Question

Let \(A = [\begin{array}{cc}-3 & 2 \\ 1 & 4\end{array}]\) and if \(A^2-2A+I = [\begin{array}{cc}18 & p \\ q & 11\end{array}]\), then \(\ldots\)

• A. \(p = -2, q = -1\)
• B. \(p = 2, q = 1\)
• C. \(p = -1, q = -2\)
• D. \(p = 1, q = 2\)

Hint:

Compute A squared, subtract 2A and add the identity, then compare entries.

Answer 1

The Correct Option is A

Step 1: Use the Cayley-Hamilton idea to cross-check:
$A^2 - 2A + I = (A - I)^2$. Compute $A - I = \begin{pmatrix}-4 & 2\\ 1 & 3\end{pmatrix}$.

Step 2: Square it:
$(A-I)^2 = \begin{pmatrix}16+2 & -8+6\\ -4+3 & 2+9\end{pmatrix} = \begin{pmatrix}18 & -2\\ -1 & 11\end{pmatrix}$.

Step 3: Read off:
The diagonal gives 18 and 11 as in the question. So $p = -2$ and $q = -1$, option (A).

Step 4: Why this works:
$A$ commutes with $I$, so $(A-I)^2 = A^2 - 2A + I$ holds with no extra terms.

Final Answer:
$p=-2$, $q=-1$, option (A). \[ \boxed{p=-2,\ q=-1 \text{ (A)}} \]