Question

If A=\(\begin{bmatrix} 2&3 \\ -1&1\end{bmatrix}\)and \(A^2-3A+kI = 0\) then the value of k is:

• A. -5
• B. 3
• C. 5
• D. -3

Answer 1

The Correct Option is C

To find the value of \( k \), we start with the given equation \( A^2 - 3A + kI = 0 \). First, calculate \( A^2 \):
Let \( A = \begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix} \). The product \( A^2 = A \times A \) is calculated as follows:
\(\begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} (2 \times 2 + 3 \times -1) & (2 \times 3 + 3 \times 1) \\ (-1 \times 2 + 1 \times -1) & (-1 \times 3 + 1 \times 1) \end{bmatrix} \)
\(= \begin{bmatrix} (4 - 3) & (6 + 3) \\ (-2 - 1) & (-3 + 1) \end{bmatrix} = \begin{bmatrix} 1 & 9 \\ -3 & -2 \end{bmatrix} \)
Next, substitute \( A^2 \) into the equation: \( \begin{bmatrix} 1 & 9 \\ -3 & -2 \end{bmatrix} - 3 \begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix} + k \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
Calculate \( 3A \):
\(3 \begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 9 \\ -3 & 3 \end{bmatrix} \).
Substitute and simplify:
\(\begin{bmatrix} 1 & 9 \\ -3 & -2 \end{bmatrix} - \begin{bmatrix} 6 & 9 \\ -3 & 3 \end{bmatrix} + \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\(= \begin{bmatrix} 1-6+k & 9-9 \\ -3+3 & -2-3+k \end{bmatrix} = \begin{bmatrix} k-5 & 0 \\ 0 & k-5 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
This results in \( k-5 = 0 \), thus \( k = 5 \). Therefore, the value of \( k \) is 5.