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  • If A = [3 -5 -4 2] then find A^2 - 5A - 14I. Hence, obtain A^3.

If A=\begin{bmatrix} 3 &-5 \\-4 & 2 \end{bmatrix} then find A^2 - 5A - 14I. Hence, obtain A^3.

Answers (1)

Given, A=\begin{bmatrix} 3 &-5 \\-4 & 2 \end{bmatrix}

\therefore A^{2}=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$
According to the rule of matrix multiplication we can write:
\\ \mathrm{A}^{2}=\left[\begin{array}{cc}3 \times 3+(-5) \times(-4) & 3 \times(-5)+2 \times(-5) \\ -4 \times 3+2 \times(-4) & (-4) \times(-5)+2 \times 2\end{array}\right]$ \\$\Rightarrow \mathrm{A}^{2}=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]_{...}(1)$
We have to find: A^{2}-5 A-14I

\begin{array}{l} \therefore A^{2}-5 A-14I=\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]-5\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right]-14\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \Rightarrow A^{2}-5 A-14 I=\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]-\left[\begin{array}{cc} 15 & -25 \\ -20 & 10 \end{array}\right]-\left[\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}\right] \\ \Rightarrow A^{2}-5 A-14 I=\left[\begin{array}{cc} 29-15-14 & -25+25+0 \\ -20+20+0 & 24-10-14 \end{array}\right] \\ A^{2}-5 A-14 I=\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right]=0 \end{array}

We need to find value of A^3 using the above equation:

Now we have,

A\textsuperscript{2} - 5A - 14I = O

\Rightarrow $ A\textsuperscript{2} = 5A + 14I

By multiplying with A both sides we get,

\\$ \Rightarrow $ A\textsuperscript{2}.A = 5A.A + 14IA \\\\$ \Rightarrow $ A\textsuperscript{3} = 5A\textsuperscript{2} + 14A

By Using equation 1 we get:

\begin{array}{l} \Rightarrow \mathrm{A}^{3}=5\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]+14\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right] \\ \Rightarrow \mathrm{A}^{3}=\left[\begin{array}{cc} 145 & -125 \\ -100 & 120 \end{array}\right]+\left[\begin{array}{cc} 42 & -70 \\ -56 & 28 \end{array}\right] \\ \Rightarrow \mathrm{A}^{3}=\left[\begin{array}{cc} 187 & -195 \\ -156 & 148 \end{array}\right] \end{array}

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