Question

If \(A= \begin{bmatrix}1 & 0 \\[0.3em]-1 & 5 \\[0.3em] \end{bmatrix}\) and \(I= \begin{bmatrix}1& 0\\[0.3em]0& 1\\[0.3em] \end{bmatrix}\) then the value of k so that \(A^2 = 6A + kI\) is given by:

• A. 5
• B. -5
• C. -6
• D. 6

Answer 1

The Correct Option is B

We need to find the value of \(k\) such that \(A^2 = 6A + kI\) where \(A= \begin{bmatrix}1 & 0 \\ -1 & 5\end{bmatrix}\) and \(I= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\).

First, calculate \(A^2\):

\[A^2 = \begin{bmatrix}1 & 0 \\ -1 & 5\end{bmatrix} \begin{bmatrix}1 & 0 \\ -1 & 5\end{bmatrix} = \begin{bmatrix}(1 \cdot 1 + 0 \cdot -1) & (1 \cdot 0 + 0 \cdot 5) \\ (-1 \cdot 1 + 5 \cdot -1) & (-1 \cdot 0 + 5 \cdot 5)\end{bmatrix} = \begin{bmatrix}1 & 0 \\ -6 & 25\end{bmatrix}\]

Now, calculate \(6A\):

\[6A = 6 \begin{bmatrix}1 & 0 \\ -1 & 5\end{bmatrix} = \begin{bmatrix}6 & 0 \\ -6 & 30\end{bmatrix}\]

Set up the equation \(A^2 = 6A + kI\):

\[\begin{bmatrix}1 & 0 \\ -6 & 25\end{bmatrix} = \begin{bmatrix}6 & 0 \\ -6 & 30\end{bmatrix} + k\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]

Then, express the right-hand side:

\[\begin{bmatrix}6 & 0 \\ -6 & 30\end{bmatrix} + \begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix} = \begin{bmatrix}6+k & 0 \\ -6 & 30+k\end{bmatrix}\]

Equate the corresponding elements to find \(k\):

• From the top-left element: \(1 = 6 + k\) gives \(k = -5\)
• Check with the bottom-right element: \(25 = 30 + k\) also gives \(k = -5\)

Therefore, the value of \(k\) is \(-5\).