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Determinants, Matrices and Solving Linear Equations

Mathematics

Determinants, Matrices and Solving Linear Equations

Overview

Overview

  • Evaluate 2×2 and 3×3 determinants and use their properties.
  • Add, multiply and transpose matrices, and find the inverse of a non-singular matrix.
  • Solve systems of linear equations by the inverse-matrix method and by Cramer's rule.

🌳 Concept tree

  • Determinants, Matrices and Linear Equations
  • ├── Determinants
  • │ ├── 2×2 rule ad − bc
  • │ ├── 3×3 expansion by minors
  • │ └── Properties of determinants
  • ├── Matrix operations
  • │ ├── Addition and scalar multiplication
  • │ ├── Matrix multiplication and transposition
  • │ └── Identity matrix
  • ├── Inverse of a matrix
  • │ ├── Condition det A ≠ 0
  • │ └── 2×2 formula with the adjugate
  • └── Solving AX = B
  • ├── X = A−1B
  • └── Cramer's rule
Determinants, Matrices and Solving Linear Equations · 1 / 7
Determinants

Determinants

The 2×2 determinant

  • For A = [[a, b], [c, d]], det A = |A| = ad − bc.
  • Example det [[3, 5], [2, 4]] = (3)(4) − (5)(2) = 12 − 10 = 2.
  • The determinant is a number, never a matrix.
Determinants, Matrices and Solving Linear: a parallelogram slab whose area is the determinant of a matrix: 3 x 2 – 1 x 1 = 5.
Determinants, Matrices and Solving Linear: a parallelogram slab whose area is the determinant of a matrix: 3 x 2 – 1 x 1 = 5.

The 3×3 determinant

Comparison aspectMain ideaKeywords
•Expand along any row or column using cofactors with alternating signs (+ − +)Expandalongcolumnusing
Expanding along the first row|A| = a(ei − fh) − b(di − fg) + c(dh − eg)
Example
  • A = [[1
  • 0]] → 1(0 − 24) − 2(0 − 20) + 3(0 − 5) = −24 + 40 − 15 = 1
Practical tipexpand along the row or column containing the most zeros — it shortens the work without changing the answerexpandalongcolumncontaining
Property Statement
Zero row or column
  • If a whole row or column is zero
  • the determinant is 0
Identical rows Two identical rows (or columns) → determinant 0
Row swap Swapping two rows changes the sign of the determinant
Scalar multiple Multiplying one row by k multiplies the determinant by k
Product rule |AB| = |A|·|B|
Transpose |Aᵀ| = |A|
Triangular matrix Determinant equals the product of the diagonal entries
Determinants, Matrices and Solving Linear Equations · 2 / 7
Matrix operations

Matrix operations

Addition and scalar multiplication

Matrices of the same order are added entry by entry(A + B)ij = aij + bij.
Scalar multiplication multiplies every entry(kA)ij = k·aij.
Different orders cannot be addeda 2×3 plus a 3×2 is undefined.
  • A (m×n) can multiply B (n×p) only when the inner dimensions agree; the product is m×p.
  • Entry (i, j) of AB = row i of A · column j of B.
  • Example [[1, 2], [3, 4]] × [[2, 0], [1, 5]] = [[1·2 + 2·1, 1·0 + 2·5], [3·2 + 4·1, 3·0 + 4·5]] = [[4, 10], [10, 20]].
  • Matrix multiplication is not commutative AB ≠ BA in general.
  • Aᵀ is formed by turning rows into columns; (AB)ᵀ = BᵀAᵀ.
  • The identity matrix I has 1s on the diagonal and 0s elsewhere, and AI = IA = A.
Determinants, Matrices and Solving Linear Equations · 3 / 7
Inverse of a matrix

