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  • Verify A(adj A) = (adj A)A = |A| I. Q : 3 Matrix 2 3 -4 -6

Q : 3           Verify \small A (adj A)=(adj A)A=|A|I.

                    \small \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}

Answers (1)

best_answer

Given the matrix: \small \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}

Let  \small A = \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}

Calculating the cofactors;

\small A_{11} = (-1)^{1+1}(-6) = -6   

\small A_{12} = (-1)^{1+2}(-4) = 4

\small A_{21} = (-1)^{2+1}(3) = -3

\small A_{22} = (-1)^{2+2}(2) = 2

Hence, \small adjA = \begin{bmatrix} -6 &-3 \\ 4& 2 \end{bmatrix}

Now,

 \small A (adj A) = \begin{bmatrix} 2 &3 \\ -4&-6 \end{bmatrix}\left ( \begin{bmatrix} -6 &-3 \\ 4 &2 \end{bmatrix} \right )

\small \begin{bmatrix} -12+12 &-6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}

aslo,

 \small (adjA)A = \begin{bmatrix} -6 &-3 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 2 &3 \\ -4& -6 \end{bmatrix}

\small = \begin{bmatrix} -12+12 &-18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}

Now, calculating |A|; 

\small |A| = -12-(-12) = -12+12 = 0 

So, \small |A|I = 0\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}

Hence we get

 \small A (adj A)=(adj A)A=|A|I

Posted by

Divya Prakash Singh

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