Question

If A =\(\begin{bmatrix}   2-k & 2 \\     1 & 3-k \end{bmatrix}\) is a singular matrix, then the value of 5k - k² is equal to

• A. -4
• B. 4
• C. 6
• D. -6

Hint:

When solving problems involving singular matrices, remember that a matrix is singular if and only if its determinant is zero. To find the determinant, use the formula for a 2x2 matrix: \( \text{det}(A) = ad - bc \), where \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \). This helps simplify the problem when solving for unknowns like \( k \) in this case.

Answer 1

The Correct Option is B

The correct answer is (B) : 4.

Answer 2

The correct answer is: (B) 4.

We are given that the matrix \( A \) is singular. The matrix \( A \) is given by:

\( A = \begin{bmatrix} 2 - k & 2 \\ 1 & 3 - k \end{bmatrix} \)

A matrix is singular if its determinant is zero. The determinant of matrix \( A \) is calculated as follows: \[ \text{det}(A) = (2 - k)(3 - k) - (2)(1) \] Expanding the terms: \[ \text{det}(A) = (2 - k)(3 - k) - 2 = 6 - 3k - 2k + k^2 - 2 = k^2 - 5k + 4 \] Since \( A \) is singular, we set the determinant equal to zero: \[ k^2 - 5k + 4 = 0 \] Solving this quadratic equation using the quadratic formula: \[ k = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)} = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm \sqrt{9}}{2} \] \[ k = \frac{5 \pm 3}{2} \] This gives two possible values for \( k \): \[ k = \frac{5 + 3}{2} = 4 \quad \text{or} \quad k = \frac{5 - 3}{2} = 1 \] Now, we are asked to find the value of \( 5k - k^2 \). Let's calculate this for both values of \( k \): - For \( k = 4 \): \[ 5k - k^2 = 5(4) - 4^2 = 20 - 16 = 4 \] - For \( k = 1 \): \[ 5k - k^2 = 5(1) - 1^2 = 5 - 1 = 4 \] In both cases, the value of \( 5k - k^2 \) is 4. Therefore, the correct answer is (B) 4.