Questions on determinants and inverses

Question 1

Find the determinant of the matrix with rows (3, 1) and (1, 2), and state what its sign and size say about the transformation.

Accepted Answer

det = 3 × 2 − 1 × 1 = 5. Areas are scaled by 5, and the positive sign means orientation is preserved, so no reflection is hiding in the transformation. B1 for the determinant, B1 for the area scale factor with orientation preserved.

Question 2

Write down the inverse of the matrix with rows (3, 1) and (1, 2), and verify it by multiplication.

Accepted Answer

Swap the diagonal, negate the off-diagonal, divide by the determinant 5: (1/5) × rows (2, −1), (−1, 3). Multiplying back gives (1/5) × rows (5, 0), (0, 5) = I, as required. M1 for swapping and negating, A1 for dividing by 5, B1 for the multiplication giving I.

Question 3

Evaluate the determinant of the 3 × 3 matrix with rows (2, 0, 0), (1, 3, 0) and (4, 5, 6).

Accepted Answer

Every entry above the leading diagonal is zero, so the determinant is the product of the diagonal entries: 2 × 3 × 6 = 36. B1 for spotting the shortcut, B1 for 36.

Question 4

Two 2 × 2 matrices A and B have det A = 3 and det B = 4. State det(AB) and det(A⁻¹).

Accepted Answer

det(AB) = det A × det B = 3 × 4 = 12. det(A⁻¹) = 1/det A = 1/3. B1 for 12, B1 for 1/3.

Question 5

Evaluate the determinant of the 3 × 3 matrix with rows (1, 2, 0), (3, 1, 4) and (2, 0, 1).

Accepted Answer

Expand along the top row: 1 × (1 × 1 − 4 × 0) − 2 × (3 × 1 − 4 × 2) + 0 = 1 − 2 × (−5) = 11. M1 for expanding along a row, A1 for the two by two minors, A1 for 11. The zero in the top row kills a third of the work; expand along whichever row or column carries the most zeros.

Question 6

A transformation has matrix with rows (3, 0) and (1, 2). A shape of area 4 is transformed. Find the area of the image.

Accepted Answer

det = 3 × 2 − 0 × 1 = 6, so areas scale by 6 and the image has area 24. B1 for the determinant, B1 for 24. The determinant is the area scale factor whatever the shape, so no information about the original outline is needed.

Question 7

Find the value of k for which the matrix with rows (k, 2) and (3, 1) is singular, and state what singular means for the transformation.

Accepted Answer

Singular means det = 0, so k − 6 = 0 and k = 6. The transformation then squashes the plane onto a line, area scale factor zero, and no inverse can exist because the squash cannot be undone. M1 for setting the determinant to zero, A1 for k = 6, B1 for the collapse onto a line.

Question 8

Use row operations to reduce the matrix with rows (1, 2, 3), (2, 5, 3) and (1, 0, 8) to upper triangular form, naming each operation, and hence evaluate its determinant.

Accepted Answer

R2 → R2 − 2R1 gives (0, 1, −3); R3 → R3 − R1 gives (0, −2, 5). Neither changes the determinant. M1 for the two operations correctly named and applied. Then R3 → R3 + 2R2 gives (0, 0, −1), also unchanged. A1 for the triangular form, rows (1, 2, 3), (0, 1, −3), (0, 0, −1). The determinant of a triangular matrix is the product of its diagonal: 1 × 1 × (−1) = −1. A1 for the diagonal product, A1 for −1. Direct expansion of the original rows agrees: 1(40 − 0) − 2(16 − 3) + 3(0 − 5) = 40 − 26 − 15 = −1.

Question 9

For A with rows (1, 2), (3, 4) and B with rows (0, 1), (−1, 2), find AB and verify that det(AB) = det A × det B.

Accepted Answer

AB has rows (−2, 5), (−4, 11): top-left 1 × 0 + 2 × (−1) = −2, and the rest follow the same way. M1 for the multiplication, A1 for AB. det(AB) = (−2)(11) − (5)(−4) = −22 + 20 = −2. Separately, det A = 4 − 6 = −2 and det B = 0 + 1 = 1, and (−2)(1) = −2 as well. M1 for det A and det B, A1 for the two values agreeing. The equality holds for this pair; det(AB) = det A × det B holds for every conformable pair of square matrices, not only invertible ones.

Question 10

A 3 × 3 matrix M has det M = 5. Find det(3M) and det(M²).

