🔲 Matrices
Matrices package whole systems of numbers into a single object — the backbone of structural analysis, robotics transformations, and solving many equations at once. Work through each tab: rules first, then the interactive tools, then worked examples and practice.
3Topics
12Worked Examples
4Eng. Cases
12Practice Qs
➕Matrix BasicsAdd, scale, multiply
🔢Determinants & Inverses2×2 and 3×3
⚖️Simultaneous EquationsThe matrix method
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Add, subtract and multiply matrices; find determinants and inverses of 2×2 and 3×3 matrices; and use the matrix method to solve systems of simultaneous equations — the same technique behind structural load analysis and circuit mesh analysis.
💡 A matrix is just a rectangular grid of numbers. The one rule that trips people up: matrix multiplication is not commutative — AB usually does not equal BA.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
➕
Matrix Basics & Operations
A grid of numbers you can add, scale and multiply — with one crucial twist on the usual rules.
🔧2×2 Matrix CalculatorEnter both matrices, choose an operation
Enter matrix A (and B, or k for scaling), choose an operation, then press Calculate. For scaling, only fill A and put k in the first B box.
📖 Key Vocabulary
Element (aij) — the entry in row i, column j.
Order — the size of a matrix, written rows × columns.
Square matrix — same number of rows and columns; required for a determinant/inverse.
Identity matrix (I) — 1's on the diagonal, 0's elsewhere; the matrix "1" — AI = A.
👁️ Visual aid: the highlighted row of A pairs with the highlighted column of B — multiply matching positions, then add.
✏️Worked Example 1 — Addition and subtractionType: Add/Subtract
Question Given A = [[3,1],[2,5]] and B = [[1,4],[0,2]], find A + B and A − B.
Add element by element:
[[3+1,1+4],[2+0,5+2]] = [[4,5],[2,7]]Subtract
[[3−1,1−4],[2−0,5−2]] = [[2,−3],[2,3]]Answer Sum = [[4,5],[2,7]], Difference = [[2,−3],[2,3]]
🔢Worked Example 2 — Scalar multiplicationType: Scale
Question Find 3A where A = [[2,−1],[0,4]].
Multiply every element by 3
[[3×2,3×−1],[3×0,3×4]]Answer
[[6,−3],[0,12]] — every entry scales, unlike addition which needs a matching matrix.✖️Worked Example 3 — Matrix multiplication (row × column)Type: Multiply
Question Find AB where A = [[1,2],[3,4]] and B = [[5,6],[7,8]].
Top-left
(1×5)+(2×7) = 5+14 = 19Top-right
(1×6)+(2×8) = 6+16 = 22Bottom-left
(3×5)+(4×7) = 15+28 = 43; Bottom-right (3×6)+(4×8) = 18+32 = 50Answer
AB = [[19,22],[43,50]]⚙️Worked Example 4 — Engineering context (nodal forces)Type: Engineering
Question A stiffness matrix K = [[8,−3],[−3,8]] (kN/mm) acts on a displacement vector d = [2, 1] (mm). Find the force vector F = Kd.
Row 1
F₁ = 8(2) + (−3)(1) = 16 − 3 = 13Row 2
F₂ = (−3)(2) + 8(1) = −6 + 8 = 2Answer F = [13, 2] kN — matrix multiplication converts displacements straight into node forces.
🧠 Common slip: assuming AB = BA. Try swapping the order in Example 3 — you get a completely different matrix. Order matters.
🔢
Determinants & Inverses
A single number that reveals whether a matrix can be "undone" — and the recipe for undoing it.
🔧Determinant & Inverse Finder (2×2)Enter a matrix, then press Solve
Enter a, b, c, d for matrix [[a,b],[c,d]], then press Solve for the determinant and inverse…
📖 Key Vocabulary
Determinant, det(A) or |A| — a single scalar summarising how A scales area/volume.
Singular matrix — det(A) = 0; no inverse exists.
