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<!DOCTYPE html>
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<title>Math 202 — Complete Solutions</title>
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</head>
<body>
<header>
<h1>Math 202 — Complete Solutions</h1>
<p>Sheets 1 · 2 · 3 | Linear Algebra</p>
</header>
<nav>
<a href="#sheet1" class="active">Sheet 1 · Matrices & Operations</a>
<a href="#sheet2">Sheet 2 · Equations & Determinants</a>
<a href="#sheet3">Sheet 3 · Inverses & Orthogonality</a>
</nav>
<main>
<!-- ═══════════════════ SHEET 1 ═══════════════════ -->
<h2 class="sheet-title" id="sheet1">Sheet 1 — Matrix Formation & Operations</h2>
<!-- Q1 -->
<div class="problem">
<div class="problem-header"><span class="num">1</span> Form A = (aᵢⱼ) of order 3×2 where aᵢⱼ = i − j + 2, then C = Aᵗ</div>
<div class="problem-body">
<ul class="steps">
<li>Indices: i ∈ {1,2,3}, j ∈ {1,2}. Fill each entry using aᵢⱼ = i − j + 2:
<span class="math">a₁₁=1−1+2=2 a₁₂=1−2+2=1
a₂₁=2−1+2=3 a₂₂=2−2+2=2
a₃₁=3−1+2=4 a₃₂=3−2+2=3</span>
</li>
<li>Write matrix A (3×2) and its transpose C (2×3):
<span class="math"> ⎡2 1⎤ ⎡2 3 4⎤
A = ⎢3 2⎥ → C = Aᵗ = ⎣1 2 3⎦
⎣4 3⎦</span>
</li>
</ul>
<span class="answer">A is 3×2 ; C = Aᵗ is 2×3</span>
</div>
</div>
<!-- Q2 -->
<div class="problem">
<div class="problem-header"><span class="num">2</span> A = (aₓᵧ) for x,y ∈ {1,2,3} with aₓᵧ = y − x</div>
<div class="problem-body">
<ul class="steps">
<li>Fill the 3×3 grid (row = x, col = y):
<span class="math">a₁₁=0 a₁₂=1 a₁₃=2
a₂₁=−1 a₂₂=0 a₂₃=1
a₃₁=−2 a₃₂=−1 a₃₃=0</span>
</li>
<li>Write A and its transpose:
<span class="math"> ⎡ 0 1 2⎤ ⎡ 0 −1 −2⎤
A = ⎢−1 0 1⎥ Aᵗ = ⎢ 1 0 −1⎥
⎣−2 −1 0⎦ ⎣ 2 1 0⎦</span>
</li>
</ul>
<div class="note">Notice: aₓᵧ = y−x = −(x−y) = −aᵧₓ, so A is a <strong>skew-symmetric</strong> matrix (Aᵗ = −A).</div>
</div>
</div>
<!-- Q3 -->
<div class="problem">
<div class="problem-header"><span class="num">3</span> Matrix computations (A,B,C,D,E,F as given)</div>
<div class="problem-body">
<div class="note">Given sizes: A(2×3), B(3×2), C(3×3), D(2×2), E(3×3), F(2×2)</div>
<div class="part">
<div class="part-label">a) C + E</div>
Both are 3×3 — add entry-by-entry:
<span class="math">C + E = ⎡3+2 −1+(−4) 3+5⎤ = ⎡5 −5 8⎤
⎢4+0 1+1 5+4⎥ ⎢4 2 9⎥
⎣2+3 1+2 3+1⎦ ⎣5 3 4⎦</span>
</div>
<div class="part">
<div class="part-label">b) AB, BA, FD, DF — and what we deduce</div>
<ul class="steps">
<li>AB: A(2×3)×B(3×2) → 2×2
<span class="math">AB = ⎡1·1+2·2+3·3 1·0+2·1+3·2⎤ = ⎡14 8⎤
⎣2·1+1·2+4·3 2·0+1·1+4·2⎦ ⎣16 9⎦</span>
</li>
<li>BA: B(3×2)×A(2×3) → 3×3
<span class="math">BA = ⎡1 2 3⎤
⎢4 5 10⎥
⎣7 8 17⎦</span>
</li>
<li>FD: F(2×2)×D(2×2) → 2×2
<span class="math">FD = ⎡−4·3+5·2 −4·(−2)+5·4⎤ = ⎡−2 28⎤
⎣ 2·3+3·2 2·(−2)+3·4⎦ ⎣12 8⎦</span>
</li>
<li>DF: D(2×2)×F(2×2) → 2×2
<span class="math">DF = ⎡3·(−4)+(−2)·2 3·5+(−2)·3⎤ = ⎡−16 9⎤
⎣2·(−4)+ 4·2 2·5+ 4·3⎦ ⎣ 0 22⎦</span>
</li>
</ul>
<div class="note">Deduction: AB ≠ BA and FD ≠ DF — matrix multiplication is in general <strong>not commutative</strong>.</div>
</div>
<div class="part">
<div class="part-label">c) 2D − 3F</div>
<span class="math">2D = ⎡6 −4⎤ 3F = ⎡−12 15⎤
⎣4 8⎦ ⎣ 6 9⎦
2D − 3F = ⎡18 −19⎤
⎣−2 −1⎦</span>
</div>
<div class="part">
<div class="part-label">d) CB + D</div>
<ul class="steps">
<li>CB: C(3×3)×B(3×2) → 3×2</li>
<li>D is 2×2. Since 3×2 ≠ 2×2, we <strong>cannot add</strong> CB + D.</li>
</ul>
<div class="warn">Not defined — dimension mismatch (3×2 + 2×2 is impossible).</div>
</div>
<div class="part">
<div class="part-label">e) EF + 2A</div>
E is 3×3 and F is 2×2 → EF requires 3 = 2 (columns of E = rows of F) — impossible.
