Q119 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
$\text { If } A=\left[\begin{array}{cc}
1 & -2 \\
4 & 5
\end{array}\right] \text { and } f(t)=t^2-3 t+7 \text { then } f(A)+\left[\begin{array}{cc}
3 & 6 \\
-12 & -9
\end{array}\right] \text { is }$
- A$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$
- B$\left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$
- Correct answerC$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$
- D$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
Solution
Given $A = \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix}$ and $f(t) = t^2 - 3t + 7$.
First, compute $A^2$:
$A^2 = \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 1(1) + (-2)(4) & 1(-2) + (-2)(5) \\ 4(1) + 5(4) & 4(-2) + 5(5) \end{bmatrix} = \begin{bmatrix} -7 & -12 \\ 24 & 17 \end{bmatrix}$
Next, compute $f(A) = A^2 - 3A + 7I$:
$f(A) = \begin{bmatrix} -7 & -12 \\ 24 & 17 \end{bmatrix} - 3 \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$f(A) = \begin{bmatrix} -7 & -12 \\ 24 & 17 \end{bmatrix} - \begin{bmatrix} 3 & -6 \\ 12 & 15 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$
$f(A) = \begin{bmatrix} -7 - 3 + 7 & -12 + 6 + 0 \\ 24 - 12 + 0 & 17 - 15 + 7 \end{bmatrix} = \begin{bmatrix} -3 & -6 \\ 12 & 9 \end{bmatrix}$
Finally, add the given matrix to $f(A)$:
$f(A) + \begin{bmatrix} 3 & 6 \\ -12 & -9 \end{bmatrix} = \begin{bmatrix} -3 & -6 \\ 12 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ -12 & -9 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
Answer: $\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$
First, compute $A^2$:
$A^2 = \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 1(1) + (-2)(4) & 1(-2) + (-2)(5) \\ 4(1) + 5(4) & 4(-2) + 5(5) \end{bmatrix} = \begin{bmatrix} -7 & -12 \\ 24 & 17 \end{bmatrix}$
Next, compute $f(A) = A^2 - 3A + 7I$:
$f(A) = \begin{bmatrix} -7 & -12 \\ 24 & 17 \end{bmatrix} - 3 \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$f(A) = \begin{bmatrix} -7 & -12 \\ 24 & 17 \end{bmatrix} - \begin{bmatrix} 3 & -6 \\ 12 & 15 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$
$f(A) = \begin{bmatrix} -7 - 3 + 7 & -12 + 6 + 0 \\ 24 - 12 + 0 & 17 - 15 + 7 \end{bmatrix} = \begin{bmatrix} -3 & -6 \\ 12 & 9 \end{bmatrix}$
Finally, add the given matrix to $f(A)$:
$f(A) + \begin{bmatrix} 3 & 6 \\ -12 & -9 \end{bmatrix} = \begin{bmatrix} -3 & -6 \\ 12 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ -12 & -9 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
Answer: $\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$
Hence, the correct answer is option: 'C'.