Question:

Let \( A \) be a \( 2 \times 2 \) real matrix and \( I \) be the identity matrix of order 2. If the roots of the equation \[ |A - xI| = 0 \] be \( -1 \) and \( 3 \), then the sum of the diagonal elements of the matrix \( A^2 \) is .....

Updated On: Nov 17, 2024
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Correct Answer: 10

Solution and Explanation

We are given a \(2 \times 2\) matrix \(A\) whose eigenvalues are \(-1\) and \(3\). We aim to determine the sum of the diagonal elements of \(A^2\), which is equivalent to the trace of \(A^2\).

The eigenvalues of a matrix provide useful information:  
- The sum of the eigenvalues is equal to the trace of the matrix \(A\):  
 \(\text{Sum of roots (eigenvalues)} = \text{tr}(A) = -1 + 3 = 2.\)

- The product of the eigenvalues is equal to the determinant of the matrix \(A\):  
 \(\text{Product of roots (eigenvalues)} = |\det(A)| = (-1)(3) = -3.\)

Thus, the matrix \(A\) satisfies:  
\(a + d = 2, \quad ad - bc = -3,\)
where \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\).

The trace of a matrix is the sum of its diagonal elements. For \(A^2\), the trace is:  
\(\text{tr}(A^2) = (A^2)_{11} + (A^2)_{22}.\)

Using matrix multiplication, compute \(A^2\): 
\(A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}^2 = \begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & d^2 + bc \end{bmatrix}.\)

The diagonal elements of \(A^2\) are:  
\((A^2)_{11} = a^2 + bc, \quad (A^2)_{22} = d^2 + bc.\)

Thus, the trace of \(A^2\) is:  
\(\text{tr}(A^2) = (A^2)_{11} + (A^2)_{22} = a^2 + bc + d^2 + bc = a^2 + d^2 + 2bc.\)

Using the properties of the matrix:
The trace of \(A\) is \(a + d = 2\). From this, express \(a^2 + d^2\) using the square of the sum:  
\((a + d)^2 = a^2 + d^2 + 2ad \implies a^2 + d^2 = (a + d)^2 - 2ad.\)

Substitute \(a + d = 2\):
\(a^2 + d^2 = 2^2 - 2ad = 4 - 2ad.\)

The determinant of \(A\) is \(ad - bc = -3\), which implies:  
\(ad = -3 + bc.\)

Substitute \(ad = -3 + bc\) into \(a^2 + d^2\):  
\(a^2 + d^2 = 4 - 2(-3 + bc) = 4 + 6 - 2bc = 10 - 2bc.\)

Thus:  
\(\text{tr}(A^2) = a^2 + d^2 + 2bc = (10 - 2bc) + 2bc = 10.\)

The Correct answer is: 10

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