<div class="warn">Not defined — EF cannot be computed (3×3 × 2×2).</div>
</div>
<div class="part">
<div class="part-label">f) A(BD) and (AB)D — verify associativity</div>
<ul class="steps">
<li>BD: B(3×2)×D(2×2) → 3×2
<span class="math">BD = ⎡1·3+0·2 1·(−2)+0·4⎤ = ⎡ 3 −2⎤
⎢2·3+1·2 2·(−2)+1·4⎥ ⎢ 8 0⎥
⎣3·3+2·2 3·(−2)+2·4⎦ ⎣13 2⎦</span>
</li>
<li>A(BD): A(2×3)×BD(3×2) → 2×2
<span class="math">A(BD) = ⎡1·3+2·8+3·13 1·(−2)+2·0+3·2⎤ = ⎡58 4⎤
⎣2·3+1·8+4·13 2·(−2)+1·0+4·2⎦ ⎣66 4⎦</span>
</li>
<li>(AB)D: AB(2×2)×D(2×2) → 2×2
<span class="math">(AB)D = ⎡14·3+8·2 14·(−2)+8·4⎤ = ⎡58 4⎤
⎣16·3+9·2 16·(−2)+9·4⎦ ⎣66 4⎦</span>
</li>
</ul>
<div class="note">A(BD) = (AB)D — confirms matrix multiplication is <strong>associative</strong>.</div>
</div>
<div class="part">
<div class="part-label">g) A(C+E) and AC + AE — verify distributivity</div>
<ul class="steps">
<li>C + E (from part a) = [[5,−5,8],[4,2,9],[5,3,4]]</li>
<li>A(C+E):
<span class="math">A(C+E) = ⎡1·5+2·4+3·5 1·(−5)+2·2+3·3 1·8+2·9+3·4⎤ = ⎡28 8 38⎤
⎣2·5+1·4+4·5 2·(−5)+1·2+4·3 2·8+1·9+4·4⎦ ⎣34 4 41⎦</span>
</li>
<li>AC:
<span class="math">AC = ⎡17 4 22⎤
⎣18 3 23⎦</span>
</li>
<li>AE:
<span class="math">AE = ⎡11 4 16⎤
⎣16 1 18⎦</span>
</li>
<li>AC + AE = [[28,8,38],[34,4,41]] ✓</li>
</ul>
<div class="note">A(C+E) = AC + AE — confirms <strong>distributive law</strong>.</div>
</div>
<div class="part">
<div class="part-label">h) Aᵗ and (Aᵗ)ᵗ</div>
<span class="math">A = ⎡1 2 3⎤ Aᵗ = ⎡1 2⎤ (Aᵗ)ᵗ = A = ⎡1 2 3⎤
⎣2 1 4⎦ ⎢2 1⎥ ⎣2 1 4⎦
⎣3 4⎦</span>
<div class="note">Double transposing returns the original matrix: <strong>(Aᵗ)ᵗ = A</strong>.</div>
</div>
<div class="part">
<div class="part-label">i) (2C + E)ᵗ and 2Cᵗ + Eᵗ</div>
<ul class="steps">
<li>2C + E = [[8,−6,11],[8,3,14],[7,4,7]]</li>
<li>(2C+E)ᵗ = [[8,8,7],[−6,3,4],[11,14,7]]</li>
<li>2Cᵗ + Eᵗ = [[8,8,7],[−6,3,4],[11,14,7]] ✓</li>
</ul>
<div class="note">Confirms: <strong>(αC + E)ᵗ = αCᵗ + Eᵗ</strong> (transpose is linear).</div>
</div>
<div class="part">
<div class="part-label">j) (AB)ᵗ and BᵗAᵗ — verify transpose of product</div>
<ul class="steps">
<li>(AB)ᵗ = [[14,16],[8,9]]</li>
<li>Bᵗ(2×3) × Aᵗ(3×2): BᵗAᵗ = [[14,16],[8,9]] ✓</li>
</ul>
<div class="note">Confirms the rule: <strong>(AB)ᵗ = BᵗAᵗ</strong>.</div>
</div>
<div class="part">
<div class="part-label">k) (2BC − D)ᵗ</div>
B is 3×2, C is 3×3 → BC requires columns of B (2) = rows of C (3): impossible.