Inverse of a matrix

When an inverse exists

  • A square matrix A has an inverse A−1 only if det A ≠ 0; such a matrix is called non-singular.
  • If det A = 0 the matrix is singular and has no inverse.
  • Definition AA−1 = A−1A = I.
Comparison aspectMain ideaKeywords
For A = [[a, b], [c, d]] with det A ≠ 0
  • A−1 = (1/det A) × [[d
  • −b]
  • [−c
  • a]]
•
  • The matrix [[d
  • −b]
  • [−c
  • a]] is the adjugate (transpose of the cofactor matrix)
matrixadjugatetransposecofactor
Example
  • A = [[3
  • 4]]
  • det A = 2 → A−1 = ½ [[4
  • −5]
  • [−2
  • 3]] = [[2
  • −2.5]
  • [−1
  • 1.5]]
Check
  • A−1 = [[3·2 + 5·(−1)
  • 3·(−2.5) + 5·1.5]
  • [2·2 + 4·(−1)
  • 2·(−2.5) + 4·1.5]] = [[1
  • 1]] = I
  • (A−1)−1 = A  ·  (AB)−1 = B−1A−1  ·  (Aᵀ)−1 = (A−1)ᵀ  ·  det(A−1) = 1/det A.
Determinants, Matrices and Solving Linear Equations · 4 / 7
Solving systems of linear equations

Solving systems of linear equations

The inverse-matrix method

  • Write the system as AX = B, then X = A−1B, provided det A ≠ 0.
  • Example 2x + 3y = 7 and x − 2y = −7.A = [[2, 3], [1, −2]], det A = −4 − 3 = −7, A−1 = (−1/7)[[−2, −3], [−1, 2]].
  • X = A−1B = (−1/7)[[−2·7 + (−3)(−7)], [−1·7 + 2(−7)]] = (−1/7)[[7], [−21]] = [[−1], [3]], so x = −1 and y = 3.
  • For AX = B with det A ≠ 0 x = Dx/D, y = Dy/D, z = Dz/D, where D = det A and Dx is D with the x-column replaced by B.
  • Same system D = −7, Dx = |[[7, 3], [−7, −2]]| = −14 + 21 = 7 → x = 7/(−7) = −1.
  • Dy = |[[2, 7], [1, −7]]| = −14 − 7 = −21 → y = (−21)/(−7) = 3.
  • Both methods give the same answer; Cramer's rule is faster for a single unknown, the inverse method is better when several right-hand sides share the same A.
det A System AX = B
≠ 0 Unique solution
= 0 and consistent Infinitely many solutions
= 0 and inconsistent No solution
Determinants, Matrices and Solving Linear Equations · 5 / 7
Practice with full answer key

Practice with full answer key

  1. det [[5, 2], [3, 1]] → 5 − 6 = −1
  2. det [[1, 0, 2], [3, 1, 0], [0, 4, 1]] → 1(1 − 0) − 0 + 2(12 − 0) = 25
  3. For which k is [[k, 4], [2, k]] singular? → k2 − 8 = 0 → k = ±2√2
  4. A = [[1, 2], [3, 4]]; find A−1 → det = −2 → [[−2, 1], [1.5, −0.5]]
  5. If |A| = 3 and |B| = 5, find |AB| and |2A| for 2×2 matrices → 15 and 12
  6. Solve 3x + y = 9, x − y = −1 by Cramer → D = −4, Dx = −8, Dy = −12 → x = 2, y = 3
  7. Solve x + 2y + z = 4, 2x − y + 3z = 9, 3x + y − z = 2 → x = 1, y = 1, z = 1
  8. If AB = I, what is B in terms of A? → B = A−1
Determinants, Matrices and Solving Linear Equations · 6 / 7
Exam traps and key takeaways

Exam traps and key takeaways

💡 High-yield traps

  • The adjugate swaps the diagonal entries and changes the sign of the off-diagonal ones: [[d, −b], [−c, a]], not [[a, −c], [−b, d]].
  • In Cramer's rule, the column of the unknown being found is the one replaced — replacing the wrong column is the classic error.
  • |kA| = kn|A| for an n×n matrix, so for a 2×2 matrix |2A| = 4|A|, not 2|A|.
  • AB ≠ BA; never reorder a product.
  • det A = 0 → no inverse; writing A−1 anyway is an automatic zero on the item.
  • Three unknowns need all three of D, Dx, Dy, Dz; a missing D gives an unverifiable answer.
Determinants, Matrices and Solving Linear Equations · 7 / 7