Question

If \( A = \begin{pmatrix} 1 & -2 4 & 5 \end{pmatrix} \); \( f(t) = t^2 - 3t + 7 \) then \( f(A) + \begin{pmatrix 3 & 6 -12 & -9 \end{pmatrix} = \)}

• A. \( \begin{pmatrix} 0 & 1 1 & 0 \end{pmatrix} \)
• B. \( \begin{pmatrix} 1 & 1 0 & 0 \end{pmatrix} \)
• C. \( \begin{pmatrix} 0 & 0 0 & 0 \end{pmatrix} \)
• D. \( \begin{pmatrix} 1 & 0 0 & 1 \end{pmatrix} \)

Hint:

For matrix polynomials, substitute \(A\) carefully and compute step-by-step using matrix multiplication.

Answer 1

The Correct Option is C

Step 1: Understand the matrix function.
Given:
\[ f(t) = t^2 - 3t + 7. \]
So,
\[ f(A) = A^2 - 3A + 7I. \]

Step 2: Compute \( A^2 \).
\[ A^2 = \begin{pmatrix} 1 & -2 4 & 5 \end{pmatrix} \begin{pmatrix} 1 & -2 4 & 5 \end{pmatrix}. \]
\[ = \begin{pmatrix} 1+(-8) & -2-10 4+20 & -8+25 \end{pmatrix} = \begin{pmatrix} -7 & -12 24 & 17 \end{pmatrix}. \]

Step 3: Compute \( -3A \).
\[ -3A = \begin{pmatrix} -3 & 6 -12 & -15 \end{pmatrix}. \]

Step 4: Compute \( 7I \).
\[ 7I = \begin{pmatrix} 7 & 0 0 & 7 \end{pmatrix}. \]

Step 5: Find \( f(A) \).
\[ f(A) = A^2 - 3A + 7I. \]
\[ = \begin{pmatrix} -7 & -12 24 & 17 \end{pmatrix} + \begin{pmatrix} -3 & 6 -12 & -15 \end{pmatrix} + \begin{pmatrix} 7 & 0 0 & 7 \end{pmatrix}. \]
\[ = \begin{pmatrix} -3 & -6 12 & 9 \end{pmatrix}. \]

Step 6: Add the given matrix.
\[ f(A) + \begin{pmatrix} 3 & 6 -12 & -9 \end{pmatrix} = \begin{pmatrix} -3+3 & -6+6 12-12 & 9-9 \end{pmatrix}. \]
\[ = \begin{pmatrix} 0 & 0 0 & 0 \end{pmatrix}. \]

Step 7: Final conclusion.
Thus, the result is the null matrix.
Final Answer:
\[ \boxed{\begin{pmatrix} 0 & 0 0 & 0 \end{pmatrix}}. \]