<div class="warn">Not defined — BC cannot be computed (dimensions 3×2 and 3×3 are incompatible for multiplication in this order).</div>
</div>
</div>
</div>
<!-- ═══════════════════ SHEET 2 ═══════════════════ -->
<h2 class="sheet-title" id="sheet2">Sheet 2 — Matrix Equations & Determinants</h2>
<!-- Q1 -->
<div class="problem">
<div class="problem-header"><span class="num">1</span> Find X such that 2Xᵗ = A² + (BC)ᵗ</div>
<div class="problem-body">
Given: A=[[13,7],[−1,8]], B=[[−3,4,5],[9,0,7]], C=[[1,0],[−1,4],[2,5]]
<ul class="steps">
<li>Compute A²:
<span class="math">A² = ⎡13·13+7·(−1) 13·7+7·8 ⎤ = ⎡162 147⎤
⎣(−1)·13+8·(−1) (−1)·7+8·8⎦ ⎣−21 57⎦</span>
</li>
<li>Compute BC (2×3 × 3×2 → 2×2):
<span class="math">BC = ⎡−3+4·(−1)+5·2 0+16+25⎤ = ⎡3 41⎤
⎣ 9+0+7·2 0+0+35 ⎦ ⎣23 35⎦</span>
</li>
<li>Transpose BC:
<span class="math">(BC)ᵗ = ⎡3 23⎤
⎣41 35⎦</span>
</li>
<li>Sum:
<span class="math">A² + (BC)ᵗ = ⎡165 170⎤
⎣ 20 92⎦</span>
</li>
<li>Solve for Xᵗ then X:
<span class="math">2Xᵗ = ⎡165 170⎤ → Xᵗ = ⎡165/2 85⎤ → X = ⎡165/2 10⎤
⎣ 20 92⎦ ⎣ 10 46⎦ ⎣ 85 46⎦</span>
</li>
</ul>
<span class="answer">X = ⎡82.5 10⎤
⎣85 46⎦</span>
</div>
</div>
<!-- Q2 -->
<div class="problem">
<div class="problem-header"><span class="num">2</span> Solve the matrix equation: 4[X + [[5,−1],[0,−3]]] = 2X + [[12,−4],[5,−6]]</div>
<div class="problem-body">
<ul class="steps">
<li>Expand the left side:
<span class="math">4X + ⎡20 −4⎤ = 2X + ⎡12 −4⎤
⎣ 0 −12⎦ ⎣ 5 −6⎦</span>
</li>
<li>Isolate X:
<span class="math">4X − 2X = ⎡12 −4⎤ − ⎡20 −4⎤ = ⎡−8 0⎤
⎣ 5 −6⎦ ⎣ 0 −12⎦ ⎣ 5 6⎦</span>
</li>
<li>Divide by 2:
<span class="math">2X = ⎡−8 0⎤ → X = ⎡−4 0 ⎤
⎣ 5 6⎦ ⎣5/2 3 ⎦</span>
</li>
</ul>
<span class="answer">X = ⎡−4 0⎤
⎣5/2 3⎦</span>
</div>
</div>
<!-- Q3 -->
<div class="problem">
<div class="problem-header"><span class="num">3</span> Given A and B, find X such that AB − |A|X + tr(B)A − diag(2,3) = 4I</div>
<div class="problem-body">
Given: A=[[1,−1],[2,4]], B=[[3,5],[−2,−12]]
<ul class="steps">
<li>Compute |A| and tr(B):
<span class="math">|A| = 1·4 − (−1)·2 = 4 + 2 = 6
tr(B) = 3 + (−12) = −9</span>
</li>
<li>Compute AB:
<span class="math">AB = ⎡1·3+(−1)(−2) 1·5+(−1)(−12)⎤ = ⎡ 5 17⎤
⎣2·3+4·(−2) 2·5+4·(−12) ⎦ ⎣−2 −38⎦</span>
</li>
<li>Compute tr(B)·A = −9·A:
<span class="math">−9A = ⎡−9 9⎤
⎣−18 −36⎦</span>
</li>
<li>Assemble right-hand side (move 6X to the other side):
<span class="math">6X = AB + tr(B)A − diag(2,3) − 4I
= ⎡5 17⎤ + ⎡−9 9⎤ − ⎡2 0⎤ − ⎡4 0⎤
⎣−2 −38⎦ ⎣−18 −36⎦ ⎣0 3⎦ ⎣0 4⎦
= ⎡5−9−2−4 17+9−0−0⎤ = ⎡−10 26⎤
⎣−2−18−0−0 −38−36−3−4⎦ ⎣−20 −81⎦</span>
</li>
<li>Divide by 6:
<span class="math">X = ⎡−5/3 13/3 ⎤
⎣−10/3 −27/2⎦</span>
Wait — let me recheck row 2, col 2: −38 − 36 − 3 − 4 = −81. −81/6 = −27/2. ✓
</li>
</ul>
<span class="answer">X = ⎡−5/3 13/3 ⎤
⎣−10/3 −27/2⎦</span>
</div>
</div>
<!-- Q4 -->
<div class="problem">
<div class="problem-header"><span class="num">4</span> Find k such that AB = BA</div>
<div class="problem-body">
Given: A=[[2,3],[−1,1]], B=[[1,9],[−3,k]]
<ul class="steps">
<li>Compute AB:
<span class="math">AB = ⎡2·1+3·(−3) 2·9+3k ⎤ = ⎡−7 18+3k⎤
⎣(−1)·1+1·(−3) (−1)·9+1·k⎦ ⎣−4 −9+k ⎦</span>
</li>
<li>Compute BA:
<span class="math">BA = ⎡1·2+9·(−1) 1·3+9·1 ⎤ = ⎡ −7 12 ⎤
⎣(−3)·2+k·(−1) (−3)·3+k·1⎦ ⎣−6−k −9+k⎦</span>
</li>
<li>Set AB = BA, compare entries:
<span class="math">Entry (1,2): 18 + 3k = 12 → 3k = −6 → k = −2
Entry (2,1): −4 = −6 − k → k = −2 ✓
Entry (2,2): −9 + k = −9 + k ✓ (always true)</span>
</li>
</ul>
<span class="answer">k = −2</span>
</div>
</div>
