Question

If \( A = \begin{bmatrix} x & 1 & -x \\ 0 & 1 & -1 \\ x & 0 & 7 \end{bmatrix} \) and \( \det(A) = \begin{vmatrix} 3 & 0 & 1 \\ 2 & -1 & 2 \\ 0 & 0 & 3 \end{vmatrix} \), then the value of \( x \) is:

• A. $-3$
• B. $3$
• C. $2$
• D. $-8$
• E. $-2$

Hint:

When equating two determinants, solve the constant side first to determine the target value for the variable side.

Answer 1

The Correct Option is A

Concept: Equate the determinant of matrix $A$ to the numerical value of the second determinant to solve for $x$.
• Evaluate the constant determinant first.
• Expand $\det(A)$ along a row or column with a zero to simplify.

Step 1: Evaluate the numerical determinant.
Let $\Delta = \begin{vmatrix} 3 & 0 & 1
2 & -1 & 2
0 & 0 & 3 \end{vmatrix}$. Expanding along $R_3$: \[ \Delta = 3 \begin{vmatrix} 3 & 0 2 & -1 \end{vmatrix} = 3(-3 - 0) = -9 \]

Step 2: Expand $\det(A)$.
$|A| = \begin{vmatrix} x & 1 & -x
0 & 1 & -1
x & 0 & 7 \end{vmatrix}$. Expanding along $C_1$: \[ |A| = x \begin{vmatrix} 1 & -1 0 & 7 \end{vmatrix} - 0( \dots ) + x \begin{vmatrix} 1 & -x 1 & -1 \end{vmatrix} \] \[ |A| = x(7 - 0) + x(-1 - (-x)) = 7x + x(x - 1) \] \[ |A| = 7x + x^2 - x = x^2 + 6x \]

Step 3: Equate and solve for $x$.
\[ x^2 + 6x = -9 \Rightarrow x^2 + 6x + 9 = 0 \] \[ (x + 3)^2 = 0 \Rightarrow x = -3 \]