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Multiplication of Two Matrices

Mathematics · FE Reference Handbook section

Mathematics
5 formulas
10 exam-style examples
~55 min
All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • In order for multiplication to be possible, the number of columns in A must equal the number of rows in B.

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 — Multiplication of Two Matrices

For A = [[9, 1], [2, 4]], compute det A and solve A·x = {10, 6}ᵀ.

Given

  • A=[[9,1],[2,4]]A = [[9, 1], [2, 4]]
  • b=10,6Tb = {10, 6}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)−(1)(2)=34det A = (9)(4) - (1)(2) = 34

  3. Inspection

    x=1,1Tsatisfiesbothrows:9(1)+1(1)=10✓and2(1)+4(1)=6✓x = {1, 1}ᵀ satisfies both rows: 9(1) + 1(1) = 10 ✓ and 2(1) + 4(1) = 6 ✓

  4. Uniqueness

    detA=34≠0,so1,1Tistheonlysolutiondet A = 34 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=34;x=1,1Tdet A = 34; x = {1, 1}ᵀ

Why the other options are there

  • det = 38 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Multiplication of Two Matrices

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

For A = [[9, 4], [6, 7]], compute det A and solve A·x = {13, 13}ᵀ.

Given

  • A=[[9,4],[6,7]]A = [[9, 4], [6, 7]]
  • b=13,13Tb = {13, 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=(9)(7)−(4)(6)=39det A = (9)(7) - (4)(6) = 39

  3. Inspection

    x=1,1Tsatisfiesbothrows:9(1)+4(1)=13✓and6(1)+7(1)=13✓x = {1, 1}ᵀ satisfies both rows: 9(1) + 4(1) = 13 ✓ and 6(1) + 7(1) = 13 ✓

  4. Uniqueness

    detA=39≠0,so1,1Tistheonlysolutiondet A = 39 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=39;x=1,1Tdet A = 39; x = {1, 1}ᵀ

Why the other options are there

  • det = 87 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Multiplication of Two Matrices

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

For A = [[3, 1], [2, 4]], compute det A and solve A·x = {4, 6}ᵀ.

Given

  • A=[[3,1],[2,4]]A = [[3, 1], [2, 4]]
  • b=4,6Tb = {4, 6}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)(4)−(1)(2)=10det A = (3)(4) - (1)(2) = 10

  3. Inspection

    x=1,1Tsatisfiesbothrows:3(1)+1(1)=4✓and2(1)+4(1)=6✓x = {1, 1}ᵀ satisfies both rows: 3(1) + 1(1) = 4 ✓ and 2(1) + 4(1) = 6 ✓

  4. Uniqueness

    detA=10≠0,so1,1Tistheonlysolutiondet A = 10 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=10;x=1,1Tdet A = 10; x = {1, 1}ᵀ

Why the other options are there

  • det = 14 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Multiplication of Two Matrices

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

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

Given

  • A=[[7,5],[1,4]]A = [[7, 5], [1, 4]]
  • b=12,5Tb = {12, 5}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)(4)−(5)(1)=23det A = (7)(4) - (5)(1) = 23

  3. Inspection

    x=1,1Tsatisfiesbothrows:7(1)+5(1)=12✓and1(1)+4(1)=5✓x = {1, 1}ᵀ satisfies both rows: 7(1) + 5(1) = 12 ✓ and 1(1) + 4(1) = 5 ✓

  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 → Multiplication of Two Matrices

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

For A = [[5, 7], [7, 6]], compute det A and solve A·x = {12, 13}ᵀ.

Given

  • A=[[5,7],[7,6]]A = [[5, 7], [7, 6]]
  • b=12,13Tb = {12, 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=(5)(6)−(7)(7)=−19det A = (5)(6) - (7)(7) = -19

  3. Inspection

    x=1,1Tsatisfiesbothrows:5(1)+7(1)=12✓and7(1)+6(1)=13✓x = {1, 1}ᵀ satisfies both rows: 5(1) + 7(1) = 12 ✓ and 7(1) + 6(1) = 13 ✓

  4. Uniqueness

    detA=−19≠0,so1,1Tistheonlysolutiondet A = -19 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=−19;x=1,1Tdet A = -19; x = {1, 1}ᵀ

Why the other options are there

  • det = 79 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Multiplication of Two Matrices

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

For A = [[3, 1], [5, 7]], compute det A and solve A·x = {4, 12}ᵀ.

