If $A = \begin{bmatrix} 1 & -3 \\ 2 & -1 \end{bmatrix}$ and $A^2 - 4A + 10I = A$, then $k$ is equal to
Step-by-Step Solution
Key Concept: Matrix equation solving by computing $A^2$ and equating corresponding entries of the resulting matrix equation.
Calculate $A^2 = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}^2 = \begin{bmatrix} -5 & -3-3k \\ 2+2k & -6+k^2 \end{bmatrix}$. Substituting into $A^2 - 4A + 10I = A$: $\begin{bmatrix} -5 & -3-3k \\ 2+2k & -6+k^2 \end{bmatrix} - \begin{bmatrix} 4 & -12 \\ 8 & 4k \end{bmatrix} + \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$. From the equation $9 - 3k = -3$, we get $k = 4$. From $4 + k^2 - 4k = k$, we have $k^2 - 5k + 4 = 0$, giving $k = 4$ or $k = 1$. Only $k = 1$ satisfies both conditions.
Correct Answer: 1