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

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