Matrices & determinants: practice problems with step-by-step solutions

Free Matrices & determinants practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: 2×2 determinant, 3×3 and products, eigenvalues and systems.

Free practice problemEasy
A =9−6−8−5

Find the determinant det A of the matrix A shown.

A = [9 −6; −8 −5]

Exact answer: a whole number, no rounding.

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What you practise

  1. Easy2×2 determinant
  2. Medium3×3 and products
  3. Hardeigenvalues and systems

Key formulas

2×2 determinant
det = ad − bc
2×2 inverse
A⁻¹ = (1/det)·[d −b; −c a]
Product
(AB)ij = Σk AikBkj
Eigenvalues
det(A − λI) = 0
Cramer’s rule
xi = det(Ai)/det(A)

Learn Matrices & determinants from scratch

1Matrices and the 2×2 determinant

A matrix is a table of numbers in rows and columns. A matrix with 2 rows and 3 columns is called a 2×3 matrix: rows first, then columns.

Engineers use matrices to write many equations at once. The system 2x + y = 5 and x − y = 1 is stored as the matrix [2 1; 1 −1] of its coefficients.

A square matrix has one special number, its determinant, written det A. For a 2×2 matrix it is the product of one diagonal minus the product of the other.

abcdA =+ad−bc
det A = a·d − b·c

For [3 1; 2 4] the determinant is 3·4 − 1·2 = 10.

If det A = 0, the matrix has no inverse, and the system of equations has no single solution.

Self-check What is the determinant of [4 2; 3 5]? (Row 1 is 4 2, row 2 is 3 5.)
  1. 14
  2. 26
  3. 9
  4. −14

Answer: 14. a·d − b·c = 4·5 − 2·3 = 20 − 6 = 14. Adding the two products gives 26, and taking bc − ad gives −14.

23×3 determinants and products

For a 3×3 matrix, expand along the first row. Each entry is multiplied by its minor: the 2×2 determinant that is left when you cross out that entry’s row and column.

det A = a11·M11 − a12·M12 + a13·M13

The signs go + − + along the first row. This sign pattern is the easiest part to forget.

A matrix product AB is built from rows of A and columns of B. The entry in row i, column j is row i of A times column j of B: multiply pairs and add.

[1 2; 3 4] · [5; 6] = [1·5 + 2·6; 3·5 + 4·6] = [17; 39]

The product only exists when A has as many columns as B has rows. In general AB is not the same as BA.

Self-check A = [2 1; 0 3] and B = [1 4; 2 1]. What is the entry in row 1, column 2 of AB?
  1. 13
  2. 4
  3. 6
  4. 9

Answer: 9. Row 1 of A is (2, 1) and column 2 of B is (4, 1): 2·4 + 1·1 = 9. Multiplying only the matching entries gives 1·4 = 4, and 13 is the entry of BA.

3Eigenvalues and systems

An eigenvalue λ of A is a number for which A·v = λ·v for some vector v ≠ 0. The matrix only stretches that vector, it does not turn it.

You find the eigenvalues from the characteristic equation. Here I is the identity matrix [1 0; 0 1].

det(A − λI) = 0

For a 2×2 matrix this becomes a quadratic. Its middle coefficient is the trace a + d, and its constant is det A:

λ² − (a + d)·λ + (ad − bc) = 0

For [4 1; 2 3]: λ² − 7λ + 10 = 0, so (λ − 2)(λ − 5) = 0 and the eigenvalues are 2 and 5.

Cramer’s rule solves a system with determinants. Replace column i of A by the right-hand side to get Ai, then xi = det Ai / det A.

For 2x + y = 5 and x − y = 1: det A = −3 and det Ax = 5·(−1) − 1·1 = −6, so x = −6/−3 = 2.

Self-check What is the larger eigenvalue of [2 1; 1 2]?
  1. 1
  2. 2
  3. 3
  4. 4

Answer: 3. λ² − 4λ + 3 = 0 gives (λ − 1)(λ − 3) = 0, so the eigenvalues are 1 and 3. 4 is the trace, which is the sum of the eigenvalues, not one of them.

4The method step by step

  1. Copy the matrix carefully, with every minus sign.
  2. 2×2 determinant: ad − bc.
  3. 3×3 determinant: for each entry of the first row, cross out its row and column and find the 2×2 minor.
  4. Combine with the signs + − +: a11M11 − a12M12 + a13M13.
  5. Eigenvalues: write λ² − (a + d)λ + det A = 0 and solve the quadratic.
  6. Check: a row of zeros, or two equal rows, always gives det A = 0.

Tip: expand along a row or column with many zeros. Every zero entry saves you a minor.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

Find the determinant of the 3×3 matrix A = [2 1 0; 1 3 1; 0 2 4].