Adjugate (adj A) — for 2×2, swap the leading diagonal and negate the other two entries.
Cofactor — a signed minor, used to build determinants and inverses for 3×3 matrices and larger.
👁️ Visual aid: swap the two diagonal entries, flip the sign of the other two, then scale by 1/det(A).
✏️Worked Example 1 — Finding a determinantType: Determinant
Question Find det(A) for A = [[4,3],[2,5]].
Formula
det = ad − bc = (4)(5) − (3)(2)Answer
= 20 − 6 = 14 — nonzero, so an inverse exists.🔁Worked Example 2 — Finding a 2×2 inverseType: Inverse
Question Find A⁻¹ for A = [[2,1],[5,3]].
Determinant
det = (2)(3) − (1)(5) = 1Swap & negate
[[3,−1],[−5,2]]Answer
A⁻¹ = (1/1)[[3,−1],[−5,2]] = [[3,−1],[−5,2]]🕵️Worked Example 3 — Recognising a singular matrixType: Singular check
Question Show that A = [[4,2],[6,3]] has no inverse.
Determinant
det = (4)(3) − (2)(6) = 12 − 12 = 0Answer det(A) = 0 → A is singular, no inverse exists. (Notice row 2 is exactly 1.5 × row 1 — the rows are not independent.)
⚙️Worked Example 4 — Engineering context (checking a stiffness matrix)Type: Engineering
Question A structure's stiffness matrix is K = [[10,−4],[−4,10]] kN/mm. Confirm it is invertible and find K⁻¹.
Determinant
det = (10)(10) − (−4)(−4) = 100 − 16 = 84 — nonzero, structure is stable and solvable.Swap & negate
[[10,4],[4,10]]Answer
K⁻¹ = (1/84)[[10,4],[4,10]] — this inverse converts any applied force vector directly into displacements.🧠 Common slip: forgetting to divide by det(A) after building the swapped/negated matrix — that scaling step is what actually makes A·A⁻¹ = I.
⚖️
Solving Simultaneous Equations — the Matrix Method
Write the system as Ax = b, then solve with x = A⁻¹b — one clean method for any size of system.
🔧Simultaneous Equation Solver (2×2 system)Enter ax+by=e, cx+dy=f, then Solve
Enter the coefficients of ax+by=e and cx+dy=f, then press Solve to see the matrix method step by step…
📖 Key Vocabulary
Coefficient matrix (A) — the grid of x and y coefficients from the equations.
Unknown vector (x) — the column of variables being solved for.
Constant vector (b) — the column of right-hand-side values.
Unique solution — exists only when det(A) ≠ 0.
👁️ Visual aid: the same information, repackaged — solve by finding A⁻¹ and computing x = A⁻¹b.
✏️Worked Example 1 — Full matrix-method solutionType: Full method
Question Solve 2x + y = 8, 5x + 3y = 19.
Set up
A = [[2,1],[5,3]], b = [8,19]Determinant
det = (2)(3)−(1)(5) = 1Inverse
A⁻¹ = [[3,−1],[−5,2]] (det = 1, so no division needed)Solve
x = A⁻¹b: x = 3(8)+(−1)(19) = 24−19 = 5; y = −5(8)+2(19) = −40+38 = −2Answer x = 5, y = −2
🔍Worked Example 2 — Checking the solutionType: Verification
Question Verify the solution x = 5, y = −2 found above satisfies both original equations.
Equation 1
2(5) + (−2) = 10 − 2 = 8 ✓Equation 2
5(5) + 3(−2) = 25 − 6 = 19 ✓Answer Both equations balance — always verify by substitution; it costs seconds and catches almost every arithmetic slip.
🚫Worked Example 3 — Recognising no unique solutionType: No solution
Question Attempt to solve 2x + 4y = 6, 3x + 6y = 9 by the matrix method.
Determinant
det = (2)(6) − (4)(3) = 12 − 12 = 0Interpret A is singular — no inverse exists, so x = A⁻¹b cannot be formed.
