Question

The matrix \(\begin{bmatrix} -2&1&0\\3&4&1\\-4&\lambda&0 \end{bmatrix}\) is non-singular for \(\lambda\ne\)

• A. 2
• B. -2
• C. 4
• D. -4
• E. 0

Answer 1

The Correct Option is A

Given matrix:

The matrix is: \(\begin{bmatrix} -2 & 1 & 0 \\ 3 & 4 & 1 \\ -4 & \lambda & 0 \end{bmatrix}\)

Goal: Determine the value of \(\lambda\) for which the matrix is non-singular.

Explanation: 

A matrix is non-singular if its determinant is non-zero. So, we need to find the determinant of the given matrix and solve for \(\lambda\) when the determinant is non-zero.

Step 1: Compute the determinant of the matrix.

We will use cofactor expansion along the first row to find the determinant:

Determinant = \(\text{det}\begin{bmatrix} -2 & 1 & 0 \\ 3 & 4 & 1 \\ -4 & \lambda & 0 \end{bmatrix}\) =

\(-2 \times \text{det}\begin{bmatrix} 4 & 1 \\ \lambda & 0 \end{bmatrix} - 1 \times \text{det}\begin{bmatrix} 3 & 1 \\ -4 & 0 \end{bmatrix} + 0 \times \text{det}\begin{bmatrix} 3 & 4 \\ -4 & \lambda \end{bmatrix}\)

Step 2: Compute the 2x2 determinants.

\(\text{det}\begin{bmatrix} 4 & 1 \\ \lambda & 0 \end{bmatrix} = (4)(0) - (1)(\lambda) = -\lambda\)

\(\text{det}\begin{bmatrix} 3 & 1 \\ -4 & 0 \end{bmatrix} = (3)(0) - (1)(-4) = 4\)

Step 3: Substitute these values into the determinant formula.

Determinant = \(-2 \times (-\lambda) - 1 \times 4\)

Determinant = \(2\lambda - 4\)

Step 4: Set the determinant to non-zero for the matrix to be non-singular.

For the matrix to be non-singular, we need the determinant to be non-zero:

2\lambda - 4 \neq 0

Step 5: Solve for \(\lambda\).

\(2\lambda - 4 = 0 \quad ⟹ \quad 2\lambda = 4 \quad ⟹ \quad \lambda = 2\)

Conclusion:

The matrix is non-singular for \(\lambda \neq 2\).