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
\[
l^3-m^3=
\]
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
\[
l^3-m^3=
\]
Show Hint
While evaluating determinants of \(3 \times 3\) matrices, expanding along the row or column containing simpler entries reduces computation.
Updated On: Jun 22, 2026
- \(35\)
- \(-35\)
- \(19\)
- \(-19\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Compute the determinant of the matrix.
Given, \[ A= \begin{bmatrix} -2 & x & 1\\ x & 1 & 1\\ 2 & 3 & -1 \end{bmatrix} \] We calculate \[ \det(A) = \begin{vmatrix} -2 & x & 1\\ x & 1 & 1\\ 2 & 3 & -1 \end{vmatrix} \] Expanding along the first row,
\[ \det(A) = -2 \begin{vmatrix} 1 & 1\\ 3 & -1 \end{vmatrix} - x \begin{vmatrix} x & 1\\ 2 & -1 \end{vmatrix} + 1 \begin{vmatrix} x & 1\\ 2 & 3 \end{vmatrix} \]
Step 2: Evaluate the minors.
First minor: \[ \begin{vmatrix} 1 & 1\\ 3 & -1 \end{vmatrix} =(1)(-1)-(1)(3) =-1-3=-4 \] Second minor: \[ \begin{vmatrix} x & 1\\ 2 & -1 \end{vmatrix} =x(-1)-2 =-x-2 \] Third minor: \[ \begin{vmatrix} x & 1\\ 2 & 3 \end{vmatrix} =3x-2 \] Substituting these values,
\[ \det(A) = -2(-4)-x(-x-2)+(3x-2) \] \[ =8+x^2+2x+3x-2 \] \[ =x^2+5x+6 \] \[ =(x+2)(x+3) \]
Step 3: Find the roots.
Since \[ \det(A)=0, \] we get \[ (x+2)(x+3)=0 \] Thus the roots are \[ l=-2,\qquad m=-3 \]
Step 4: Compute \(l^3-m^3\).
\[ l^3-m^3 = (-2)^3-(-3)^3 \] \[ =-8-(-27) \] \[ =-8+27 \] \[ =19 \]
Step 5: Final conclusion.
Hence, \[ \boxed{19} \] which corresponds to option (3).
Given, \[ A= \begin{bmatrix} -2 & x & 1\\ x & 1 & 1\\ 2 & 3 & -1 \end{bmatrix} \] We calculate \[ \det(A) = \begin{vmatrix} -2 & x & 1\\ x & 1 & 1\\ 2 & 3 & -1 \end{vmatrix} \] Expanding along the first row,
\[ \det(A) = -2 \begin{vmatrix} 1 & 1\\ 3 & -1 \end{vmatrix} - x \begin{vmatrix} x & 1\\ 2 & -1 \end{vmatrix} + 1 \begin{vmatrix} x & 1\\ 2 & 3 \end{vmatrix} \]
Step 2: Evaluate the minors.
First minor: \[ \begin{vmatrix} 1 & 1\\ 3 & -1 \end{vmatrix} =(1)(-1)-(1)(3) =-1-3=-4 \] Second minor: \[ \begin{vmatrix} x & 1\\ 2 & -1 \end{vmatrix} =x(-1)-2 =-x-2 \] Third minor: \[ \begin{vmatrix} x & 1\\ 2 & 3 \end{vmatrix} =3x-2 \] Substituting these values,
\[ \det(A) = -2(-4)-x(-x-2)+(3x-2) \] \[ =8+x^2+2x+3x-2 \] \[ =x^2+5x+6 \] \[ =(x+2)(x+3) \]
Step 3: Find the roots.
Since \[ \det(A)=0, \] we get \[ (x+2)(x+3)=0 \] Thus the roots are \[ l=-2,\qquad m=-3 \]
Step 4: Compute \(l^3-m^3\).
\[ l^3-m^3 = (-2)^3-(-3)^3 \] \[ =-8-(-27) \] \[ =-8+27 \] \[ =19 \]
Step 5: Final conclusion.
Hence, \[ \boxed{19} \] which corresponds to option (3).
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