Question

\( A = \begin{bmatrix} 0 & k & k k & -4 & -6 k & -3 & -5 \end{bmatrix} \) is a singular matrix for:

• A. \( k = 2 \) only
• B. \( k = \pm 2 \) only
• C. no real value of \( k \)
• D. all real values of \( k \)

Hint:

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.

Answer 1

The Correct Option is D

To determine when the matrix \( A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix} \) is singular, we must consider the determinant of this matrix. A matrix is singular if and only if its determinant is zero.

The determinant of a 3x3 matrix \(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\) is given by:

\(\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\)

Applying this formula to the matrix \( A \), we have:

\(det(A) = 0((-4)(-5) - (-6)(-3)) - k(k(-5) - (-6)k) + k(k(-3) - (-4)k)\)

Let's calculate each part:

• The first term is: \(0\), as the first element is zero.
• The second term is: \(-k(k(-5) - (-6)k) = -k(-5k + 6k) = -k(-11k) = 11k^2\).
• The third term is: \(k(k(-3) - (-4)k) = k(-3k + 4k) = k(1k) = k^2\).

So, the determinant becomes:

\(\det(A) = 11k^2 + k^2 = 12k^2\)

For the matrix to be singular, we set the determinant to zero:

\(12k^2 = 0\)

Solving for \( k \), we get:

\(k^2 = 0 \implies k = 0\)

Since the given determinant \( 12k^2 \) is zero regardless of the value of \( k = 0\), the matrix is singular for all real values of \( k\).

Thus, the correct option is: all real values of \( k \).