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Matrices

Mathematics · FE Reference Handbook section

Mathematics
1 formulas
10 exam-style examples
~47 min
All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • A matrix is an ordered rectangular array of numbers with m rows and n columns. The element aij refers to row i and column j.
  • The rank of a matrix is equal to the number of rows that are linearly independent.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Determinant and inverse of a 2×2 system — Matrices

For A = [[8, 8], [5, 3]], compute det A and solve A·x = {16, 8}ᵀ.

Given

  • A=[[8,8],[5,3]]A = [[8, 8], [5, 3]]
  • b=16,8Tb = {16, 8}ᵀ

Find

det A and the solution vector

Start with the thinking

  • A non-zero determinant guarantees a unique solution.
  • Cramer's rule is fastest for 2×2.

Step-by-step solution

  1. Determinant

    detA=ad−bcdet A = ad - bc

  2. Substituting

    detA=(8)(3)−(8)(5)=−16det A = (8)(3) - (8)(5) = -16

  3. Inspection

    x=1,1Tsatisfiesbothrows:8(1)+8(1)=16✓and5(1)+3(1)=8✓x = {1, 1}ᵀ satisfies both rows: 8(1) + 8(1) = 16 ✓ and 5(1) + 3(1) = 8 ✓

  4. Uniqueness

    detA=−16≠0,so1,1Tistheonlysolutiondet A = -16 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=−16;x=1,1Tdet A = -16; x = {1, 1}ᵀ

Why the other options are there

  • det = 64 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrices

Example 2
Determinant and inverse of a 2×2 system — Matrices (2)

For A = [[2, 2], [7, 7]], compute det A and solve A·x = {4, 14}ᵀ.

Given

  • A=[[2,2],[7,7]]A = [[2, 2], [7, 7]]
  • b=4,14Tb = {4, 14}ᵀ

Find

det A and the solution vector

Start with the thinking

  • A non-zero determinant guarantees a unique solution.
  • Cramer's rule is fastest for 2×2.

Step-by-step solution

  1. Determinant

    detA=ad−bcdet A = ad - bc

  2. Substituting

    detA=(2)(7)−(2)(7)=0det A = (2)(7) - (2)(7) = 0

  3. Inspection

    x=1,1Tsatisfiesbothrows:2(1)+2(1)=4✓and7(1)+7(1)=14✓x = {1, 1}ᵀ satisfies both rows: 2(1) + 2(1) = 4 ✓ and 7(1) + 7(1) = 14 ✓

  4. Uniqueness

    detA=0≠0,so1,1Tistheonlysolutiondet A = 0 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=0;x=1,1Tdet A = 0; x = {1, 1}ᵀ

Why the other options are there

  • det = 28 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrices

Example 3
Determinant and inverse of a 2×2 system — Matrices (3)

For A = [[7, 4], [1, 6]], compute det A and solve A·x = {11, 7}ᵀ.

Given

  • A=[[7,4],[1,6]]A = [[7, 4], [1, 6]]
  • b=11,7Tb = {11, 7}ᵀ

Find

det A and the solution vector

Start with the thinking

  • A non-zero determinant guarantees a unique solution.
  • Cramer's rule is fastest for 2×2.

Step-by-step solution

  1. Determinant

    detA=ad−bcdet A = ad - bc

  2. Substituting

    detA=(7)(6)−(4)(1)=38det A = (7)(6) - (4)(1) = 38

  3. Inspection

    x=1,1Tsatisfiesbothrows:7(1)+4(1)=11✓and1(1)+6(1)=7✓x = {1, 1}ᵀ satisfies both rows: 7(1) + 4(1) = 11 ✓ and 1(1) + 6(1) = 7 ✓

  4. Uniqueness

    detA=38≠0,so1,1Tistheonlysolutiondet A = 38 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=38;x=1,1Tdet A = 38; x = {1, 1}ᵀ

Why the other options are there

  • det = 46 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrices

Example 4
Determinant and inverse of a 2×2 system — Matrices (4)

For A = [[9, 4], [5, 4]], compute det A and solve A·x = {13, 9}ᵀ.

