Matrices | Criticalyx Mathematics

🎯 Learning Objectives🎯 Objektif Pembelajaran

Understand matrix operations: addition, scalar multiplication, matrix multiplicationFahami operasi matriks: penambahan, pendaraban skalar, pendaraban matriks

Find the inverse of a 2×2 matrix and understand when a matrix is singularCari songsangan matriks 2×2 dan fahami apabila matriks adalah singular

Solve simultaneous equations using matricesSelesaikan persamaan serentak menggunakan matriks

Verify matrix inverse properties (M × M⁻¹ = I)Sahihkan sifat songsangan matriks (M × M⁻¹ = I)

📖 Key Concepts📖 Konsep Utama

Matrix Addition & Scalar Multiplication (Form 4)Penambahan & Pendaraban Skalar Matriks (Tingkatan 4)

Same-order matrices add element by element. Scalar multiplies each element.Matriks susunan sama ditambah unsur demi unsur. Skalar mendarab setiap unsur.

A + B = [a_ij + b_ij]
kA = [k × a_ij]
Only same-order matrices can be added!

Matrix Multiplication (Form 4)Pendaraban Matriks (Tingkatan 4)

Row of first × column of second. Not commutative (AB ≠ BA)!Baris pertama × lajur kedua. Tidak komutatif (AB ≠ BA)!

🟣 Row 1 × 🔴 Col 2 = (1×0 + 2×5) = 10🟣 Baris 1 × 🔴 Lajur 2 = (1×0 + 2×5) = 10

AB = [∑(a_ik × b_kj)]
Columns of A must = Rows of B
Order of (m×n)(n×p) = m×p

Matrix Inverse (Form 5)Songsangan Matriks (Tingkatan 5)

Inverse exists if determinant ≠ 0. Swap a,d, negate b,c, divide by determinant.Songsangan wujud jika determinant ≠ 0. Tukar a,d, nafi b,c, bahagi dengan determinant.

M⁻¹ = 1/(ad−bc) × [d −b; −c a]
Singular if ad−bc = 0
Identity: I = [1 0; 0 1]

Solving Simultaneous Equations (Form 5)Penyelesaian Persamaan Serentak (Tingkatan 5)

Write as AX = B, then X = A⁻¹B.Tulis sebagai AX = B, kemudian X = A⁻¹B.

AX = B → X = A⁻¹B
Must check determinant ≠ 0 first!
If singular, no unique solution

✏️ Worked Examples✏️ Contoh Penyelesaian

Find A + B where A = [3 1; 2 4], B = [1 5; 3 2]Cari A + B di mana A = [3 1; 2 4], B = [1 5; 3 2]

A + B = [3+1, 1+5; 2+3, 4+2]

A + B = [4 6; 5 6]

Multiply [1 2; 0 3] by [4 0; 1 5]Darab [1 2; 0 3] dengan [4 0; 1 5]

Row 1 × Col 1: 1(4) + 2(1) = 6

Row 1 × Col 2: 1(0) + 2(5) = 10

Row 2 × Col 1: 0(4) + 3(1) = 3

Row 2 × Col 2: 0(0) + 3(5) = 15

AB = [6 10; 3 15]

Find the inverse of M = [3 2; 1 4]Cari songsangan M = [3 2; 1 4]

det = (3)(4) − (2)(1) = 12 − 2 = 10

Swap diagonal: 3↔4, 4↔3

Negate off-diagonal: 2↔−2, 1↔−1

M⁻¹ = ¹⁄₁₀ [4 −2; −1 3] = [²⁄₅ −¹⁄₅; −¹⁄₁₀ ³⁄₁₀]

M⁻¹ = [²⁄₅ −¹⁄₅; −¹⁄₁₀ ³⁄₁₀]

Solve using matrices: 2x + y = 7, x + 3y = 11Selesaikan menggunakan matriks: 2x + y = 7, x + 3y = 11

Write as AX = B: [2 1; 1 3][x; y] = [7; 11]

det = 2(3) − 1(1) = 5

A⁻¹ = ¹⁄₅ [3 −1; −1 2]

X = ¹⁄₅ [3 −1; −1 2][7; 11] = ¹⁄₅ [10; 15] = [2; 3]

x = 2, y = 3

⚠️ Common Mistakes⚠️ Kesilapan Biasa

❌ Wrong multiplication order — not commutative! AB ≠ BASusunan darab salah — tidak komutatif! AB ≠ BA

Always multiply in the correct order. Matrix multiplication is NOT like number multiplication.Sentiasa darab dalam susunan yang betul. Pendaraban matriks BUKAN seperti pendaraban nombah.

