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

Answer: (C)

45

Answer 2

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.