Given

  • A=[[9,4],[5,4]]A = [[9, 4], [5, 4]]
  • b=13,9Tb = {13, 9}ᵀ

Find

det A and the solution vector

Start with the thinking

  • A non-zero determinant guarantees a unique solution.
  • Cramer's rule is fastest for 2×2.

Step-by-step solution

  1. Determinant

    detA=ad−bcdet A = ad - bc

  2. Substituting

    detA=(9)(4)−(4)(5)=16det A = (9)(4) - (4)(5) = 16

  3. Inspection

    x=1,1Tsatisfiesbothrows:9(1)+4(1)=13✓and5(1)+4(1)=9✓x = {1, 1}ᵀ satisfies both rows: 9(1) + 4(1) = 13 ✓ and 5(1) + 4(1) = 9 ✓

  4. Uniqueness

    detA=16≠0,so1,1Tistheonlysolutiondet A = 16 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=16;x=1,1Tdet A = 16; x = {1, 1}ᵀ

Why the other options are there

  • det = 56 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrices

Example 5
Determinant and inverse of a 2×2 system — Matrices (5)

For A = [[4, 1], [5, 7]], compute det A and solve A·x = {5, 12}ᵀ.

Given

  • A=[[4,1],[5,7]]A = [[4, 1], [5, 7]]
  • b=5,12Tb = {5, 12}ᵀ

Find

det A and the solution vector

Start with the thinking

  • A non-zero determinant guarantees a unique solution.
  • Cramer's rule is fastest for 2×2.

Step-by-step solution

  1. Determinant

    detA=ad−bcdet A = ad - bc

  2. Substituting

    detA=(4)(7)−(1)(5)=23det A = (4)(7) - (1)(5) = 23

  3. Inspection

    x=1,1Tsatisfiesbothrows:4(1)+1(1)=5✓and5(1)+7(1)=12✓x = {1, 1}ᵀ satisfies both rows: 4(1) + 1(1) = 5 ✓ and 5(1) + 7(1) = 12 ✓

  4. Uniqueness

    detA=23≠0,so1,1Tistheonlysolutiondet A = 23 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=23;x=1,1Tdet A = 23; x = {1, 1}ᵀ

Why the other options are there

  • det = 33 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrices

Example 6
Determinant and inverse of a 2×2 system — Matrices (6)

For A = [[6, 7], [1, 6]], compute det A and solve A·x = {13, 7}ᵀ.

Given

  • A=[[6,7],[1,6]]A = [[6, 7], [1, 6]]
  • b=13,7Tb = {13, 7}ᵀ

Find

det A and the solution vector

Start with the thinking

  • A non-zero determinant guarantees a unique solution.
  • Cramer's rule is fastest for 2×2.

Step-by-step solution

  1. Determinant

    detA=ad−bcdet A = ad - bc

  2. Substituting

    detA=(6)(6)−(7)(1)=29det A = (6)(6) - (7)(1) = 29

  3. Inspection

    x=1,1Tsatisfiesbothrows:6(1)+7(1)=13✓and1(1)+6(1)=7✓x = {1, 1}ᵀ satisfies both rows: 6(1) + 7(1) = 13 ✓ and 1(1) + 6(1) = 7 ✓

  4. Uniqueness

    detA=29≠0,so1,1Tistheonlysolutiondet A = 29 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=29;x=1,1Tdet A = 29; x = {1, 1}ᵀ

Why the other options are there

  • det = 43 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrices

Example 7
Determinant and inverse of a 2×2 system — Matrices (7)

For A = [[2, 5], [3, 7]], compute det A and solve A·x = {7, 10}ᵀ.

Given

  • A=[[2,5],[3,7]]A = [[2, 5], [3, 7]]
  • b=7,10Tb = {7, 10}ᵀ

Find

det A and the solution vector

Start with the thinking

  • A non-zero determinant guarantees a unique solution.
  • Cramer's rule is fastest for 2×2.