<!-- Q5a -->
<div class="problem">
<div class="problem-header"><span class="num">5a</span> Equality of matrices — find x, y, z, w</div>
<div class="problem-body">
D = [[8,15],[−1, z+y]] = [[w³, x²−1],[y, 8]]
<ul class="steps">
<li>Entry (1,1): 8 = w³ → <strong>w = 2</strong></li>
<li>Entry (1,2): 15 = x²−1 → x² = 16 → <strong>x = ±4</strong></li>
<li>Entry (2,1): −1 = y → <strong>y = −1</strong></li>
<li>Entry (2,2): z + y = 8 → z + (−1) = 8 → <strong>z = 9</strong></li>
</ul>
<span class="answer">w = 2, x = ±4, y = −1, z = 9</span>
</div>
</div>
<!-- Q5b -->
<div class="problem">
<div class="problem-header"><span class="num">5b</span> Find x − y from matrix equality</div>
<div class="problem-body">
[[x³−y³, 6],[x²+xy+y², 12]] = 3·[[3,2],[1,4]] = [[9,6],[3,12]]
<ul class="steps">
<li>From entry (2,1): x² + xy + y² = 3</li>
<li>From entry (1,1): x³ − y³ = 9</li>
<li>Factor: x³ − y³ = (x − y)(x² + xy + y²) → 9 = (x − y)·3</li>
<li>Therefore: x − y = 3</li>
</ul>
<span class="answer">x − y = 3</span>
</div>
</div>
<!-- Q6 -->
<div class="problem">
<div class="problem-header"><span class="num">6</span> Find x, y, z for symmetric matrix A</div>
<div class="problem-body">
A = [[5, 2y, 3y+x],[z+1, −1, z+3],[2+3x, 8, 0]]
<div class="note">A symmetric means Aᵗ = A, i.e., aᵢⱼ = aⱼᵢ for all i,j.</div>
<ul class="steps">
<li>a₂₃ = a₃₂: z + 3 = 8 → <strong>z = 5</strong></li>
<li>a₁₂ = a₂₁: 2y = z + 1 = 6 → <strong>y = 3</strong></li>
<li>a₁₃ = a₃₁: 3y + x = 2 + 3x → 9 + x = 2 + 3x → 7 = 2x → <strong>x = 7/2</strong></li>
</ul>
<span class="answer">x = 7/2, y = 3, z = 5</span>
</div>
</div>
<!-- Q7 -->
<div class="problem">
<div class="problem-header"><span class="num">7</span> Find a, b, c, d, e, f for skew-symmetric matrix B</div>
<div class="problem-body">
B = [[a, a+d, b−e],[1, b, −2],[2, f, c]]
<div class="note">B skew-symmetric means Bᵗ = −B, so bᵢⱼ = −bⱼᵢ and all diagonal entries = 0.</div>
<ul class="steps">
<li>Diagonal: a = 0, b = 0, c = 0</li>
<li>b₁₂ = −b₂₁: a + d = −1 → 0 + d = −1 → <strong>d = −1</strong></li>
<li>b₁₃ = −b₃₁: b − e = −2 → 0 − e = −2 → <strong>e = 2</strong></li>
<li>b₂₃ = −b₃₂: −2 = −f → <strong>f = 2</strong></li>
</ul>
<span class="answer">a=0, b=0, c=0, d=−1, e=2, f=2</span>
</div>
</div>
<!-- Q8 -->
<div class="problem">
<div class="problem-header"><span class="num">8</span> Trace properties — tr(A) = 4, tr(B) = −3</div>
<div class="problem-body">
<div class="note">Key properties: tr(A+B)=tr(A)+tr(B), tr(αA)=α·tr(A), tr(Aᵗ)=tr(A)</div>
<div class="part">
<div class="part-label">a) tr(A+B)</div>
tr(A+B) = tr(A) + tr(B) = 4 + (−3) = <strong>1</strong>
</div>
<div class="part">
<div class="part-label">b) tr(2Aᵗ − B)</div>
= 2·tr(Aᵗ) − tr(B) = 2·4 − (−3) = 8 + 3 = <strong>11</strong>
</div>
<div class="part">
<div class="part-label">c) tr((4A + B)ᵗ)</div>
tr((4A+B)ᵗ) = tr(4A+B) = 4·tr(A) + tr(B) = 16 − 3 = <strong>13</strong>
</div>
</div>
</div>
<!-- Q9 -->
<div class="problem">
<div class="problem-header"><span class="num">9</span> Compute determinants</div>
<div class="problem-body">
<div class="part">
<div class="part-label">a) |[[1,0,−3],[−2,2,3],[7,5,2]]|</div>
Expand along the first row:
<span class="math">= 1·|⎡2 3⎤| − 0 + (−3)·|⎡−2 2⎤|
|⎣5 2⎦| |⎣ 7 5⎦|
= 1·(4−15) − 0 + (−3)·(−10−14)
= −11 + (−3)(−24) = −11 + 72 = 61</span>
<span class="answer">Determinant = 61</span>
</div>
<div class="part">
<div class="part-label">b) |[[1,0,−3],[−2,0,3],[7,0,2]]|</div>
The entire second column is zero → determinant = 0 (column of zeros).
