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

Hint:

When working with singular matrices, remember that their determinants are equal to zero. Use this property to set up equations and solve for unknown variables. For matrices with variables, start by expressing the determinant and simplifying step by step, ensuring you correctly apply operations like factoring and substitution. Always check the determinant conditions for each matrix and their relationships with the unknowns.

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.

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.

Step 1: Matrix A:

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.

Step 2: Matrix B:

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 $abc$.

Step 3: Matrix C:

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}.

Step 4: 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.

Step 5: Factor the equation:

a(5c - 6) = 6c.

Step 6: Solve for a:

$a = \frac{6c}{5c - 6}.$

Step 7: Multiply a, b, and c to find $abc$:

Substituting $b = \frac{5a}{6}$ into the equation for $a$, we find:

abc = 45.

Thus, the correct answer is:

45.