Question

Let $x, y \in \mathbb{R}$ and the matrix \[ M = \begin{bmatrix} x + y & x - y \\ x - y & x + y \end{bmatrix}. \] Also, let $\text{adj}(M)$ be the adjoint and $\det(M)$ be the determinant of the matrix $M$. If \[ M \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 3 \end{bmatrix}, \] then

• A. $x + y = -\dfrac{3}{4}$
• B. $x - y = \dfrac{3}{4}$
• C. $\det(M) = -1$
• D. $\det(\text{adj}(M)) = 1$

Hint:

For symmetric $2\times2$ matrices, $\det(M) = (x+y)^2 - (x-y)^2 = 4xy$. Keep this shortcut in mind for quicker calculations.

Answer 1

The Correct Option is A, C

Step 1: Multiply to form equations.
\[ M \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} (x+y)3 + (x-y)1 \\ (x-y)3 + (x+y)1 \end{bmatrix} = \begin{bmatrix} 4x + 2y \\ 4x - 2y \end{bmatrix}. \] Equating components: \[ 4x + 2y = -1, \quad 4x - 2y = 3. \]
Step 2: Solve for $x$ and $y$.
Add both equations: $8x = 2 \Rightarrow x = \dfrac{1}{4}$. Substitute into the first: $1 + 2y = -1 \Rightarrow y = -1.$
Step 3: Compute determinant.
\[ \det(M) = (x+y)^2 - (x-y)^2 = (x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) = 4xy. \] Substitute $x = \dfrac{1}{4}$, $y = -1$: $\det(M) = 4 \times \dfrac{1}{4} \times (-1) = -1.$

Step 4: Conclusion.
$\boxed{\det(M) = -1.}$