Given matrix $A = \begin{bmatrix} 1 & x & 1 \\ x & 2 & y \\ 1 & y & 3 \end{bmatrix}; B = \begin{bmatrix} 3 & -3 & z \\ -3 & 2 & -3 \\ z & -3 & 1 \end{bmatrix}$. Obtain $x, y, z$ if the matrix $AB$ is symmetric.
Step-by-Step Solution
Key Concept: For a matrix product to be symmetric, the resulting matrix must equal its transpose, which requires equating each off-diagonal element $(AB)_{ij}$ with its corresponding element $(AB)_{ji}$.
<div>$AB = \begin{bmatrix} 1 & x & 1 \\ x & 2 & y \\ 1 & y & 3 \end{bmatrix} \begin{bmatrix} 3 & -3 & z \\ -3 & 2 & -3 \\ z & -3 & 1 \end{bmatrix} = \begin{bmatrix} 3-3x+z & -3+2x-3 & z-3x+1 \\ 3x-6+yz & -3x+4-3y & xz-6+y \\ 3-3y+3z & -3+2y-9 & z-3y+3 \end{bmatrix}$<br>$AB$ is symmetric matrix,<br>So, $-6 + 2x = 3x - 6 + yz$ ...(i)<br>$\Rightarrow 3 - 3y + 3z = z - 3x + 1$ ...(ii)<br>$\Rightarrow -3 + 2y - 9 = xz - 6 + y$ ...(iii)<br>So possible set of values of $x, y, z$<br>$\Rightarrow \left( -\frac{4\sqrt{2}}{3}, \frac{2}{3}, 2\sqrt{2} \right), \left( \frac{4\sqrt{2}}{3}, \frac{2}{3}, -2\sqrt{2} \right), (3, 3, -1)$</div>
Correct Answer: \left( -\frac{4\sqrt{2}}{3}, \frac{2}{3}, 2\sqrt{2} \right), \left( \frac{4\sqrt{2}}{3}, \frac{2}{3}, -2\sqrt{2} \right), (3, 3, -1)