Question

If \(A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\), such that \(A^2 - 4A + \lambda I = 0\), where \(I\) is identity matrix of order 2, then '\(\lambda\)' is equal to

• A. \(-1\)
• B. \(1\)
• C. \(-2\)
• D. \(2\)

Hint:

Use Cayley-Hamilton: \(A^2-(\operatorname{tr}A)A+(\det A)I=0\). Here \(\det A = 1\).

Answer 1

The Correct Option is B

Step 1: Use the characteristic equation.
For a 2 by 2 matrix, the Cayley-Hamilton theorem says $A^2 - (\operatorname{tr}A)A + (\det A)I = 0$.

Step 2: Find trace and determinant.
$\operatorname{tr}A = 2+2 = 4$. $\det A = 2\cdot 2 - 3\cdot 1 = 1$.

Step 3: Compare with the given equation.
The theorem gives $A^2 - 4A + 1\cdot I = 0$. The question has $A^2-4A+\lambda I=0$. Matching the coefficient of $I$ gives $\lambda = 1$.

Step 4: Pick the option.
$\lambda=1$ is option 2.

Step 5: Check by entries.
We can also use one entry only. The top left entry of $A^2$ is $2\cdot 2+3\cdot 1=7$. The top left entry of $4A$ is 8. So the top left entry of the equation is $7-8+\lambda=0$, giving $\lambda=1$. The off diagonal entries are $12-12=0$ and $4-4=0$, which are already zero. Nothing else can change $\lambda$.

Final Answer:
The value is $\lambda = 1$. \[ \boxed{\lambda = 1} \]