<span class="answer">Determinant = 0</span>
</div>
<div class="part">
<div class="part-label">c) |[[1,−1,4,15],[0,2,−100,12],[0,0,−1,0],[0,0,0,4]]|</div>
Upper triangular matrix → determinant = product of diagonal entries.
<span class="math">det = 1 × 2 × (−1) × 4 = −8</span>
<span class="answer">Determinant = −8</span>
</div>
</div>
</div>
<!-- Q10 -->
<div class="problem">
<div class="problem-header"><span class="num">10</span> Find x such that A is singular (|A| = 0)</div>
<div class="problem-body">
A = [[x,2,4],[1,1,x],[3,7,9]]
<ul class="steps">
<li>Expand along first row:
<span class="math">|A| = x·(9−7x) − 2·(9−3x) + 4·(7−3)
= 9x − 7x² − 18 + 6x + 16
= −7x² + 15x − 2</span>
</li>
<li>Set |A| = 0:
<span class="math">7x² − 15x + 2 = 0</span>
</li>
<li>Quadratic formula:
<span class="math">x = (15 ± √(225 − 56)) / 14 = (15 ± √169) / 14 = (15 ± 13) / 14</span>
</li>
<li>Two solutions:
<span class="math">x = 28/14 = 2 OR x = 2/14 = 1/7</span>
</li>
</ul>
<span class="answer">x = 2 or x = 1/7</span>
</div>
</div>
<!-- Q11 -->
<div class="problem">
<div class="problem-header"><span class="num">11</span> Determinant properties — |A|=−3 (order 3), |B|=4 (order 3)</div>
<div class="problem-body">
<div class="note">Key rules: |AB|=|A||B|; |Aᵗ|=|A|; |kA|=kⁿ|A| for n×n matrix; |A⁻¹|=1/|A|.</div>
<div class="part">
<div class="part-label">a) |AB|</div>
|AB| = |A|·|B| = (−3)(4) = <strong>−12</strong>
</div>
<div class="part">
<div class="part-label">b) |ABᵗ|</div>
|ABᵗ| = |A|·|Bᵗ| = |A|·|B| = (−3)(4) = <strong>−12</strong>
</div>
<div class="part">
<div class="part-label">c) |(AB)ᵗ|</div>
|(AB)ᵗ| = |AB| = <strong>−12</strong>
</div>
<div class="part">
<div class="part-label">d) |−3Bᵗ|</div>
<span class="math">|−3Bᵗ| = (−3)³ · |Bᵗ| = −27 · |B| = −27 · 4 = −108</span>
<strong>−108</strong>
</div>
</div>
</div>
<!-- ═══════════════════ SHEET 3 ═══════════════════ -->
<h2 class="sheet-title" id="sheet3">Sheet 3 — Inverses, Determinant Properties & Orthogonal Matrices</h2>
<!-- Q1 -->
<div class="problem">
<div class="problem-header"><span class="num">1</span> Find the inverses of the following matrices</div>
<div class="problem-body">
<div class="note">Method: A⁻¹ = (1/|A|) · adj(A). A is invertible iff |A| ≠ 0.</div>
<div class="part">
<div class="part-label">a) A = [[12,7],[2,1]]</div>
<ul class="steps">
<li>|A| = 12·1 − 7·2 = 12 − 14 = <strong>−2 ≠ 0</strong> → invertible</li>
<li>adj of 2×2: swap diagonal, negate off-diagonal:
<span class="math">adj(A) = ⎡ 1 −7⎤
⎣−2 12⎦</span>
</li>
<li>A⁻¹ = (1/−2)·adj(A):
<span class="math">A⁻¹ = ⎡−1/2 7/2⎤
⎣ 1 −6 ⎦</span>
</li>
</ul>
<span class="answer">A⁻¹ = ⎡−1/2 7/2⎤
