 + 2) = 6 \\ A_{32} = (- 1)^{3 + 2} |\begin{array}{cc} - 1 & - 2 \\ 2 & - 2 \end{array}| = - (2 + 4) = - 6 \\ A_{33} = (- 1)^{3 + 3} |\begin{array}{cc} - 1 & - 2 \\ 2 & 1 \end{array}| = (- 1 + 4) = 3 \end{array}$
Clearly, the adjoint of the matrix A is given by
$adj A = [\begin{array}{lll} A_{11} A_{21} A_{31} \\ A_{12} A_{22} A_{32} \\ A_{13} A_{23} A_{33} \end{array}] = [\begin{array}{ccc} - 3 & 6 & 6 \\ - 6 & 3 & - 6 \\ - 6 & - 6 & 3 \end{array}]$
Now, $| A | = |\begin{array}{ccc} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{array}|$
$\begin{array}{l} = - 1 (1 - 4) + 2 (2 + 4) - 2 (- 4 - 2) \\ = - 1 (- 3) + 2 (6) - 2 (- 6) \\ = 3 + 12 + 12 = 27 \end{array}$
and $A \cdot (adj A) = [\begin{array}{ccc} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{array}] [\begin{array}{ccc} - 3 & 6 & 6 \\ - 6 & 3 & - 6 \\ - 6 & - 6 & 3 \end{array}]$
$\begin{array}{l} = [\begin{array}{lcc} 3 + 12 + 12 & - 6 - 6 + 12 & - 6 + 12 - 6 \\ - 6 - 6 + 12 & 12 + 3 + 12 & 12 - 6 - 6 \\ - 6 + 12 - 6 & 12 - 6 - 6 & 12 + 12 + 3 \end{array}] \\ = [\begin{array}{ccc} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{array}] = 27 [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \end{array}$
$= 27 I_{3} = | A | I_{3} Hence proved .$