Chapter 7: Matrices
Determinants, adjoint and inverse of a matrix, matrix identities, inverse-matrix method and Cramer’s rule — reconstructed from the supplied 61-page handwritten notes.
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Basic Concepts and Formulae
Determinant of a 2 × 2 Matrix
Adjoint of a 2 × 2 Matrix
For a 2 × 2 matrix, interchange the elements of the main diagonal and change the signs of the other two elements.
Identity Matrix
Singular and Non-Singular Matrix
Singular matrix: a matrix whose determinant is 0.
Non-singular matrix: a matrix whose determinant is not 0.
Inverse of a Matrix
A square matrix A has an inverse only if it is non-singular. Its inverse A−1 satisfies:
The formula used throughout the source is:
Worked Example
For A = [1 2; 3 4]:
|A| = 1×4 − 2×3 = −2
adj A = [4 −2; −3 1]
A−1 = (1/−2)[4 −2; −3 1]
∴ A−1 = [−2 1; 3/2 −1/2]
Exercise 7.1 – Determinants, Adjoint and Inverse
Finding Determinants
(a) A = [3 2; 4 5]
|A| = 3×5 − 4×2 = 15−8
∴ |A| = 7
(b) B = [−2 3; 4 5]
|B| = (−2)×5 − 4×3
∴ |B| = −22
(c) C = [2 5; −3 6]
|C| = 2×6 − (−3×5)
∴ |C| = 27
(d) D = [3 5; 0 6]
|D| = 3×6 − 0×5
∴ |D| = 18
(e) E = [sin θ cos θ; −cos θ sin θ]
|E| = sin²θ + cos²θ
∴ |E| = 1
Additional determinant examples
|3 2; 4 −5| = −23
|2 4; 1 2| = 0
|0 4; 5 2| = −20
|a+b 2b; −b a−b| = a²+b²
Solving Unknowns Using Determinants
(a) |2n 5; 1 3| = 7
6n − 5 = 7
6n = 12
∴ n = 2
(b) |4 n; −3 5| = 8
20 + 3n = 8
3n = −12
∴ n = −4
|2 3; n 4| = 0
8 − 3n = 0
∴ n = 8/3
Another determinant equation
Page 9 sets a determinant involving n equal to zero and simplifies it to a linear equation.
The readable source result is n = 8.
Question 7
Determinant Properties and Matrix Expressions
Verify |AB| = |A||B|
For A = [3 1; 4 0] and B = [4 0; 2 5]:
|A| = −4
|B| = 20
|A||B| = −80
AB = [14 5; 16 0]
|AB| = −80
∴ |AB| = |A||B|
The source verifies the same property again on pages 12–13 using another pair of matrices.
Matrix-expression determinants
Using matrices A, B and C given on pages 13–16, the handwritten source evaluates:
|A+B| = 16
|A−C| = 131
|BA| = −104
|3A−2B| = 254
|2CB| = −576
|B+C−A| = 147
|A²−5A+2I| = −299 (source result)
|2A−3B+5I| = 234 (source result)
Finding Inverse Matrices
A = [9 5; 7 4]
|A| = 9×4 − 7×5 = 1
adj A = [4 −5; −7 9]
∴ A−1 = [4 −5; −7 9]
A = [0 −1; 4 4]
|A| = 4
adj A = [4 1; −4 0]
∴ A−1 = [1 1/4; −1 0]
Pages 18–22 continue with inverses of additional matrices and inverse expressions such as D−1, (B+C)−1 and (BD+AC)−1.
Inverse Matrix Properties
Verify AA−1 = A−1A = I
For the matrix used on pages 22–24, the source first obtains its inverse and then multiplies in both orders.
AA−1 = I
A−1A = I
Verify (AB)−1 = B−1A−1
Pages 24–27 calculate A−1, B−1, AB and (AB)−1 and show that both sides are equal.
Identity-matrix relation
Pages 27–29 contain a proof involving a matrix A, its inverse and the identity matrix I.
Exercise 7.2 – Solving Simultaneous Equations by Matrix Methods
Inverse Square Matrix Method
For a system AX=B:
Example: x + 2y = 4, 3x + 4y = 7
A = [1 2; 3 4], X=[x;y], B=[4;7]
|A| = −2
A−1 = (1/−2)[4 −2; −3 1]
X=A−1B
∴ x=−1, y=5/2
[4 −3; 3 7] [x;y] = [11;−1]
D = 37
The source obtains:
∴ x=2, y=−1
[5 2; 3 2] [x;y] = [3;5]
|A|=4
The source obtains:
∴ x=−1, y=4
AX=B, A=[1 −2; −3 5], B=[4;−7]
|A| = −1
A−1 = [−5 −2; −3 −1]
∴ x=−6, y=−5
Long-Type Questions – Inverse Matrix Method
5x + 2y = 4, 7x + 3y = 5
The coefficient matrix has determinant 1.
∴ x=2, y=−3
3x + 4y = −1, 2x + 5y = 4
∴ x=−3, y=2
x + 2y = 1, 3x + y = 4
∴ x=7/5, y=−1/5
3x + 4y = 7, x + y = 3
∴ x=5, y=−2
(3/2)x + 2y = 1, (1/3)x − (1/3)y = 1
∴ x=2, y=−1
x/6 − y/3 = 4, x/12 − 2y/3 = 4
Pages 49–53 continue with additional systems. Each is rewritten as AX=B, A−1 is found, and X=A−1B is evaluated.
Exercise 7.3 – Cramer’s Rule
Solving Simultaneous Equations Using Cramer’s Rule
For:
a₂x+b₂y=c₂
Use:
3x + 2y = −15, 2x − y = −3
D = −7
D₁ = 21
D₂ = 21
∴ x=−3, y=−3
2x + 3y = 5, 3x − 2y = 1
D = −13
D₁ = −13
D₂ = −13
∴ x=1, y=1
Pages 56–57 contain further fractional simultaneous equations solved with the same determinant procedure.
Cramer’s Rule after Substitution
8/x + 9/y = 7, 2/x + 3/y = 2
Let:
8u + 9v = 7
2u + 3v = 2
D = 6
D₁ = 3 ⇒ u=1/2
D₂ = 2 ⇒ v=1/3
Therefore:
x=2, y=3
40/(x+y) + 2/(x−y) = 5
25/(x+y) − 3/(x−y) = 1
Let:
40a + 2b = 5
25a − 3b = 1
The source obtains:
a = 1/10, b = 1/2
Therefore:
x+y = 10
x−y = 2
Adding: 2x=12 ⇒ x=6
Then y=4.
∴ x=6, y=4
Discussion
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