Question and Core Idea

Given

A = | 2 1 |
| 1 1 |,
B = | 2 -5 |
| 0 1 |,

and the matrix equation

AX + 3B = 0,

find det(X) where the options are: -18, -6, 6, 18.

The key idea is to isolate X using matrix inverses and then use determinant properties to avoid computing every entry of X.

Step-by-Step Solution

1. Express X in terms of A and B

From AX + 3B = 0 ⇒ AX = -3B.

Since A is a 2×2 matrix with det(A) ≠ 0, it is invertible, so

X = A^-1(-3B) = -3A^-1B.

2. Use determinant rules

Use:

• det(kM) = kⁿdet(M) for an n×n matrix (here n=2).
• det(AB) = det(A)det(B).
• det(A^-1) = 1/det(A).

So det(X) = det(-3A^-1B) = det(-3I) det(A^-1) det(B).

For a 2×2 matrix, -3I has determinant (-3)² = 9. Hence

det(X) = 9 ⋅ det(A^-1) ⋅ det(B) = 9 ⋅ det(B)/det(A).

3. Compute det(A) and det(B)

For A = | 2 1 |
| 1 1 |,
det(A) = 2·1 – 1·1 = 1.

For B = | 2 -5 |
| 0 1 |,
det(B) = 2·1 – (-5)·0 = 2.

Thus det(X) = 9 · 2/1 = 18.

Option-by-Option Check

• (A) -18
  Sign error: scaling by -3 in a 2×2 matrix multiplies the determinant by (-3)² = 9, which is positive, so det(X) must be positive, not negative.
• (B) -6
  Wrong both in magnitude and sign; it ignores the factor 9 from -3I and misuses det(A^-1).
• (C) 6
  Has correct sign but wrong magnitude; taking 3 instead of 3² in the determinant scaling gives 3 ⋅ det(B)/det(A) = 3 · 2 = 6, which is a common mistake.
• (D) 18
  Correct: det(X) = 9 ⋅ det(B)/det(A) = 9 · 2 = 18.