Question

If $f (x) = x^2 + 4x - 5$ and $A = \begin{bmatrix}1&2\\ 4&-3\end{bmatrix} $ then f (A) is equal to

• A. $\begin{bmatrix} 0 & - 4 \\ 8 & 8 \end{bmatrix} $
• B. $\begin{bmatrix} 2 & 1 \\ 2 & 0 \end{bmatrix} $
• C. $\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} $
• D. $\begin{bmatrix} 8 & 4 \\ 8 & 0 \end{bmatrix} $

Answer 1

Answer: (D)

Given : $A = \begin{bmatrix}1&2\\ 4&-3\end{bmatrix} $
$\therefore A^{2} = A.A = \begin{bmatrix}1&2\\ 4&-3\end{bmatrix}\begin{bmatrix}1&2\\ 4&-3\end{bmatrix}$
$ = \begin{bmatrix}1+8&2-6\\ 4-12&8+9\end{bmatrix} = \begin{bmatrix}9&-4\\ -8&17\end{bmatrix}$
Now, $ f \left(x\right) = x^{2} + 4x - 5$
$ \therefore f \left(A\right) = A^{2} + 4A - 5$
$ = A^{2} + 4A - 5 I $ (I is a $2 \times 2$unit matrix)
$ = \begin{bmatrix}9&-4\\ -8&17\end{bmatrix} + 4 \begin{bmatrix}1&2\\ 4&-3\end{bmatrix}- 5 \begin{bmatrix}1&0\\ 0&1\end{bmatrix} $
$ = \begin{bmatrix}9&-4\\ -8&17\end{bmatrix} + \begin{bmatrix}4&8\\ 16&-12\end{bmatrix}+ \begin{bmatrix}-5&0\\ 0&-5\end{bmatrix}$
$ = \begin{bmatrix}8&4\\ 8&0\end{bmatrix} $