Question

Let \[ A = \begin{bmatrix} 2 & -3 \\ 1 & -2 \end{bmatrix}, \quad B = \begin{bmatrix} 14 & 21 \\ 7 & 10 \end{bmatrix}. \] If \[ (A^{4} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, \] find (x, y).

• A. (0, 0)
• B. (1, -1)
• C. (2, -2)
• D. (-1, 1)

Hint:

If the determinant of the coefficient matrix of a homogeneous system is non-zero, the system has only the trivial solution.

Answer 1

The Correct Option is A

To solve the problem, we need to find the values of \(x\) and \(y\) such that:

\[ (A^4 + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \]

Given matrices:

\[ A = \begin{bmatrix} 2 & -3 \\ 1 & -2 \end{bmatrix}, \quad B = \begin{bmatrix} 14 & 21 \\ 7 & 10 \end{bmatrix}. \]

First, we calculate \(A^4\) by computing powers of matrix \(A\):

• First, \(A^2\):

\[ A^2 = A \times A = \begin{bmatrix} 2 & -3 \\ 1 & -2 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} (2 \cdot 2) + (-3 \cdot 1) & (2 \cdot -3) + (-3 \cdot -2) \\ (1 \cdot 2) + (-2 \cdot 1) & (1 \cdot -3) + (-2 \cdot -2) \end{bmatrix} = \begin{bmatrix} 4 - 3 & -6 + 6 \\ 2 - 2 & -3 + 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \]

• Thus, \(A^2 = I\), the identity matrix, and therefore \(A^4 = (A^2)^2 = I^2 = I\).

Next, we compute the sum \(A^4 + B\):

\[ A^4 + B = I + B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 14 & 21 \\ 7 & 10 \end{bmatrix} = \begin{bmatrix} 1+14 & 0+21 \\ 0+7 & 1+10 \end{bmatrix} = \begin{bmatrix} 15 & 21 \\ 7 & 11 \end{bmatrix}. \]

We then solve the equation:

\[ \begin{bmatrix} 15 & 21 \\ 7 & 11 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \]

• This represents a homogeneous system of linear equations with the form:

\[ \begin{align*} 15x + 21y &= 0, \\ 7x + 11y &= 0. \end{align*} \]

To find the solution, observe that one obvious solution for this homogeneous system is \(x = 0\) and \(y = 0\), which satisfies both equations. This is known as the trivial solution.

Since the matrices involved don't lead to any nontrivial solutions (as determinant considerations suggest full rank), the only solution is indeed the trivial solution.

Thus, the correct answer is:

\[ (x, y) = (0, 0). \]