1

First minor

210131024A =

Cross out row 1 and column 1 of the entry 2. What is left is [3 1; 2 4].

M11 = 3·4 − 1·2 = 10
2

Second minor

210131024A =

Cross out row 1 and column 2 of the entry 1. What is left is [1 1; 0 4].

M12 = 1·4 − 1·0 = 4
3

Third minor

The entry a13 is 0, so its term is 0 whatever the minor is. You can skip it.

4

Combine with + − +

det A = 2·10 − 1·4 + 0 = 20 − 4 = 16
5

Check the answer

For 3×3 matrices, the rule of Sarrus gives the same value: three “down” diagonal products minus three “up” diagonal products.

(2·3·4 + 1·1·0 + 0·1·2) − (0·3·0 + 2·1·2 + 1·1·4) = 24 − 8 = 16
Answer

det A = 16

6Common mistakes

  • Forgetting the minus sign for the middle term in the 3×3 expansion.
  • Computing bc − ad instead of ad − bc.
  • Losing a minus sign when an entry is negative: −2 · −3 is +6.
  • Multiplying matrices entry by entry instead of row times column.
  • Taking the trace or a diagonal entry as an eigenvalue instead of solving the characteristic equation.
Self-check What is the determinant of [1 2 0; 3 1 0; 0 0 2]?
  1. 10
  2. 14
  3. −10
  4. −14

Answer: −10. det = 1·(1·2 − 0) − 2·(3·2 − 0) + 0 = 2 − 12 = −10. Using + for the middle term gives 2 + 12 = 14, the most common slip.

7How exam questions look

  • Easy: the determinant of a 2×2 matrix, often with negative entries. Use ad − bc.
  • Medium: a 3×3 determinant or a matrix product. Expand along the first row with + − +.
  • Hard: eigenvalues of a 2×2 matrix, or one unknown of a 2×2 system with Cramer’s rule.

A quick check for eigenvalues: they must add up to the trace a + d and multiply to det A.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
A =1−8−10

Find the determinant det A of the matrix A shown.

A = [1 −8; −1 0]

Find det A

Show hints
  1. For a 2×2 matrix, det = ad − bc.
  2. a and d are on the main diagonal, b and c on the other diagonal.
  3. ad = 0.
Show the answer and the step-by-step solution

Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.

Diagonal products

ad − bc = 1·0 − (−8)·(−1) = 0 − 8 = −8

Answer

det A = −8
Example 2Medium
A =−153−5−1−3−513

Find the determinant det A of the 3×3 matrix A shown.

A = [−1 5 3; −5 −1 −3; −5 1 3]

Find det A

Show hints
  1. Expand along the first row with signs + − +.
  2. Each term is an entry times the 2×2 determinant left when you cross out its row and column.
  3. minors: 0, −30, −10.
Show the answer and the step-by-step solution

Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.

2×2 minors of the first row

M₁₁ = 0, M₁₂ = −30, M₁₃ = −10

Cofactor expansion

det A = (−1)·0 − 5·(−30) + 3·(−10) = 120

Answer

det A = 120
Example 3Hard
A =−3−5−2−4b =19

Solve the system −3x − 5y = 1 and −2x − 4y = 9. Find x.

−3x − 5y = 1−2x − 4y = 9

Find x

Show hints
  1. Cramer's rule: x = det(Aₓ)/det(A), where Aₓ has the first column replaced by the right-hand side.
  2. det A = 2.
  3. det Aₓ = 41.
Show the answer and the step-by-step solution

Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.

Determinant of the coefficient matrix

det A = (−3)·(−4) − (−5)·(−2) = 2

Replace the x column

det Aₓ = 1·(−4) − (−5)·9 = 41
x = 41/2 = 20.5

Answer

x = 20.5
y = −12.5

Questions and answers

What is Matrices & determinants?

A matrix is a table of numbers in rows and columns. A matrix with 2 rows and 3 columns is called a 2×3 matrix: rows first, then columns.

Engineers use matrices to write many equations at once. The system 2x + y = 5 and x − y = 1 is stored as the matrix [2 1; 1 −1] of its coefficients.

How do I solve Matrices & determinants problems step by step?

Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.

What do the Matrices & determinants exercises cover?
  • Easy: 2×2 determinant
  • Medium: 3×3 and products
  • Hard: eigenvalues and systems

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Calculators

Matrix determinant and inverse calculatorEigenvalue and eigenvector calculatorSystem of linear equations solver (2×2 and 3×3)

Courses with this topic

Engineering MathematicsEngineering Mathematics 2Numerical Methods

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Printable worksheet

Matrices & determinants worksheet with answers
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