Ian Kim ✏️
10.5 — INVERSE MATRICES & SYSTEMS OF EQUATIONS
Standards: HSA.REI.C.8 · HSA.REI.C.9 · HSA.CED.A.3 | Self-Study Worksheet
📘 Concept Review
AX = B → X = A⁻¹B
A = Coefficient Matrix · X = Variable Matrix · B = Constant Matrix
🔑 Quick keyword: "ACE" — Answer = Coefficients Explain everything!
📌 Example — Worked Solution
Solve the system: \(2x + y = 6\), \(-x - 3y = 2\)
1
Write as matrix equation AX = B:
\[\begin{bmatrix}2 & 1\\-1 & -3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}6\\2\end{bmatrix}\]
\[\begin{bmatrix}2 & 1\\-1 & -3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}6\\2\end{bmatrix}\]
2
Find A⁻¹ using det(A):
\(\det(A) = (2)(-3)-(1)(-1) = -6+1 = -5\)
For a 2×2 matrix \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\), \(A^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)
\[A^{-1} = \frac{1}{-5}\begin{bmatrix}-3&-1\\1&2\end{bmatrix} = \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\-\frac{1}{5}&-\frac{2}{5}\end{bmatrix}\]
\(\det(A) = (2)(-3)-(1)(-1) = -6+1 = -5\)
For a 2×2 matrix \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\), \(A^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)
\[A^{-1} = \frac{1}{-5}\begin{bmatrix}-3&-1\\1&2\end{bmatrix} = \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\-\frac{1}{5}&-\frac{2}{5}\end{bmatrix}\]
3
Multiply X = A⁻¹B:
\[\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\-\frac{1}{5}&-\frac{2}{5}\end{bmatrix}\begin{bmatrix}6\\2\end{bmatrix} = \begin{bmatrix}4\\-2\end{bmatrix}\]
✅ Answer: \((4,\ -2)\)
\[\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\-\frac{1}{5}&-\frac{2}{5}\end{bmatrix}\begin{bmatrix}6\\2\end{bmatrix} = \begin{bmatrix}4\\-2\end{bmatrix}\]
✅ Answer: \((4,\ -2)\)
⭐ MEMORY POINT
SWAP · SIGN · SCALE
To find A⁻¹ for \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\):
SWAP a ↔ d | SIGN flip b and c | SCALE by \(\tfrac{1}{\det}\)
⚡ det = ad − bc (NOT ac − bd!)
🧠 Reminders:
• A · A⁻¹ = I (Identity matrix!)
• A⁻¹ · I = A (Multiplying by I does nothing)
• If det(A) = 0, NO inverse exists! (singular matrix)
• A · A⁻¹ = I (Identity matrix!)
• A⁻¹ · I = A (Multiplying by I does nothing)
• If det(A) = 0, NO inverse exists! (singular matrix)
✏️ PRACTICE PROBLEMS
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