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: Using the Cayley-Hamilton theorem instead:
For a \(2\times2\) matrix, the characteristic equation is \(A^2-(\text{tr }A)A+(\det A)I=0\), i.e. \(A^2=(\text{tr }A)A-(\det A)I\).

Step 2: Computing trace and determinant:
\(\text{tr}(A)=3+(-2)=1\). \(\det(A)=3(-2)-(-2)(4)=-6+8=2\).

Step 3: Matching to the given form:
Cayley-Hamilton gives \(A^2=1\cdot A-2I\), i.e. \(A^2=(1)A-2I\), matching the given \(A^2=KA-2I\) directly.

Final Answer:
So \(K=\boxed{1}\), the same as trace(A), confirming the entrywise method.