Answer The equations are actually the same line scaled by 1.5 — infinitely many solutions, not a unique pair.
⚙️Worked Example 4 — Engineering context (two-node truss)Type: Engineering
Question Two coupled beam displacements satisfy 10x − 4y = 60 and −4x + 10y = 20 (forces in kN). Find x and y (mm).
Set up
A = [[10,−4],[−4,10]], b = [60,20]. From the earlier example det(A) = 84, A⁻¹ = (1/84)[[10,4],[4,10]]Solve
x = (10(60)+4(20))/84 = (600+80)/84 = 680/84Continue
y = (4(60)+10(20))/84 = (240+200)/84 = 440/84Answer x ≈ 8.10 mm, y ≈ 5.24 mm — the same inverse from the Determinants tab now solves an entire structural system in one multiplication.
🧠 Common slip: multiplying A⁻¹ by b in the wrong order or on the wrong side. Since matrix multiplication is not commutative, always compute A⁻¹b, never bA⁻¹.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🏗️ Structural Stiffness — Force from Displacement
For a 2-DOF spring system, F = Kd links nodal forces to displacements through the stiffness matrix K.
F = K d (matrix multiplication)
📟 Try it — Nodal Force Calculator
Enter K and d, then press Calculate…
🔌 Circuit Mesh Analysis
Kirchhoff's voltage law for a two-mesh circuit gives a 2×2 system solved directly by the matrix method for the mesh currents.
R·I = V → I = R⁻¹V
📟 Try it — Mesh Current Solver
Enter R and V, then press Calculate…
🤖 2D Rotation — Robotics Transformation
A rotation matrix transforms a point's coordinates by angle θ — the foundation of robot-arm kinematics and CAD transformations.
[x'] = [cosθ −sinθ][x]
[y'] [sinθ cosθ][y]
[y'] [sinθ cosθ][y]
📟 Try it — Rotate a Point
Enter x, y and the rotation angle, then press Calculate…
📡 Sensor Fusion — Checking a System is Solvable
Before trusting any multi-sensor least-squares solution, engineers check the determinant of the normal-equations matrix isn't (numerically) zero — otherwise the system is ill-conditioned.
det(A) ≈ 0 → ill-conditioned; small errors in data cause huge errors in the solution
📟 Try it — Conditioning Check
Enter a, b, c, d, then press Calculate…
🎮
Quick-Fire Matrices Game
12 questions — operations, determinants and solving mixed. How many can you get?
⚡ Matrices Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
Add A = [[2,3],[1,4]] and B = [[5,0],[2,1]].
[[2+5,3+0],[1+2,4+1]] = [[7,3],[3,5]]
Question 2 🥉2 marks
Find 4A where A = [[1,−2],[3,0]].
[[4,−8],[12,0]] = [[4,−8],[12,0]]
Question 3 🥉2 marks
Find det(A) for A = [[3,2],[1,5]].
(3)(5) − (2)(1) = 15 − 2 = 13
Question 4 🥉2 marks
State the order of the product when a 2×3 matrix multiplies a 3×4 matrix.
Inner dimensions (3 and 3) match, outer dimensions give the result: 2×4
Question 5 🥈3 marks
Find AB where A = [[2,0],[1,3]] and B = [[4,1],[2,5]].
Top row: (2×4+0×2, 2×1+0×5) = (8,2)
Bottom row: (1×4+3×2, 1×1+3×5) = (10,16)
AB = [[8,2],[10,16]]
Bottom row: (1×4+3×2, 1×1+3×5) = (10,16)
AB = [[8,2],[10,16]]
Question 6 🥈4 marks
Find the inverse of A = [[3,2],[4,3]].
det = 3(3)−2(4) = 9−8 = 1
Swap & negate: [[3,−2],[−4,3]]
A⁻¹ = [[3,−2],[−4,3]]
Swap & negate: [[3,−2],[−4,3]]
A⁻¹ = [[3,−2],[−4,3]]
Question 7 🥈4 marks
Solve using the matrix method: 3x + 2y = 16, x + 4y = 18.
