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Matrix Transpose

Mathematics · FE Reference Handbook section

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

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • Rows become columns. Columns become rows.

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
Two-equation truss reaction solve

Equilibrium gives 2A + 3B = 34 and 4A − B = 12. Solve for A and B using determinants.

Given

  • 2A+3B=342A + 3B = 34
  • 4A−B=124A - B = 12

Find

A and B

Start with the thinking

  • Cramer's rule is fastest for a 2×2 system on the exam.
  • Check by substituting into the equation you did not use last.

Step-by-step solution

  1. System determinant

    D=(2)(−1)−(3)(4)=−14.0D = (2)(-1) - (3)(4) = -14.0

  2. Numerator for A

    DA=(34)(−1)−(3)(12)=−70.0D_A = (34)(-1) - (3)(12) = -70.0

  3. Solve

    A=DA/D=−70.0/−14.0=5.00A = D_A/D = -70.0/-14.0 = 5.00

  4. Back-substitute

    4(5.00)−B=12,soB=8.004(5.00) - B = 12, so B = 8.00

  5. Check

    2(5)+3(8)=34✓2(5) + 3(8) = 34 ✓
Answer:
A=5.00,B=8.00A = 5.00, B = 8.00

Why the other options are there

  • A = 8, B = 5 (variables swapped)
  • A = −5 (determinant sign dropped)

Reference: FE Reference Handbook — Mathematics — Linear algebra

Example 2
Determinant and inverse of a 2×2 system — Matrix Transpose

For A = [[2, 5], [1, 3]], compute det A and solve A·x = {7, 4}ᵀ.

Given

  • A=[[2,5],[1,3]]A = [[2, 5], [1, 3]]
  • b=7,4Tb = {7, 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=(2)(3)−(5)(1)=1det A = (2)(3) - (5)(1) = 1

  3. Inspection

    x=1,1Tsatisfiesbothrows:2(1)+5(1)=7✓and1(1)+3(1)=4✓x = {1, 1}ᵀ satisfies both rows: 2(1) + 5(1) = 7 ✓ and 1(1) + 3(1) = 4 ✓

  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 = 11 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrix Transpose

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

For A = [[5, 6], [4, 4]], compute det A and solve A·x = {11, 8}ᵀ.

Given

  • A=[[5,6],[4,4]]A = [[5, 6], [4, 4]]
  • b=11,8Tb = {11, 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=(5)(4)−(6)(4)=−4det A = (5)(4) - (6)(4) = -4

  3. Inspection

    x=1,1Tsatisfiesbothrows:5(1)+6(1)=11✓and4(1)+4(1)=8✓x = {1, 1}ᵀ satisfies both rows: 5(1) + 6(1) = 11 ✓ and 4(1) + 4(1) = 8 ✓

  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 = 44 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrix Transpose

Example 4
Determinant and inverse of a 2×2 system — Matrix Transpose (3)

For A = [[7, 5], [2, 2]], compute det A and solve A·x = {12, 4}ᵀ.

Given

  • A=[[7,5],[2,2]]A = [[7, 5], [2, 2]]
  • b=12,4Tb = {12, 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=(7)(2)−(5)(2)=4det A = (7)(2) - (5)(2) = 4

  3. Inspection

    x=1,1Tsatisfiesbothrows:7(1)+5(1)=12✓and2(1)+2(1)=4✓x = {1, 1}ᵀ satisfies both rows: 7(1) + 5(1) = 12 ✓ 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 = 24 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrix Transpose

Example 5
Determinant and inverse of a 2×2 system — Matrix Transpose (4)

For A = [[7, 1], [2, 3]], compute det A and solve A·x = {8, 5}ᵀ.

Given

  • A=[[7,1],[2,3]]A = [[7, 1], [2, 3]]
  • b=8,5Tb = {8, 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)(3)−(1)(2)=19det A = (7)(3) - (1)(2) = 19

  3. Inspection

    x=1,1Tsatisfiesbothrows:7(1)+1(1)=8✓and2(1)+3(1)=5✓x = {1, 1}ᵀ satisfies both rows: 7(1) + 1(1) = 8 ✓ and 2(1) + 3(1) = 5 ✓

  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 = 23 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrix Transpose

Example 6
Determinant and inverse of a 2×2 system — Matrix Transpose (5)

For A = [[9, 8], [6, 4]], compute det A and solve A·x = {17, 10}ᵀ.

