A is a 3 × 3 matrix with |A| = 4. The value of |2A−1| is
- |A−1| = 1/|A| = 1/4.
- A−1 is 3 × 3, so |2A−1| = 23 × |A−1| = 8 × 1/4.
- 8 × 1/4 = 2.
Answer: 2
Matrices and determinants
NUST NET (Engineering) · Maths · Matrices and determinants. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.
The idea
A matrix is a rectangular table of numbers with an order, rows × columns. The product AB exists only when the number of columns of A equals the number of rows of B, and in general AB ≠ BA.
Every square matrix has a determinant, a single number. If it is zero the matrix is singular and has no inverse. Many test questions are answered by a property of determinants, not by a long expansion, so look first for equal or proportional rows.
A system of linear equations can be written AX = B. The determinant of A tells you at once whether there is exactly one solution or not.
Rules to remember
- (m × n)(n × p) gives an m × p matrix; the inner numbers must match
- For A with rows [a b] and [c d]: |A| = ad − bc, and A−1 = (1/|A|) × the matrix with rows [d −b] and [−c a]
- (AB)−1 = B−1A−1 and (AB)t = BtAt (the order reverses)
- |At| = |A|, |AB| = |A| × |B|, |A−1| = 1/|A|, and |kA| = kn|A| for an n × n matrix
- A determinant is 0 if a row is all zeros or two rows (or columns) are equal or proportional; swapping two rows changes its sign
- AX = B: if |A| ≠ 0 there is one solution, X = A−1B; if |A| = 0 there is either no solution (inconsistent) or infinitely many
- Homogeneous system AX = O: only x = y = z = 0 if |A| ≠ 0; non-trivial solutions if |A| = 0
Common mistake
Writing |kA| = k|A|. The factor k comes out of every row, so for an n × n matrix |kA| = kn|A|: k2 for a 2 × 2 matrix and k3 for a 3 × 3 matrix.
Worked example
Practice questions
Try each one, then open the answer.
1. A is a 2 × 2 matrix with |A| = −3. The value of |2At| is
- A−6
- B−12
- C12
- D−24
Show answer
Answer: B. Transposing does not change a determinant, and multiplying a 2 × 2 matrix by 2 multiplies its determinant by 22 = 4. So |2At| = 4(−3) = −12.
2. The system x + 2y = 3, 2x + ky = 5 has no solution when k is
- A−4
- B4
- C2
- D6
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Answer: B. The coefficient determinant k − 4 must be 0, so k = 4. Then doubling the first equation gives 2x + 4y = 6, which contradicts 2x + 4y = 5, so there is no solution.
3. Every element of a 3 × 3 determinant of value Δ is multiplied by −1. The new value is
- AΔ
- B−3Δ
- C3Δ
- D−Δ
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Answer: D. Taking −1 out of each of the 3 rows gives a factor (−1)3 = −1, so the value becomes −Δ. For a 2 × 2 determinant the value would not change.
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