Given

  • A=[[3,1],[5,7]]A = [[3, 1], [5, 7]]
  • b=4,12Tb = {4, 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=(3)(7)−(1)(5)=16det A = (3)(7) - (1)(5) = 16

  3. Inspection

    x=1,1Tsatisfiesbothrows:3(1)+1(1)=4✓and5(1)+7(1)=12✓x = {1, 1}ᵀ satisfies both rows: 3(1) + 1(1) = 4 ✓ and 5(1) + 7(1) = 12 ✓

  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 = 26 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Multiplication of Two Matrices

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

For A = [[3, 2], [1, 6]], compute det A and solve A·x = {5, 7}ᵀ.

Given

  • A=[[3,2],[1,6]]A = [[3, 2], [1, 6]]
  • b=5,7Tb = {5, 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=(3)(6)−(2)(1)=16det A = (3)(6) - (2)(1) = 16

  3. Inspection

    x=1,1Tsatisfiesbothrows:3(1)+2(1)=5✓and1(1)+6(1)=7✓x = {1, 1}ᵀ satisfies both rows: 3(1) + 2(1) = 5 ✓ and 1(1) + 6(1) = 7 ✓

  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 = 20 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Multiplication of Two Matrices

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

For A = [[6, 3], [4, 6]], compute det A and solve A·x = {9, 10}ᵀ.

Given

  • A=[[6,3],[4,6]]A = [[6, 3], [4, 6]]
  • b=9,10Tb = {9, 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=(6)(6)−(3)(4)=24det A = (6)(6) - (3)(4) = 24

  3. Inspection

    x=1,1Tsatisfiesbothrows:6(1)+3(1)=9✓and4(1)+6(1)=10✓x = {1, 1}ᵀ satisfies both rows: 6(1) + 3(1) = 9 ✓ and 4(1) + 6(1) = 10 ✓

  4. Uniqueness

    detA=24≠0,so1,1Tistheonlysolutiondet A = 24 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=24;x=1,1Tdet A = 24; x = {1, 1}ᵀ

Why the other options are there

  • det = 48 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Multiplication of Two Matrices

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

For A = [[6, 4], [2, 2]], compute det A and solve A·x = {10, 4}ᵀ.

Given

  • A=[[6,4],[2,2]]A = [[6, 4], [2, 2]]
  • b=10,4Tb = {10, 4}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)(2)−(4)(2)=4det A = (6)(2) - (4)(2) = 4

  3. Inspection

    x=1,1Tsatisfiesbothrows:6(1)+4(1)=10✓and2(1)+2(1)=4✓x = {1, 1}ᵀ satisfies both rows: 6(1) + 4(1) = 10 ✓ and 2(1) + 2(1) = 4 ✓

  4. Uniqueness

    detA=4≠0,so1,1Tistheonlysolutiondet A = 4 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=4;x=1,1Tdet A = 4; x = {1, 1}ᵀ

Why the other options are there

  • det = 20 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Multiplication of Two Matrices

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

For A = [[8, 6], [4, 4]], compute det A and solve A·x = {14, 8}ᵀ.

Given

  • A=[[8,6],[4,4]]A = [[8, 6], [4, 4]]
  • b=14,8Tb = {14, 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)(4)−(6)(4)=8det A = (8)(4) - (6)(4) = 8

  3. Inspection

    x=1,1Tsatisfiesbothrows:8(1)+6(1)=14✓and4(1)+4(1)=8✓x = {1, 1}ᵀ satisfies both rows: 8(1) + 6(1) = 14 ✓ and 4(1) + 4(1) = 8 ✓

  4. Uniqueness

    detA=8≠0,so1,1Tistheonlysolutiondet A = 8 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=8;x=1,1Tdet A = 8; 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 → Multiplication of Two Matrices

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