Given, matrix $$A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}$$ and I is a unit matrix of order 2 × 2.
(a) Find the matrix, A2 and 5A
(b) value of $k$, such that $A^2 - 5A = k.I$
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(a) Calculating A2,
$$A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 \times 3 + (-5)(-4) & 3 \times (-5) + (-5)(2) \\ (-4)(3) + 2(-4) & (-4)(-5) + 2 \times 2 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 + 20 & -15 - 10 \\ -12 - 8 & 20 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}.$$
Calculating 5A,
$$5A = 5\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix}.$$
Hence, $$A^2 = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}$$ and $$5A = \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix}$$.
(b) Calculating A2 − 5A,
$$A^2 - 5A = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} - \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} \\[1em] = \begin{bmatrix} 29 - 15 & -25 - (-25) \\ -20 - (-20) & 24 - 10 \end{bmatrix} \\[1em] = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} \\[1em] = 14\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] = 14 \cdot I.$$
Comparing with $A^2 - 5A = k.I$,
$\Rightarrow k = 14.$
Hence, the value of $\bm{k}$ = 14.