Question

Let \[ A= \begin{bmatrix} -2 & x & 1\\ x & 1 & 1\\ 2 & 3 & -1 \end{bmatrix} \] If the roots of the equation \[ \det(A)=0 \] are \(l,m\), then \[ l^3-m^3= \]

• A. \(35\)
• B. \(-35\)
• C. \(19\)
• D. \(-19\)

Hint:

While evaluating determinants of \(3 \times 3\) matrices, expanding along the row or column containing simpler entries reduces computation.

Answer 1

The Correct Option is C

Step 1: Expand the determinant.
For $A=\begin{bmatrix}-2 & x & 1\\ x & 1 & 1\\ 2 & 3 & -1\end{bmatrix}$, expand $\det(A)$ along the first row.
Step 2: Compute each minor.
$\begin{vmatrix}1&1\\3&-1\end{vmatrix}=-4$, $\begin{vmatrix}x&1\\2&-1\end{vmatrix}=-x-2$, $\begin{vmatrix}x&1\\2&3\end{vmatrix}=3x-2$.
Step 3: Assemble the determinant.
$\det(A)=-2(-4)-x(-x-2)+1(3x-2)=8+x^2+2x+3x-2=x^2+5x+6$.
Step 4: Solve $\det(A)=0$.
$x^2+5x+6=0$ factors as $(x+2)(x+3)=0$, so the roots are $-2$ and $-3$.
Step 5: Assign $l$ and $m$.
Take $l=-2,\ m=-3$. Then $l^3=-8$ and $m^3=-27$.
Step 6: Compute $l^3-m^3$.
$l^3-m^3=-8-(-27)=19$.
\[ \boxed{19} \]