⎣ 1 −6 ⎦</span>
</div>
<div class="part">
<div class="part-label">b) A = [[1,2,−1],[−2,0,1],[1,−1,0]]</div>
<ul class="steps">
<li>Compute |A| (expand row 1):
<span class="math">|A| = 1·(0·0−1·(−1)) − 2·(−2·0−1·1) + (−1)·(−2·(−1)−0·1)
= 1·(1) − 2·(−1) + (−1)·(2)
= 1 + 2 − 2 = 1</span>
</li>
<li>Compute all 9 cofactors Cᵢⱼ = (−1)^(i+j)·Mᵢⱼ:
<span class="math">C₁₁=+|⎡0 1⎤|=0+1=1 C₁₂=−|⎡−2 1⎤|=−(0−1)=1 C₁₃=+|⎡−2 0⎤|=2
⎣−1 0⎦ ⎣ 1 0⎦ ⎣ 1 −1⎦
C₂₁=−|⎡ 2 −1⎤|=−(0−1)=1 C₂₂=+|⎡1 −1⎤|=0+1=1 C₂₃=−|⎡1 2⎤|=−(−1−2)=3
⎣−1 0⎦ ⎣1 0⎦ ⎣1 −1⎦
C₃₁=+|⎡ 2 −1⎤|=2−0=2 C₃₂=−|⎡1 −1⎤|=−(1−2)=1 C₃₃=+|⎡1 2⎤|=0+4=4
⎣ 0 1⎦ ⎣−2 1⎦ ⎣−2 0⎦</span>
</li>
<li>adj(A) = (cofactor matrix)ᵗ:
<span class="math">adj(A) = ⎡1 1 2⎤
⎢1 1 1⎥
⎣2 3 4⎦</span>
</li>
<li>A⁻¹ = (1/1)·adj(A) = adj(A)</li>
</ul>
<span class="answer">A⁻¹ = ⎡1 1 2⎤
⎢1 1 1⎥
⎣2 3 4⎦</span>
</div>
<div class="part">
<div class="part-label">c) A = [[1,3,5],[2,1,1],[3,4,2]]</div>
<ul class="steps">
<li>Compute |A| (expand row 1):
<span class="math">|A| = 1·(2−4) − 3·(4−3) + 5·(8−3)
= 1·(−2) − 3·(1) + 5·(5)
= −2 − 3 + 25 = 20</span>
</li>
<li>Cofactors:
<span class="math">C₁₁ = |⎡1 1⎤| = 2−4 = −2 C₁₂ = −|⎡2 1⎤| = −(4−3) = −1 C₁₃ = |⎡2 1⎤| = 8−3 = 5
⎣4 2⎦ ⎣3 2⎦ ⎣3 4⎦
C₂₁ = −|⎡3 5⎤| = −(6−20)=14 C₂₂ = |⎡1 5⎤| = 2−15=−13 C₂₃ = −|⎡1 3⎤| = −(4−9)=5
⎣4 2⎦ ⎣3 2⎦ ⎣3 4⎦
C₃₁ = |⎡3 5⎤| = 3−5 = −2 C₃₂ = −|⎡1 5⎤| = −(1−10)=9 C₃₃ = |⎡1 3⎤| = 1−6=−5
⎣1 1⎦ ⎣2 1⎦ ⎣2 1⎦</span>
</li>
<li>adj(A) = Cᵗ:
<span class="math">adj(A) = ⎡−2 14 −2⎤
⎢−1 −13 9⎥
⎣ 5 5 −5⎦</span>
</li>
</ul>
<span class="answer">A⁻¹ = (1/20)·⎡−2 14 −2⎤
⎢−1 −13 9⎥
⎣ 5 5 −5⎦</span>
</div>
<div class="part">
<div class="part-label">d) A = diag(1, 2, 3, 4)</div>
<ul class="steps">
<li>Diagonal matrix: |A| = 1·2·3·4 = 24 ≠ 0</li>
<li>Inverse of a diagonal matrix: take reciprocal of each diagonal entry.</li>
</ul>
<span class="answer">A⁻¹ = diag(1, 1/2, 1/3, 1/4)</span>
</div>
</div>
</div>
<!-- Q2 -->
<div class="problem">
<div class="problem-header"><span class="num">2</span> A=[[1,1],[3,4]], B=[[3,2],[−2,−1]] — find X, Y, Z</div>
<div class="problem-body">
<ul class="steps">
<li>Compute |A| = 4−3 = 1, so A⁻¹ = [[4,−1],[−3,1]]</li>
<li>Compute |B| = (3)(−1)−(2)(−2) = −3+4 = 1, so B⁻¹ = [[−1,−2],[2,3]]</li>
</ul>
<div class="part">
<div class="part-label">a) AX = B → X = A⁻¹B</div>
<span class="math">X = ⎡4 −1⎤·⎡ 3 2⎤ = ⎡4·3+(−1)(−2) 4·2+(−1)(−1)⎤ = ⎡14 9⎤
⎣−3 1⎦ ⎣−2 −1⎦ ⎣−3·3+1·(−2) −3·2+1·(−1)⎦ ⎣−11 −7⎦</span>
<span class="answer">X = ⎡14 9⎤
⎣−11 −7⎦</span>
</div>
<div class="part">
<div class="part-label">b) YB = |B|A = 1·A = A → Y = AB⁻¹</div>
<span class="math">Y = ⎡1 1⎤·⎡−1 −2⎤ = ⎡−1+2 −2+3⎤ = ⎡1 1⎤
