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 .....

Answer 1

Correct Answer: 10

Given a \( 2 \times 2 \) real matrix \( A \) and the identity matrix \( I \), the determinant equation \(|A - xI| = 0\) shows that the roots are eigenvalues of matrix \( A \). These eigenvalues are \(-1\) and \(3\). For a matrix, the trace (sum of diagonal elements) is equal to the sum of its eigenvalues. Therefore, the trace of \( A \) is:

\[ \text{trace}(A) = -1 + 3 = 2 \] 

For matrix \( A^2 \), another property is that the trace of \( A^2 \) equals the sum of the squares of the eigenvalues of \( A \). Thus, we calculate:

\[ (-1)^2 + 3^2 = 1 + 9 = 10 \]

The sum of the diagonal elements of \( A^2 \) is \(10\), which falls within the given range \([10, 10]\).

Answer 2

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