Question:
\( A = \begin{bmatrix} 0 & k & k k & -4 & -6 k & -3 & -5 \end{bmatrix} \) is a singular matrix for:
\( A = \begin{bmatrix} 0 & k & k k & -4 & -6 k & -3 & -5 \end{bmatrix} \) is a singular matrix for:
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Whenever a row or column expansion yields a submatrix where one row/column is a scalar multiple of another (in this case, both columns are identical to \(\begin{bmatrix} k k \end{bmatrix}\) up to a factor), its determinant is automatically zero! Recognizing linear dependency early saves you from unnecessary algebraic expansions.
Updated On: May 23, 2026
- \( k = 2 \) only
- \( k = \pm 2 \) only
- no real value of \( k \)
- all real values of \( k \)
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The Correct Option is D
Solution and Explanation
Concept:
A square matrix \( A \) is defined as a singular matrix if and only if its determinant is precisely equal to zero:
\[
|A| = 0
\]
If the determinant of a matrix containing a variable evaluates identically to zero across simplification, it indicates that the rows or columns are linearly dependent regardless of the specific value chosen for that variable.
Step 1: Setting up the condition for a singular matrix.
For matrix \( A \) to be singular, we must set its determinant to zero: \[ |A| = \begin{vmatrix} 0 & k & k k & -4 & -6 k & -3 & -5 \end{vmatrix} = 0 \]
Step 2: Applying column operations to simplify the determinant.
Notice that the second and third elements of the first row are identical (\(k\)). Let us perform the elementary column operation \(C_3 \rightarrow C_3 - C_2\): \[ |A| = \begin{vmatrix} 0 & k & k - k k & -4 & -6 - (-4) k & -3 & -5 - (-3) \end{vmatrix} = 0 \] Simplifying the elements in the third column: \[ |A| = \begin{vmatrix} 0 & k & 0 k & -4 & -2 k & -3 & -2 \end{vmatrix} = 0 \]
Step 3: Expanding the determinant along the first row.
Expanding along the first row (\(R_1\)), which now contains two zeros: \[ |A| = -k \cdot \begin{vmatrix} k & -2 k & -2 \end{vmatrix} = 0 \] Evaluating the \(2 \times 2\) determinant: \[ \begin{vmatrix} k & -2 k & -2 \end{vmatrix} = k(-2) - (-2)(k) = -2k + 2k = 0 \] Substituting this back into the expansion expression: \[ |A| = -k \cdot (0) = 0 \]
Step 4: Interpreting the mathematical result.
Since the expression simplifies to \( 0 = 0 \) independently of \(k\), the determinant of matrix \(A\) is identically zero for any value of \(k\). Therefore, the matrix is singular for all real values of \(k\), which matches option (D).
Step 1: Setting up the condition for a singular matrix.
For matrix \( A \) to be singular, we must set its determinant to zero: \[ |A| = \begin{vmatrix} 0 & k & k k & -4 & -6 k & -3 & -5 \end{vmatrix} = 0 \]
Step 2: Applying column operations to simplify the determinant.
Notice that the second and third elements of the first row are identical (\(k\)). Let us perform the elementary column operation \(C_3 \rightarrow C_3 - C_2\): \[ |A| = \begin{vmatrix} 0 & k & k - k k & -4 & -6 - (-4) k & -3 & -5 - (-3) \end{vmatrix} = 0 \] Simplifying the elements in the third column: \[ |A| = \begin{vmatrix} 0 & k & 0 k & -4 & -2 k & -3 & -2 \end{vmatrix} = 0 \]
Step 3: Expanding the determinant along the first row.
Expanding along the first row (\(R_1\)), which now contains two zeros: \[ |A| = -k \cdot \begin{vmatrix} k & -2 k & -2 \end{vmatrix} = 0 \] Evaluating the \(2 \times 2\) determinant: \[ \begin{vmatrix} k & -2 k & -2 \end{vmatrix} = k(-2) - (-2)(k) = -2k + 2k = 0 \] Substituting this back into the expansion expression: \[ |A| = -k \cdot (0) = 0 \]
Step 4: Interpreting the mathematical result.
Since the expression simplifies to \( 0 = 0 \) independently of \(k\), the determinant of matrix \(A\) is identically zero for any value of \(k\). Therefore, the matrix is singular for all real values of \(k\), which matches option (D).
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