Question

For what value of \( \alpha \), the matrix A is a singular matrix if \(A=\begin{bmatrix} 1 & 3 & \alpha+2 \\[0.3em] 2 & 4 & 8 \\[0.3em] 3 & 5 & 10 \end{bmatrix}\) ?

• A. \( 2 \)
• B. \( 3 \)
• C. \( 4 \)
• D. \( 8 \)

Hint:

A square matrix is called singular if its determinant is zero. If the determinant is non-zero, the matrix is called non-singular or invertible. Calculating the determinant is key to solving such problems.

Answer 1

The Correct Option is C

Step 1: Understand the condition for a singular matrix. A matrix is singular if its determinant is equal to zero. So, we need to find the value of \(\alpha\)such that \(\det(A) = 0\).

Step 2: Write down the matrix A.\(A = \begin{bmatrix} 1 & 3 & \alpha+2  \\[0.3em] 2 & 4 & 8  \\[0.3em] 3 & 5 & 10 \end{bmatrix}\)

Step 3: Calculate the determinant of A. We can expand the determinant along the first row: \(\det(A) = 1 \begin{vmatrix} 4 & 8 \\[0.3em] 5 & 10 \end{vmatrix} - 3 \begin{vmatrix} 2 & 8 \\[0.3em] 3 & 10 \end{vmatrix} + (\alpha+2) \begin{vmatrix} 2 & 4 \\[0.3em] 3 & 5 \end{vmatrix}\)

Step 4: Evaluate the 2x2 determinants. 

\[\begin{vmatrix} 4 & 8 \\[0.3em] 5 & 10 \end{vmatrix} = (4)(10) - (8)(5) = 40 - 40 = 0\]\[\begin{vmatrix} 2 & 8 \\[0.3em] 3 & 10 \end{vmatrix} = (2)(10) - (8)(3) = 20 - 24 = -4\]\[\begin{vmatrix} 2 & 4 \\[0.3em] 3 & 5 \end{vmatrix} = (2)(5) - (4)(3) = 10 - 12 = -2\]

Step 5: Substitute these values back into the determinant expression.
\(\det(A) = 1(0) - 3(-4) + (\alpha+2)(-2)\)
\(\det(A) = 0 + 12 - 2(\alpha+2)\)
\(\det(A) = 12 - 2\alpha - 4\)
\(\det(A) = 8 - 2\alpha\)

Step 6: Set the determinant to zero and solve for \(\alpha\).
\(8 - 2\alpha = 0\)
\(8 = 2\alpha\)
\(\alpha = \frac{8}{2} = 4\)

Step 7: Compare the result with the given options. The value \(\alpha = 4\) makes the matrix singular. This matches option (C).