A=[[3,2],[1,4]], det=12−2=10
A⁻¹=(1/10)[[4,−2],[−1,3]]
x=(4×16−2×18)/10=(64−36)/10=2.8
y=(−16+3×18)/10=(−16+54)/10=x=2.8, y=3.8
A⁻¹=(1/10)[[4,−2],[−1,3]]
x=(4×16−2×18)/10=(64−36)/10=2.8
y=(−16+3×18)/10=(−16+54)/10=x=2.8, y=3.8
Question 8 🥈3 marks
Show that A = [[6,3],[2,1]] is singular and explain what this means.
det = 6(1)−3(2) = 6−6 = 0 — singular
Row 2 is exactly row 1 ÷ 3 — the rows aren't independent, so no unique inverse or solution exists.
Row 2 is exactly row 1 ÷ 3 — the rows aren't independent, so no unique inverse or solution exists.
Question 9 🥇4 marks
A rotation matrix R(θ) = [[cosθ,−sinθ],[sinθ,cosθ]] rotates point (3,0) by 90°. Find the new coordinates.
cos90°=0, sin90°=1
x' = 0(3) − 1(0) = 0
y' = 1(3) + 0(0) = 3
(0, 3) — a quarter-turn anticlockwise, as expected
x' = 0(3) − 1(0) = 0
y' = 1(3) + 0(0) = 3
(0, 3) — a quarter-turn anticlockwise, as expected
Question 10 🥇5 marks
A stiffness matrix K = [[12,−6],[−6,12]] kN/mm acts on a load vector F = [30, 6] kN. Find the displacement vector d = K⁻¹F.
det = 144−36 = 108
K⁻¹ = (1/108)[[12,6],[6,12]]
d₁ = (12×30+6×6)/108 = (360+36)/108 = 3.667
d₂ = (6×30+12×6)/108 = (180+72)/108 = 2.333
d ≈ (3.67, 2.33) mm
K⁻¹ = (1/108)[[12,6],[6,12]]
d₁ = (12×30+6×6)/108 = (360+36)/108 = 3.667
d₂ = (6×30+12×6)/108 = (180+72)/108 = 2.333
d ≈ (3.67, 2.33) mm
Question 11 🥇5 marks
Verify (AB)⁻¹ = B⁻¹A⁻¹ is at least dimensionally sensible by checking orders, given A and B are both 3×3.
AB is 3×3 (3×3 × 3×3 = 3×3), so (AB)⁻¹ must be 3×3.
B⁻¹ is 3×3, A⁻¹ is 3×3, so B⁻¹A⁻¹ is 3×3 ✓
Orders match — consistent with the identity (proof requires showing AB·B⁻¹A⁻¹ = I).
B⁻¹ is 3×3, A⁻¹ is 3×3, so B⁻¹A⁻¹ is 3×3 ✓
Orders match — consistent with the identity (proof requires showing AB·B⁻¹A⁻¹ = I).
Question 12 🥇5 marks
Two circuit meshes give 20I₁ − 8I₂ = 40 and −8I₁ + 18I₂ = 10 (volts). Find I₁ and I₂ in amps to 3 s.f.
A=[[20,−8],[−8,18]], det=360−64=296
A⁻¹=(1/296)[[18,8],[8,20]]
I₁=(18×40+8×10)/296=(720+80)/296=2.703
I₂=(8×40+20×10)/296=(320+200)/296=1.757
I₁ ≈ 2.70 A, I₂ ≈ 1.76 A
A⁻¹=(1/296)[[18,8],[8,20]]
I₁=(18×40+8×10)/296=(720+80)/296=2.703
I₂=(8×40+20×10)/296=(320+200)/296=1.757
I₁ ≈ 2.70 A, I₂ ≈ 1.76 A