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: Understanding the Concept:
We compute \(A^2\) by matrix multiplication, then put it into the equation and compare entries. The equation must hold for every entry of the matrix.

Step 2: Find A squared.
\[ A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4+3 & 6+6 \\ 2+2 & 3+4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \]

Step 3: Find 4A.
\[ 4A = \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix} \]

Step 4: Form the left side.
\[ A^2-4A = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I \]
So the equation becomes \(-I+\lambda I = 0\), that is \((\lambda-1)I=0\).

Step 5: Solve for lambda.
Since \(I\) is not the zero matrix, \(\lambda - 1 = 0\), so \(\lambda = 1\).

Step 6: Check the options.
The values \(-1\), \(-2\) and \(2\) would leave a non zero matrix. For example \(\lambda=2\) gives \(I \neq 0\). Only \(\lambda=1\) works, which is option 2.

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