⎣3 4⎦ ⎣ 2 3⎦ ⎣−3+8 −6+12⎦ ⎣5 6⎦</span>
<span class="answer">Y = ⎡1 1⎤
⎣5 6⎦</span>
</div>
<div class="part">
<div class="part-label">c) AZB + I = 2Bᵗ → Z = A⁻¹(2Bᵗ − I)B⁻¹</div>
<ul class="steps">
<li>Bᵗ = [[3,−2],[2,−1]]; 2Bᵗ − I = [[6,−4],[4,−2]] − [[1,0],[0,1]] = [[5,−4],[4,−3]]</li>
<li>Let M = 2Bᵗ − I. Compute A⁻¹·M:
<span class="math">A⁻¹M = ⎡4 −1⎤·⎡5 −4⎤ = ⎡20−4 −16+3⎤ = ⎡16 −13⎤
⎣−3 1⎦ ⎣4 −3⎦ ⎣−15+4 12−3⎦ ⎣−11 9⎦</span>
</li>
<li>Compute (A⁻¹M)·B⁻¹:
<span class="math">Z = ⎡16 −13⎤·⎡−1 −2⎤ = ⎡−16−26 −32−39⎤ = ⎡−42 −71⎤
⎣−11 9⎦ ⎣ 2 3⎦ ⎣ 11+18 22+27⎦ ⎣ 29 49⎦</span>
</li>
</ul>
<span class="answer">Z = ⎡−42 −71⎤
⎣ 29 49⎦</span>
</div>
</div>
</div>
<!-- Q3 -->
<div class="problem">
<div class="problem-header"><span class="num">3</span> Find X such that AX + tr(B)A − diag(2,3) = |A|I</div>
<div class="problem-body">
Given: A=[[1,−1],[−2,4]], B=[[3,5],[−2,−12]]
<ul class="steps">
<li>Compute |A| = (1)(4)−(−1)(−2) = 4−2 = <strong>2</strong></li>
<li>tr(B) = 3+(−12) = <strong>−9</strong></li>
<li>Rearrange: AX = |A|I − tr(B)A + diag(2,3)
<span class="math">= 2·⎡1 0⎤ − (−9)·⎡1 −1⎤ + ⎡2 0⎤
⎣0 1⎦ ⎣−2 4⎦ ⎣0 3⎦
= ⎡2 0⎤ + ⎡9 −9⎤ + ⎡2 0⎤ = ⎡13 −9⎤
⎣0 2⎦ ⎣−18 36⎦ ⎣0 3⎦ ⎣−18 41⎦</span>
</li>
<li>Compute A⁻¹: |A|=2, adj(A) = [[4,1],[2,1]]
<span class="math">A⁻¹ = (1/2)·⎡4 1⎤ = ⎡2 1/2⎤
⎣2 1⎦ ⎣1 1/2⎦</span>
</li>
<li>X = A⁻¹·[[13,−9],[−18,41]]:
<span class="math">X = ⎡2·13+(1/2)(−18) 2·(−9)+(1/2)(41)⎤ = ⎡26−9 −18+20.5⎤ = ⎡17 5/2⎤
⎣1·13+(1/2)(−18) 1·(−9)+(1/2)(41)⎦ ⎣13−9 −9+20.5⎦ ⎣4 23/2⎦</span>
</li>
</ul>
<span class="answer">X = ⎡17 5/2 ⎤
⎣4 23/2⎦</span>
</div>
</div>
<!-- Q4 -->
<div class="problem">
<div class="problem-header"><span class="num">4</span> If A² + 2A − I = 0, find A⁻¹</div>
<div class="problem-body">
<ul class="steps">
<li>Rearrange the equation:
<span class="math">A² + 2A = I</span>
</li>
<li>Factor out A on the left:
<span class="math">A(A + 2I) = I</span>
</li>
<li>By definition of inverse, A⁻¹ = A + 2I:
<span class="math">A · (A + 2I) = I → A⁻¹ = A + 2I</span>
</li>
</ul>
<span class="answer">A⁻¹ = A + 2I</span>
</div>
</div>
<!-- Q5 -->
<div class="problem">
<div class="problem-header"><span class="num">5</span> |A|=2, |B|=3 (both 3×3) — compute determinant expressions</div>
<div class="problem-body">
<div class="note">Key rules for n=3: |kA|=k³|A|; |A⁻¹|=1/|A|; |Aᵐ|=|A|ᵐ; |Aᵗ|=|A|</div>
<div class="part">
<div class="part-label">a) |A⁻¹|</div>
<span class="math">|A⁻¹| = 1/|A| = 1/2</span>
<span class="answer">1/2</span>
</div>
<div class="part">
<div class="part-label">b) |5B⁻¹|</div>
<span class="math">|5B⁻¹| = 5³·|B⁻¹| = 125·(1/3) = 125/3</span>
<span class="answer">125/3</span>