Step-by-step solution

  1. Determinant

    detA=ad−bcdet A = ad - bc

  2. Substituting

    detA=(2)(7)−(5)(3)=−1det A = (2)(7) - (5)(3) = -1

  3. Inspection

    x=1,1Tsatisfiesbothrows:2(1)+5(1)=7✓and3(1)+7(1)=10✓x = {1, 1}ᵀ satisfies both rows: 2(1) + 5(1) = 7 ✓ and 3(1) + 7(1) = 10 ✓

  4. Uniqueness

    detA=−1≠0,so1,1Tistheonlysolutiondet A = -1 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=−1;x=1,1Tdet A = -1; x = {1, 1}ᵀ

Why the other options are there

  • det = 29 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrices

Example 8
Determinant and inverse of a 2×2 system — Matrices (8)

For A = [[6, 1], [3, 5]], compute det A and solve A·x = {7, 8}ᵀ.

Given

  • A=[[6,1],[3,5]]A = [[6, 1], [3, 5]]
  • b=7,8Tb = {7, 8}ᵀ

Find

det A and the solution vector

Start with the thinking

  • A non-zero determinant guarantees a unique solution.
  • Cramer's rule is fastest for 2×2.

Step-by-step solution

  1. Determinant

    detA=ad−bcdet A = ad - bc

  2. Substituting

    detA=(6)(5)−(1)(3)=27det A = (6)(5) - (1)(3) = 27

  3. Inspection

    x=1,1Tsatisfiesbothrows:6(1)+1(1)=7✓and3(1)+5(1)=8✓x = {1, 1}ᵀ satisfies both rows: 6(1) + 1(1) = 7 ✓ and 3(1) + 5(1) = 8 ✓

  4. Uniqueness

    detA=27≠0,so1,1Tistheonlysolutiondet A = 27 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=27;x=1,1Tdet A = 27; x = {1, 1}ᵀ

Why the other options are there

  • det = 33 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrices

Example 9
Determinant and inverse of a 2×2 system — Matrices (9)

For A = [[3, 8], [4, 6]], compute det A and solve A·x = {11, 10}ᵀ.

Given

  • A=[[3,8],[4,6]]A = [[3, 8], [4, 6]]
  • b=11,10Tb = {11, 10}ᵀ

Find

det A and the solution vector

Start with the thinking

  • A non-zero determinant guarantees a unique solution.
  • Cramer's rule is fastest for 2×2.

Step-by-step solution

  1. Determinant

    detA=ad−bcdet A = ad - bc

  2. Substituting

    detA=(3)(6)−(8)(4)=−14det A = (3)(6) - (8)(4) = -14

  3. Inspection

    x=1,1Tsatisfiesbothrows:3(1)+8(1)=11✓and4(1)+6(1)=10✓x = {1, 1}ᵀ satisfies both rows: 3(1) + 8(1) = 11 ✓ and 4(1) + 6(1) = 10 ✓

  4. Uniqueness

    detA=−14≠0,so1,1Tistheonlysolutiondet A = -14 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=−14;x=1,1Tdet A = -14; x = {1, 1}ᵀ

Why the other options are there

  • det = 50 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrices

Example 10
Determinant and inverse of a 2×2 system — Matrices (10)

For A = [[7, 4], [7, 6]], compute det A and solve A·x = {11, 13}ᵀ.

Given

  • A=[[7,4],[7,6]]A = [[7, 4], [7, 6]]
  • b=11,13Tb = {11, 13}ᵀ

Find

det A and the solution vector

Start with the thinking

  • A non-zero determinant guarantees a unique solution.
  • Cramer's rule is fastest for 2×2.

Step-by-step solution

  1. Determinant

    detA=ad−bcdet A = ad - bc

  2. Substituting

    detA=(7)(6)−(4)(7)=14det A = (7)(6) - (4)(7) = 14

  3. Inspection

    x=1,1Tsatisfiesbothrows:7(1)+4(1)=11✓and7(1)+6(1)=13✓x = {1, 1}ᵀ satisfies both rows: 7(1) + 4(1) = 11 ✓ and 7(1) + 6(1) = 13 ✓

  4. Uniqueness

    detA=14≠0,so1,1Tistheonlysolutiondet A = 14 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=14;x=1,1Tdet A = 14; x = {1, 1}ᵀ

Why the other options are there

  • det = 70 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrices

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