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Determinants and inverses questions
The determinant gives the area scale factor of a matrix transformation and determines whether the matrix is invertible. A non-zero determinant permits an inverse, and this lesson finds it in the 2 × 2 and 3 × 3 cases.
7 original questions · 24 marks · the determinants and inverses notes · Matrices
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Find the determinant of the matrix with rows (3, 1) and (1, 2), and state what its sign and size say about the transformation.
Worked 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.Write down the inverse of the matrix with rows (3, 1) and (1, 2), and verify it by multiplication.
Worked 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.Evaluate the determinant of the 3 × 3 matrix with rows (1, 2, 0), (3, 1, 4) and (2, 0, 1).
Worked 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.A transformation has matrix with rows (3, 0) and (1, 2). A shape of area 4 is transformed. Find the area of the image.
Worked 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.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.
Worked 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.For invertible matrices A and B, prove that (AB)⁻¹ = B⁻¹A⁻¹, and explain the order reversal in one sentence.
Worked 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.N is the matrix with rows (1, 2, 0), (3, 1, 4) and (2, 0, 1). Find N⁻¹.
Worked 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, which is the check worth doing. The transpose is the step candidates omit, and it only shows up in the answer when the matrix is not symmetric.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise determinants and inverses one question at a time
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