</div>
<div class="part">
<div class="part-label">c) |(AB)⁻¹|</div>
<span class="math">|(AB)⁻¹| = 1/|AB| = 1/(|A|·|B|) = 1/(2·3) = 1/6</span>
<span class="answer">1/6</span>
</div>
<div class="part">
<div class="part-label">d) |−3A³B⁻¹|</div>
<span class="math">= (−3)³·|A³|·|B⁻¹|
= −27 · |A|³ · (1/|B|)
= −27 · 8 · (1/3)
= −27 · 8/3 = −72</span>
<span class="answer">−72</span>
</div>
</div>
</div>
<!-- Q6 -->
<div class="problem">
<div class="problem-header"><span class="num">6</span> Prove A is orthogonal and find A⁻¹</div>
<div class="problem-body">
<span class="math">A = ⎡ 1/2 √3/2 0⎤
⎢−√3/2 1/2 0⎥
⎣ 0 0 1⎦</span>
<div class="note">A matrix is orthogonal iff AAᵗ = I, and then A⁻¹ = Aᵗ.</div>
<ul class="steps">
<li>Write Aᵗ:
<span class="math">Aᵗ = ⎡ 1/2 −√3/2 0⎤
⎢ √3/2 1/2 0⎥
⎣ 0 0 1⎦</span>
</li>
<li>Compute AAᵗ (the rows of A dotted with the cols of Aᵗ, which are the rows of A):
<span class="math">Row1·Col1: (1/2)² + (√3/2)² + 0 = 1/4 + 3/4 = 1 ✓
Row1·Col2: (1/2)(−√3/2) + (√3/2)(1/2) + 0 = −√3/4 + √3/4 = 0 ✓
Row1·Col3: 0 ✓
Row2·Col1: (−√3/2)(1/2) + (1/2)(√3/2) + 0 = 0 ✓
Row2·Col2: (−√3/2)² + (1/2)² + 0 = 3/4 + 1/4 = 1 ✓
Row2·Col3: 0 ✓
Row3·Col3: 0² + 0² + 1² = 1 ✓</span>
</li>
<li>Therefore AAᵗ = I → A is <strong>orthogonal</strong>.</li>
</ul>
<span class="answer">A is orthogonal; A⁻¹ = Aᵗ = ⎡ 1/2 −√3/2 0⎤
⎢ √3/2 1/2 0⎥
⎣ 0 0 1⎦</span>
</div>
</div>
<!-- Q7 -->
<div class="problem">
<div class="problem-header"><span class="num">7</span> A = (1/√2)[[1],[−1]] — prove B = I − 2AAᵗ is orthogonal</div>
<div class="problem-body">
<ul class="steps">
<li>Compute AAᵗ (column vector × row vector):
<span class="math">AAᵗ = (1/√2)⎡1⎤·(1/√2)[1 −1] = (1/2)⎡ 1 −1⎤
⎣−1⎦ ⎣−1 1⎦</span>
</li>
<li>Compute 2AAᵗ and then B:
<span class="math">2AAᵗ = ⎡ 1 −1⎤
⎣−1 1⎦
B = I − 2AAᵗ = ⎡1 0⎤ − ⎡ 1 −1⎤ = ⎡0 1⎤
⎣0 1⎦ ⎣−1 1⎦ ⎣1 0⎦</span>
</li>
<li>Check BBᵗ: Bᵗ = B (B is symmetric), so BBᵗ = B²:
<span class="math">B² = ⎡0 1⎤·⎡0 1⎤ = ⎡1 0⎤ = I ✓
⎣1 0⎦ ⎣1 0⎦ ⎣0 1⎦</span>
</li>
</ul>
<span class="answer">BBᵗ = I → B is orthogonal. (B is a reflection matrix.)</span>
</div>
</div>
<!-- Q8 -->
<div class="problem">
<div class="problem-header"><span class="num">8</span> Prove: the determinant of any orthogonal matrix equals ±1</div>
<div class="problem-body">
<ul class="steps">
<li>Let A be orthogonal, so by definition: AAᵗ = I</li>
<li>Take the determinant of both sides:
<span class="math">|AAᵗ| = |I| = 1</span>
</li>
<li>Use the multiplicative property of determinants and |Aᵗ| = |A|:
<span class="math">|A|·|Aᵗ| = 1 → |A|·|A| = 1 → |A|² = 1</span>
</li>
<li>Therefore:
<span class="math">|A| = ±1 □</span>
</li>