Question

If \(A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\) and \(I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\) and \(A^2=KA-2I\), then find the value of \(K\).

Hint:

Compute \(A^2\) and \(KA-2I\), then compare corresponding entries (or use Cayley-Hamilton: \(K=\text{tr}(A)\)).

Answer 1

Step 1: Understanding the Concept:
Compute \(A^2\) directly, compute \(KA-2I\) symbolically in terms of \(K\), then match corresponding entries to solve for \(K\).

Step 2: Computing A squared:
\(A^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}\).

Step 3: Writing KA - 2I symbolically:
\(KA-2I=\begin{bmatrix}3K-2&-2K\\4K&-2K-2\end{bmatrix}\).

Step 4: Matching entries:
Comparing (1,1): \(3K-2=1\Rightarrow K=1\). Comparing (2,1): \(4K=4\Rightarrow K=1\). Both (and the remaining two entries) agree.

Final Answer:
\(K=\boxed{1}\).