IB · Mathematics: Applications & Interpretation SL & HL · Number & algebra

Matrices & eigenvalues: IB Mathematics: Applications & Interpretation questions with answers

Free Matrices & eigenvalues questions for IB Mathematics: Applications & Interpretation (SL & HL), HL only, part of Number & algebra. Each question has the answer and the marking points an examiner would look for.

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Matrices & eigenvalues practice questions

Question 1Structured · 7 marksA* challenge

Let M = 5 −26 −3. M has eigenvalues 3 and −1.

(a)

Find an eigenvector for each eigenvalue.

[3]
(b)

Hence find M6, giving the exact entries.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)11, 13
(M − λI)v = 0 M1
11 A1
13 A1
3
(b)1093 −3641092 −363
Mn = PDnP−1 M1
D6 = 729 00 1 A1
multiplying out (M1)
1093 −3641092 −363 A1
4
Question 2Structured · 5 marks

Let M = 11 −424 −9.

(a)

Find the eigenvalues of M.

[3]
(b)

Find an eigenvector corresponding to λ = 3.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)λ = 3, λ = −1
det(M − λI) = 0 M1
λ2 − 2λ − 3 = 0 A1
λ = 3, λ = −1 A1
3
(b)12
(M − 3I)v = 0 (M1)
12 or any multiple A1
2
Question 3Structured · 5 marks

A = 2 43 2

(a)

Find det A and A−1.

[3]
(b)

Hence solve 2x + 4y = 16, 3x + 2y = 12.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)det A = −8; A−1 = 1−82 −4−3 2
det A = −8 A1
swap leading diagonal, negate other diagonal, divide by det (M1)
A−1 = 1−82 −4−3 2 A1
3
(b)x = 2, y = 3
use of inverse (M1)
x = 2, y = 3 A1
2
Question 4Structured · 5 marks

Let A = 1 21 1 and B = 7 04 1.

(a)

Find det A, and explain why A−1 exists.

[2]
(b)

Find the matrix X such that AX = B.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)−1
det A = −1 A1
det A ≠ 0 R1
2
(b)1 23 −1
X = A−1B (M1)
use of GDC (M1)
1 23 −1 A1
3
Question 5Structured · 7 marksA* challenge

Let M = 5 −26 −3. M has eigenvalues 3 and −1.

(a)

Find an eigenvector for each eigenvalue.

[3]
(b)

Hence find M5, giving the exact entries.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)11, 13
(M − λI)v = 0 M1
11 A1
13 A1
3
(b)365 −122366 −123
Mn = PDnP−1 M1
D5 = 243 00 −1 A1
multiplying out (M1)
365 −122366 −123 A1
4
Question 6Structured · 5 marks

Let M = −1 −40 3.

(a)

Find the eigenvalues of M.

[3]
(b)

Find an eigenvector corresponding to λ = 3.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)λ = 3, λ = −1
det(M − λI) = 0 M1
λ2 − 2λ − 3 = 0 A1
λ = 3, λ = −1 A1
3
(b)1−1
(M − 3I)v = 0 (M1)
1−1 or any multiple A1
2

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in