If A = […] and f(x) = 2x^2-4x+5, the f(A) is equal to

The correct answer is Option (1) → $\begin{bmatrix}19&-16\\-32&51\end{bmatrix}$

Given:

$A = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix}$, $f(x) = 2x^2 - 4x + 5$

Compute $A^2$:

$A^2 = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} = \begin{bmatrix} (1)(1)+(2)(4) & (1)(2)+(2)(-3) \\ (4)(1)+(-3)(4) & (4)(2)+(-3)(-3) \end{bmatrix} = \begin{bmatrix} 9 & -4 \\ -8 & 17 \end{bmatrix}$

Now compute:

$2A^2 = \begin{bmatrix} 18 & -8 \\ -16 & 34 \end{bmatrix}$

$-4A = \begin{bmatrix} -4 & -8 \\ -16 & 12 \end{bmatrix}$

$5I = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

Add them:

$f(A) = 2A^2 - 4A + 5I = \begin{bmatrix} 18 & -8 \\ -16 & 34 \end{bmatrix} + \begin{bmatrix} -4 & -8 \\ -16 & 12 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

$f(A) = \begin{bmatrix} 19 & -16 \\ -32 & 51 \end{bmatrix}$

Therefore, $f(A) = \begin{bmatrix} 19 & -16 \\ -32 & 51 \end{bmatrix}$.

Note: The question has been reproduced as per the NTA paper. In matrix polynomials, the constant term is implicitly understood as a scalar multiple of the identity matrix of the same order. Therefore, “5” is interpreted as “5I”, where I denotes the identity matrix.