Given

  • A=[[9,8],[6,4]]A = [[9, 8], [6, 4]]
  • b=17,10Tb = {17, 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)(4)−(8)(6)=−12det A = (9)(4) - (8)(6) = -12

  3. Inspection

    x=1,1Tsatisfiesbothrows:9(1)+8(1)=17✓and6(1)+4(1)=10✓x = {1, 1}ᵀ satisfies both rows: 9(1) + 8(1) = 17 ✓ and 6(1) + 4(1) = 10 ✓

  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 = 84 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrix Transpose

Example 7
Determinant and inverse of a 2×2 system — Matrix Transpose (6)

For A = [[3, 1], [2, 8]], compute det A and solve A·x = {4, 10}ᵀ.

Given

  • A=[[3,1],[2,8]]A = [[3, 1], [2, 8]]
  • b=4,10Tb = {4, 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)(8)−(1)(2)=22det A = (3)(8) - (1)(2) = 22

  3. Inspection

    x=1,1Tsatisfiesbothrows:3(1)+1(1)=4✓and2(1)+8(1)=10✓x = {1, 1}ᵀ satisfies both rows: 3(1) + 1(1) = 4 ✓ and 2(1) + 8(1) = 10 ✓

  4. Uniqueness

    detA=22≠0,so1,1Tistheonlysolutiondet A = 22 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=22;x=1,1Tdet A = 22; 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 → Matrix Transpose

Example 8
Determinant and inverse of a 2×2 system — Matrix Transpose (7)

For A = [[9, 2], [6, 2]], compute det A and solve A·x = {11, 8}ᵀ.

Given

  • A=[[9,2],[6,2]]A = [[9, 2], [6, 2]]
  • b=11,8Tb = {11, 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=(9)(2)−(2)(6)=6det A = (9)(2) - (2)(6) = 6

  3. Inspection

    x=1,1Tsatisfiesbothrows:9(1)+2(1)=11✓and6(1)+2(1)=8✓x = {1, 1}ᵀ satisfies both rows: 9(1) + 2(1) = 11 ✓ and 6(1) + 2(1) = 8 ✓

  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 = 30 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrix Transpose

Example 9
Determinant and inverse of a 2×2 system — Matrix Transpose (8)

For A = [[7, 5], [2, 9]], compute det A and solve A·x = {12, 11}ᵀ.

Given

  • A=[[7,5],[2,9]]A = [[7, 5], [2, 9]]
  • b=12,11Tb = {12, 11}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)(9)−(5)(2)=53det A = (7)(9) - (5)(2) = 53

  3. Inspection

    x=1,1Tsatisfiesbothrows:7(1)+5(1)=12✓and2(1)+9(1)=11✓x = {1, 1}ᵀ satisfies both rows: 7(1) + 5(1) = 12 ✓ and 2(1) + 9(1) = 11 ✓

  4. Uniqueness

    detA=53≠0,so1,1Tistheonlysolutiondet A = 53 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=53;x=1,1Tdet A = 53; x = {1, 1}ᵀ

Why the other options are there

  • det = 73 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrix Transpose

Example 10
Determinant and inverse of a 2×2 system — Matrix Transpose (9)

For A = [[7, 4], [4, 2]], compute det A and solve A·x = {11, 6}ᵀ.

Given

  • A=[[7,4],[4,2]]A = [[7, 4], [4, 2]]
  • b=11,6Tb = {11, 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=(7)(2)−(4)(4)=−2det A = (7)(2) - (4)(4) = -2

  3. Inspection

    x=1,1Tsatisfiesbothrows:7(1)+4(1)=11✓and4(1)+2(1)=6✓x = {1, 1}ᵀ satisfies both rows: 7(1) + 4(1) = 11 ✓ and 4(1) + 2(1) = 6 ✓

  4. Uniqueness

    detA=−2≠0,so1,1Tistheonlysolutiondet A = -2 \ne 0, so {1, 1}ᵀ is the only solution
Answer:
detA=−2;x=1,1Tdet A = -2; x = {1, 1}ᵀ

Why the other options are there

  • det = 30 (terms added)
  • No unique solution (determinant treated as zero)

Reference: FE Reference Handbook — Mathematics → Matrix Transpose

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