Accepted Answer

Scaling a 3 × 3 matrix by k scales its determinant by k³, since all three rows are scaled: det(3M) = 3³ × 5 = 135. B1 for 135. det(M²) = det M × det M = 5² = 25, from det(AB) = det A × det B with A = B = M. B1 for 25, B1 for citing det(AB) = det A × det B as the reason.

Question 11

A matrix M has det M = 7. State the determinant of the matrix obtained from M by (i) swapping its first two rows, (ii) multiplying every entry of its first row by 4, and (iii) adding twice its second row to its first row.

Accepted Answer

(i) Swapping two rows reverses the sign: −7. (ii) Scaling one row by k scales the determinant by k: 4 × 7 = 28. (iii) Adding a multiple of one row to another leaves the determinant unchanged: 7. B1 B1 B1 for the three values. Only the third operation leaves the value unchanged, which is why it is the one used to produce zeros without disturbing it.

Question 12

For invertible matrices A and B, prove that (AB)⁻¹ = B⁻¹A⁻¹, and explain the order reversal in one sentence.

Accepted Answer

Multiply: (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AA⁻¹ = I, and likewise on the other side, so B⁻¹A⁻¹ is the inverse of AB. The order reverses because undoing a sequence means undoing the last action first, like unwrapping layers in the reverse order they went on. M1 for forming the product, A1 for reducing it to I, A1 for the other side as well, B1 for the reason for the reversal.

Question 13

N is the matrix with rows (1, 2, 0), (3, 1, 4) and (2, 0, 1). Find N⁻¹.

Accepted Answer

det N = 11 from the earlier expansion, so the inverse exists. The matrix of minors has rows (1, −5, −2), (2, 1, −4) and (8, 4, −5). Applying the alternating sign pattern gives the cofactors, rows (1, 5, −2), (−2, 1, 4) and (8, −4, −5). Transposing gives the adjugate, rows (1, −2, 8), (5, 1, −4) and (−2, 4, −5), and dividing by 11 gives N⁻¹ = (1/11) × rows (1, −2, 8), (5, 1, −4), (−2, 4, −5). B1 for det N = 11, M1 for the matrix of minors, A1 for the minors, M1 for the alternating signs, M1 for transposing, A1 for the adjugate, A1 for the inverse. Multiplying N by it returns 11I before the division. The adjugate is the transpose of the cofactor matrix, and the transpose changes the answer whenever the matrix is not symmetric.

Question 14

M is the matrix with rows (1, 0, k), (2, 1, 0) and (0, k, 1), for a real number k. Show that det M = 1 + 2k², and explain why M has an inverse for every real value of k.

Accepted Answer

Expanding along the top row: det M = 1(1 × 1 − 0 × k) − 0(2 × 1 − 0 × 0) + k(2 × k − 1 × 0) = 1 + 2k². M1 for expanding along the top row, A1 for the three cofactor terms, A1 for 1 + 2k². Since k² ≥ 0 for every real k, 1 + 2k² ≥ 1, so it is never zero. M1 for the inequality, A1 for the conclusion that M is never singular, so an inverse exists for every real k.

Question 15

Prove that if A is a singular n × n matrix then AB is singular for every conformable matrix B. Give a specific 2 × 2 example with A ≠ 0 and B invertible, and verify your example directly.

Accepted Answer

det(AB) = det A × det B for square matrices of the same size. If A is singular, det A = 0, so det(AB) = 0 × det B = 0 whatever det B is, and AB is singular too. M1 for quoting det(AB) = det A × det B, A1 for the argument that a zero factor forces det(AB) = 0. Take A with rows (1, 2), (2, 4), singular since det A = 4 − 4 = 0 while A itself is not the zero matrix, and B with rows (1, 1), (0, 1), with det B = 1 so B is invertible. B1 for a valid singular A that is not the zero matrix, with an invertible B. Then AB has rows (1, 3), (2, 6): top-left 1 × 1 + 2 × 0 = 1, top-right 1 × 1 + 2 × 1 = 3, and the bottom row follows the same way. M1 for the product AB. det(AB) = 1 × 6 − 3 × 2 = 0, confirming AB is singular, agreeing with det A × det B = 0 × 1 = 0. A1 for det(AB) = 0. Being invertible is a property B cannot rescue A from: whatever B does, a singular factor collapses the product's determinant to zero. B1 for that closing statement linking the specific example back to the general argument.