❌ Calculating determinant incorrectly — it's ad−bc, not ac−bdMengira determinant dengan salah — ia ad−bc, bukan ac−bd

For [a b; c d]: determinant = ad − bc. Cross-multiply top-left×bottom-right MINUS top-right×bottom-left.Bagi [a b; c d]: determinant = ad − bc. Darab silang kiri-atas×kanan-bawah TOLAK kanan-atas×kiri-bawah.

❌ Forgetting to check if inverse existsLupa menyemak jika songsangan wujud

If determinant = 0, the matrix is singular and has NO inverse. Always check first!Jika determinant = 0, matriks adalah singular dan TIADA songsangan. Sentiasa semak dahulu!

📝 SPM Practice📝 Amalan SPM

Paper 1 — Multiple ChoiceKertas 1 — Pilihan Berganda

Find the determinant of [3 2; 1 4]Cari determinant bagi [3 2; 1 4]

Which matrix has no inverse?Matriks manakah yang tiada songsangan?

[1 2; 3 4]

[2 4; 1 2]

[3 1; 2 1]

[4 2; 2 1]

Multiply [1 2; 0 3] by [4 0; 1 5]Darab [1 2; 0 3] dengan [4 0; 1 5]

[6 10; 3 15]

[4 10; 0 15]

[6 0; 3 5]

[4 2; 1 3]

Solve 2x + y = 7, x + 3y = 11 using matricesSelesaikan 2x + y = 7, x + 3y = 11 menggunakan matriks

x=4, y=1

x=2, y=3

x=5, y=2

x=3, y=1

What is the inverse of [1 0; 0 1]?Apakah songsangan bagi [1 0; 0 1]?

Itself

Zero matrix

Does not exist

Negative matrix

Paper 2 — Structured QuestionsKertas 2 — Soalan Berstruktur

5 marks5 markah

Given A = [3 1; 2 4] and B = [1 5; 3 2].
(a) Find A + B. [2 marks]
(b) Find 2A − B. [3 marks]Diberi A = [3 1; 2 4] dan B = [1 5; 3 2].
(a) Cari A + B. [2 markah]
(b) Cari 2A − B. [3 markah]

Solution (a)Penyelesaian (a)

A + B = [3+1, 1+5; 2+3, 4+2]

A + B = [4 6; 5 6]

A + B = [4 6; 5 6] [2 marks]

Solution (b)Penyelesaian (b)

2A = [6 2; 4 8]

2A − B = [6−1, 2−5; 4−3, 8−2]

2A − B = [5 −3; 1 6]

2A − B = [5 −3; 1 6] [3 marks]

6 marks6 markah

Given M = [4 3; 2 1].
(a) Find M⁻¹. [3 marks]
(b) Verify that M × M⁻¹ = I. [3 marks]Diberi M = [4 3; 2 1].
(a) Cari M⁻¹. [3 markah]
(b) Sahihkan bahawa M × M⁻¹ = I. [3 markah]

Solution (a)Penyelesaian (a)

det = 4(1) − 3(2) = 4 − 6 = −2

M⁻¹ = 1/(−2) [1 −3; −2 4]

M⁻¹ = [−½ ³⁄₂; 1 −2]

M⁻¹ = [−½ ³⁄₂; 1 −2] [3 marks]

Solution (b)Penyelesaian (b)

M × M⁻¹ = [4 3; 2 1][−½ ³⁄₂; 1 −2]

Row 1: 4(−½)+3(1) = −2+3 = 1, 4(³⁄₂)+3(−2) = 6−6 = 0

Row 2: 2(−½)+1(1) = −1+1 = 0, 2(³⁄₂)+1(−2) = 3−2 = 1

Result = [1 0; 0 1] = I ✓

M × M⁻¹ = [1 0; 0 1] = I ✓ [3 marks]

📋 Formula Summary📋 Ringkasan Formula

Matrices:Matriks:

Addition: A + B = [a_ij + b_ij]
Scalar: kA = [k × a_ij]
Multiplication: AB = [∑(a_ik × b_kj)]
Inverse: M⁻¹ = 1/(ad−bc) × [d −b; −c a]
Singular if ad−bc = 0
Solving: AX = B → X = A⁻¹B

🔲 Matrices🔲 Matriks