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If A = [3−5−42],\begin{bmatrix}[r] 3 & -5 \ -4 & 2 \end{bmatrix}, find A2 - 5A - 14I, where I is unit matrix of order 2 × 2.

Matrices

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Answer

We need to find the value of A2 - 5A - 14I.

A2−5A−14I=[3−5−42][3−5−42]−5[3−5−42]−14[1001]=[3×3+(−5)×(−4)3×(−5)+(−5)×2(−4)×3+2×(−4)(−4)×(−5)+2×2]−[15−25−2010]−[140014]=[9+20−15−10−12−820+4]−[15−25−2010]−[140014]=[29−25−2024]−[15−25−2010]−[140014]=[29−15−14−25−(−25)−0−20−(−20)−024−10−14]=[29−29−25+25−20+2024−24]=[0000]A^2 - 5A -14I = \begin{bmatrix}[r] 3 & -5 \ -4 & 2 \end{bmatrix} \begin{bmatrix}[r] 3 & -5 \ -4 & 2 \end{bmatrix} - 5\begin{bmatrix}[r] 3 & -5 \ -4 & 2 \end{bmatrix} - 14\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 3 \times 3 + (-5) \times (-4) & 3 \times (-5) + (-5) \times 2 \ (-4) \times 3 + 2 \times (-4) & (-4) \times (-5) + 2 \times 2 \end{bmatrix} - \begin{bmatrix}[r] 15 & -25 \ -20 & 10 \end{bmatrix} - \begin{bmatrix}[r] 14 & 0 \ 0 & 14 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 9 + 20 & -15 - 10 \ -12 - 8 & 20 + 4 \end{bmatrix} - \begin{bmatrix}[r] 15 & -25 \ -20 & 10 \end{bmatrix} - \begin{bmatrix}[r] 14 & 0 \ 0 & 14 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 29 & -25 \ -20 & 24 \end{bmatrix} - \begin{bmatrix}[r] 15 & -25 \ -20 & 10 \end{bmatrix} - \begin{bmatrix}[r] 14 & 0 \ 0 & 14 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 29 - 15 - 14 & -25 - (-25) - 0 \ -20 - (-20) - 0 & 24 - 10 - 14 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 29 - 29 & -25 + 25 \ -20 + 20 & 24 - 24 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[0.5em]

Hence, the value of A2−5A−14I=[0000].A^2 - 5A - 14I = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix}.

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