Question

If $$ A = \begin{pmatrix} 3 & -2 \\ 4 & -2 \end{pmatrix} $$ satisfies the matrix equation $$ A^2 - K A + 2I = 0, $$ then the value of $ K $ is:

• A. 0
• B. -1
• C. 1
• D. 2

Hint:

To solve matrix polynomial equations, compute powers and compare coefficients.

Answer 1

The Correct Option is C

Given matrix \(A = \begin{pmatrix} 3 & -2 \\ 4 & -2 \end{pmatrix}\), let's solve the matrix equation \(A^2 - KA + 2I = 0\) to find \(K\).

First, calculate \(A^2\):

\[A^2 = A \cdot A = \begin{pmatrix} 3 & -2 \\ 4 & -2 \end{pmatrix} \cdot \begin{pmatrix} 3 & -2 \\ 4 & -2 \end{pmatrix}\]

Matrix multiplication gives:

\[\begin{aligned} &\begin{pmatrix} 3\cdot3 + (-2)\cdot4 & 3\cdot(-2) + (-2)\cdot(-2) \\ 4\cdot3 + (-2)\cdot4 & 4\cdot(-2) + (-2)\cdot(-2) \end{pmatrix}\\ &= \begin{pmatrix} 9 - 8 & -6 + 4 \\ 12 - 8 & -8 + 4 \end{pmatrix}\\ &= \begin{pmatrix} 1 & -2 \\ 4 & -4 \end{pmatrix} \end{aligned}\]

Next, substitute into the equation \(A^2 - KA + 2I = 0\):

\[\begin{pmatrix} 1 & -2 \\ 4 & -4 \end{pmatrix} - K\begin{pmatrix} 3 & -2 \\ 4 & -2 \end{pmatrix} + 2\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\]

Substitute identity matrix \(I\):

\[\begin{pmatrix} 1 & -2 \\ 4 & -4 \end{pmatrix} - \begin{pmatrix} 3K & -2K \\ 4K & -2K \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\]

Add matrices:

\[\begin{pmatrix} 1 + 2 & -2 \\ 4 & -4 + 2 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 4 & -2 \end{pmatrix}\]

Equation becomes:

\[\begin{pmatrix} 3 & -2 \\ 4 & -2 \end{pmatrix} - \begin{pmatrix} 3K & -2K \\ 4K & -2K \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\]

Match elements for equality:

\(\ \begin{aligned} &3 - 3K = 0\\ &-2 + 2K = 0 \end{aligned}\)

Solve the equations:

\(\ \begin{aligned} K = 3/3 &= 1\\ K = 2/2 &= 1 \end{aligned}\)

Both conditions confirm \(K = 1\).