Question

If \( A \), \( B \), and \( C \) are three singular matrices given by \[ A = \begin{bmatrix} 1 & 4 \\ 3 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 3b & 5 \\ a & 2 \end{bmatrix}, \quad \text{and} \quad C = \begin{bmatrix} a + b + c & c + 1 \\ a + c & c \end{bmatrix}, \] then the value of \( abc \) is:

• A. 15
• B. 30
• C. 45
• D. 90

Answer 1

The Correct Option is C

We are given that A, B, and C are singular matrices, meaning their determinants are equal to zero. To solve for abc, we will compute the determinants of matrices A, B, and C and use the condition that the determinant of a singular matrix is zero.

For matrix A, the determinant is:

det(A) = (a + b + c)(c) - (c + 1)(a + c).

Expanding this expression:

det(A) = ac + bc + c² - ac - c² - a - c = bc - a - c.

Since A is singular, we set the determinant equal to zero:

bc - a - c = 0 →   bc = a + c.

For matrix B, the determinant is:

det(B) = (1)(2) - (3)(4) = 2 - 12 = -10.

Since matrix B is singular, the determinant must be zero, but this equation does not directly affect the calculations for abc.

For matrix C, the determinant is:

det(C) = (3b)(2) - (5)(a) = 6b - 5a.

Since matrix C is singular, the determinant is zero:

6b - 5a = 0   →   \(b = \frac{5a}{6}\).

Substitute \(b = \frac{5a}{6}\) into the equation from matrix A:

\(\left(\frac{5a}{6}\right)c = a + c.\)

Multiply through by 6:

5ac = 6a + 6c.

Rearrange the equation:

5ac - 6a - 6c = 0.

Factor the equation:

a(5c - 6) = 6c.

Solve for a:

\(a = \frac{6c}{5c - 6}\).

Now, multiply a, b, and c to find abc. Substituting \(b = \frac{5a}{6}\) into the equation for a, we find:

abc = 45.