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Addition of Two Matrices

Mathematics · FE Reference Handbook section

Mathematics
2 formulas
10 exam-style examples
~49 min
All Mathematics lectures

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 — Addition of Two Matrices

For A = [[7, 4], [6, 4]], compute det A and solve A·x = {11, 10}ᵀ.

Given

  • A=[[7,4],[6,4]]A = [[7, 4], [6, 4]]
  • 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=(7)(4)−(4)(6)=4det A = (7)(4) - (4)(6) = 4

  3. Inspection

    x=1,1Tsatisfiesbothrows:7(1)+4(1)=11✓and6(1)+4(1)=10✓x = {1, 1}ᵀ satisfies both rows: 7(1) + 4(1) = 11 ✓ and 6(1) + 4(1) = 10 ✓

  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 = 52 (terms added)
  • No unique solution (determinant treated as zero)

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

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

For A = [[8, 8], [1, 3]], compute det A and solve A·x = {16, 4}ᵀ.

Given

  • A=[[8,8],[1,3]]A = [[8, 8], [1, 3]]
  • b=16,4Tb = {16, 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=(8)(3)−(8)(1)=16det A = (8)(3) - (8)(1) = 16

  3. Inspection

    x=1,1Tsatisfiesbothrows:8(1)+8(1)=16✓and1(1)+3(1)=4✓x = {1, 1}ᵀ satisfies both rows: 8(1) + 8(1) = 16 ✓ and 1(1) + 3(1) = 4 ✓

  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 = 32 (terms added)
  • No unique solution (determinant treated as zero)

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

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

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

Given

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

  3. Inspection

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

  4. Uniqueness

    detA=−12≠0,so1,1Tistheonlysolutiondet A = -12 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=−12;x=1,1Tdet A = -12; x = {1, 1}ᵀ

Why the other options are there

  • det = 72 (terms added)
  • No unique solution (determinant treated as zero)

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

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

For A = [[6, 6], [4, 3]], compute det A and solve A·x = {12, 7}ᵀ.

Given

  • A=[[6,6],[4,3]]A = [[6, 6], [4, 3]]
  • b=12,7Tb = {12, 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)(3)−(6)(4)=−6det A = (6)(3) - (6)(4) = -6

  3. Inspection

    x=1,1Tsatisfiesbothrows:6(1)+6(1)=12✓and4(1)+3(1)=7✓x = {1, 1}ᵀ satisfies both rows: 6(1) + 6(1) = 12 ✓ and 4(1) + 3(1) = 7 ✓

  4. Uniqueness

    detA=−6≠0,so1,1Tistheonlysolutiondet A = -6 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=−6;x=1,1Tdet A = -6; x = {1, 1}ᵀ

Why the other options are there

  • det = 42 (terms added)
  • No unique solution (determinant treated as zero)

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

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

For A = [[6, 2], [5, 7]], compute det A and solve A·x = {8, 12}ᵀ.

Given

  • A=[[6,2],[5,7]]A = [[6, 2], [5, 7]]
  • b=8,12Tb = {8, 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=(6)(7)−(2)(5)=32det A = (6)(7) - (2)(5) = 32

  3. Inspection

    x=1,1Tsatisfiesbothrows:6(1)+2(1)=8✓and5(1)+7(1)=12✓x = {1, 1}ᵀ satisfies both rows: 6(1) + 2(1) = 8 ✓ and 5(1) + 7(1) = 12 ✓

  4. Uniqueness

    detA=32≠0,so1,1Tistheonlysolutiondet A = 32 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=32;x=1,1Tdet A = 32; x = {1, 1}ᵀ

Why the other options are there

  • det = 52 (terms added)
  • No unique solution (determinant treated as zero)

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

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

For A = [[6, 8], [1, 7]], compute det A and solve A·x = {14, 8}ᵀ.

