Question:
If
\[
\begin{vmatrix}
0 & 3 & 2b \\
2 & 0 & 1 \\
4 & -1 & 6
\end{vmatrix}
\]
is singular, then the value of \(b\) is
If
\[
\begin{vmatrix}
0 & 3 & 2b \\
2 & 0 & 1 \\
4 & -1 & 6
\end{vmatrix}
\]
is singular, then the value of \(b\) is
Show Hint
For singular matrices, directly equate determinant to zero and solve.
Updated On: May 8, 2026
- \(-3\)
- \(3\)
- \(-6\)
- \(6\)
- \(-2\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
Singular matrix ⇒ determinant = 0
Step 1: Expand determinant
\[ 0 \begin{vmatrix} 0 & 1 \\ -1 & 6 \end{vmatrix} -3 \begin{vmatrix} 2 & 1 \\ 4 & 6 \end{vmatrix} +2b \begin{vmatrix} 2 & 0 \\ 4 & -1 \end{vmatrix} \]
Step 2: Compute minors
Second: \[ 2\cdot6 - 4\cdot1 = 12-4=8 \] Third: \[ 2(-1)-0 = -2 \]
Step 3: Substitute
\[ -3(8) + 2b(-2) \] \[ = -24 -4b \]
Step 4: Set determinant zero
\[ -24 -4b = 0 \] \[ -4b = 24 \Rightarrow b = -6 \]
Step 5: Final answer
\[ \boxed{-6} \]
Step 1: Expand determinant
\[ 0 \begin{vmatrix} 0 & 1 \\ -1 & 6 \end{vmatrix} -3 \begin{vmatrix} 2 & 1 \\ 4 & 6 \end{vmatrix} +2b \begin{vmatrix} 2 & 0 \\ 4 & -1 \end{vmatrix} \]
Step 2: Compute minors
Second: \[ 2\cdot6 - 4\cdot1 = 12-4=8 \] Third: \[ 2(-1)-0 = -2 \]
Step 3: Substitute
\[ -3(8) + 2b(-2) \] \[ = -24 -4b \]
Step 4: Set determinant zero
\[ -24 -4b = 0 \] \[ -4b = 24 \Rightarrow b = -6 \]
Step 5: Final answer
\[ \boxed{-6} \]
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