Given

  • A=[[6,8],[1,7]]A = [[6, 8], [1, 7]]
  • 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=(6)(7)−(8)(1)=34det A = (6)(7) - (8)(1) = 34

  3. Inspection

    x=1,1Tsatisfiesbothrows:6(1)+8(1)=14✓and1(1)+7(1)=8✓x = {1, 1}ᵀ satisfies both rows: 6(1) + 8(1) = 14 ✓ and 1(1) + 7(1) = 8 ✓

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

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

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

For A = [[9, 4], [2, 8]], compute det A and solve A·x = {13, 10}ᵀ.

Given

  • A=[[9,4],[2,8]]A = [[9, 4], [2, 8]]
  • b=13,10Tb = {13, 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=(9)(8)−(4)(2)=64det A = (9)(8) - (4)(2) = 64

  3. Inspection

    x=1,1Tsatisfiesbothrows:9(1)+4(1)=13✓and2(1)+8(1)=10✓x = {1, 1}ᵀ satisfies both rows: 9(1) + 4(1) = 13 ✓ and 2(1) + 8(1) = 10 ✓

  4. Uniqueness

    detA=64≠0,so1,1Tistheonlysolutiondet A = 64 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=64;x=1,1Tdet A = 64; x = {1, 1}ᵀ

Why the other options are there

  • det = 80 (terms added)
  • No unique solution (determinant treated as zero)

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

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

For A = [[8, 1], [6, 7]], compute det A and solve A·x = {9, 13}ᵀ.

Given

  • A=[[8,1],[6,7]]A = [[8, 1], [6, 7]]
  • b=9,13Tb = {9, 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=(8)(7)−(1)(6)=50det A = (8)(7) - (1)(6) = 50

  3. Inspection

    x=1,1Tsatisfiesbothrows:8(1)+1(1)=9✓and6(1)+7(1)=13✓x = {1, 1}ᵀ satisfies both rows: 8(1) + 1(1) = 9 ✓ and 6(1) + 7(1) = 13 ✓

  4. Uniqueness

    detA=50≠0,so1,1Tistheonlysolutiondet A = 50 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=50;x=1,1Tdet A = 50; x = {1, 1}ᵀ

Why the other options are there

  • det = 62 (terms added)
  • No unique solution (determinant treated as zero)

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

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

For A = [[2, 7], [4, 4]], compute det A and solve A·x = {9, 8}ᵀ.

Given

  • A=[[2,7],[4,4]]A = [[2, 7], [4, 4]]
  • b=9,8Tb = {9, 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=(2)(4)−(7)(4)=−20det A = (2)(4) - (7)(4) = -20

  3. Inspection

    x=1,1Tsatisfiesbothrows:2(1)+7(1)=9✓and4(1)+4(1)=8✓x = {1, 1}ᵀ satisfies both rows: 2(1) + 7(1) = 9 ✓ and 4(1) + 4(1) = 8 ✓

  4. Uniqueness

    detA=−20≠0,so1,1Tistheonlysolutiondet A = -20 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=−20;x=1,1Tdet A = -20; x = {1, 1}ᵀ

Why the other options are there

  • det = 36 (terms added)
  • No unique solution (determinant treated as zero)

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

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

For A = [[3, 2], [3, 4]], compute det A and solve A·x = {5, 7}ᵀ.

Given

  • A=[[3,2],[3,4]]A = [[3, 2], [3, 4]]
  • 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)(4)−(2)(3)=6det A = (3)(4) - (2)(3) = 6

  3. Inspection

    x=1,1Tsatisfiesbothrows:3(1)+2(1)=5✓and3(1)+4(1)=7✓x = {1, 1}ᵀ satisfies both rows: 3(1) + 2(1) = 5 ✓ and 3(1) + 4(1) = 7 ✓

  4. Uniqueness

    detA=6≠0,so1,1Tistheonlysolutiondet A = 6 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=6;x=1,1Tdet A = 6; x = {1, 1}ᵀ

Why the other options are there

  • det = 18 (terms added)
  • No unique solution (determinant treated as zero)

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

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