9.7Solving Matrix Equations with Inverses
Find the inverse of a 2×2 matrix using A⁻¹ = (1/det A)[d, −b; −c, a]. Use AX = B → X = A⁻¹B to solve systems. Determine invertibility using the determinant.
Matrix inverses provide an elegant solution to linear systems: AX = B → X = A⁻¹B. This approach is used in computer graphics, cryptography, and any application that requires solving the same system with multiple right-hand sides.
Essential Question
How does multiplying both sides of the matrix equation Ax = b by A⁻¹ solve the system, and when does this method fail?
Lesson Overview
Matrix equation form
Write a linear system as Ax = b, where A is the coefficient matrix, x is the variable vector, and b is the constant vector.
Solving with the inverse
If A is invertible (det(A) ≠ 0), multiply both sides on the left by A⁻¹: A⁻¹(Ax) = A⁻¹b → Ix = A⁻¹b → x = A⁻¹b.
2×2 inverse formula
For A = [[a, b], [c, d]]: det(A) = ad − bc. Then A⁻¹ = (1/det(A)) · [[d, −b], [−c, a]]. A⁻¹ exists if and only if det(A) ≠ 0.
Gauss-Jordan method (n×n)
Form the augmented matrix [A | I]. Row reduce until the left side becomes I. The right side becomes A⁻¹.
Verification
Always check: A · A⁻¹ = A⁻¹ · A = I.
When the method fails
If det(A) = 0, A is singular (not invertible). The system Ax = b has no unique solution.
Key advantage
Once A⁻¹ is found, solving Ax = b for different b vectors requires only a single matrix-vector multiplication each time.
Worked Examples
Find A⁻¹ for A = [[3, 1], [5, 2]].
det = 3·2 − 1·5 = 6 − 5 = 1
A⁻¹ = (1/1)·[[2, −1], [−5, 3]] = [[2, −1], [−5, 3]]
Verify: A·A⁻¹ = [[3·2+1·(−5), 3·(−1)+1·3], [5·2+2·(−5), 5·(−1)+2·3]] = [[1, 0], [0, 1]] ✓
Solve Ax = b using A⁻¹: A = [[3, 1], [5, 2]], b = [[7], [11]].
x = A⁻¹b = [[2, −1], [−5, 3]] · [[7], [11]]
x₁ = 2·7 + (−1)·11 = 14 − 11 = 3
x₂ = (−5)·7 + 3·11 = −35 + 33 = −2
Find A⁻¹ for A = [[1, 2], [3, 4]] using Gauss-Jordan.
[A|I] = [[1, 2 | 1, 0], [3, 4 | 0, 1]]
R₂ → R₂ − 3R₁: [[1, 2 | 1, 0], [0, −2 | −3, 1]]
R₂ → R₂ ÷ (−2): [[1, 2 | 1, 0], [0, 1 | 3/2, −1/2]]
R₁ → R₁ − 2R₂: [[1, 0 | −2, 1], [0, 1 | 3/2, −1/2]]
Determine if A = [[2, 4], [1, 2]] is invertible.
det(A) = 2·2 − 4·1 = 4 − 4 = 0
det = 0 → A is singular, not invertible
Solve 2x + y = 5, 3x + 2y = 8 using the inverse method.
A = [[2, 1], [3, 2]]; det = 4 − 3 = 1
A⁻¹ = (1/1)·[[2, −1], [−3, 2]] = [[2, −1], [−3, 2]]
x = A⁻¹b = [[2, −1], [−3, 2]] · [[5], [8]] = [[10−8], [−15+16]] = [[2], [1]]
Guided Practice
Find A⁻¹ for A = [[4, 3], [3, 2]].
Hint: Use the 2×2 formula: A⁻¹ = (1/det)·[[d, −b], [−c, a]]. Compute det first.
Solve Ax = b where A = [[4, 3], [3, 2]] and b = [[1], [2]].
Hint: Use your answer from problem 1: x = A⁻¹b.
Find A⁻¹ for A = [[1, 0, 2], [0, 1, 3], [0, 0, 1]] using Gauss-Jordan.
Hint: Form [A|I] and row reduce. The upper-triangular structure makes this straightforward.
Determine if A = [[3, 6], [1, 2]] is invertible. If not, explain what this means for Ax = b.
Hint: Compute det(A) = 3·2 − 6·1.
A company uses the same coefficient matrix A for three different right-hand sides b₁, b₂, b₃. Why is the inverse method more efficient than Gaussian elimination in this case?
Hint: Think about how many times you need to compute A⁻¹ vs. how many row reductions you need.
Key Vocabulary
Invertible matrix
A square matrix A with det(A) ≠ 0; has a unique inverse A⁻¹.
Singular matrix
A square matrix with det(A) = 0; has no inverse.
Determinant
For A = [[a, b], [c, d]], det(A) = ad − bc; measures whether A is invertible.
Identity matrix I
The matrix satisfying AI = IA = A.
Inverse matrix A⁻¹
The matrix satisfying A · A⁻¹ = A⁻¹ · A = I.
Gauss-Jordan method for inverse
Augment A with I, row reduce to get [I | A⁻¹].
Matrix equation
Ax = b, solved as x = A⁻¹b when A is invertible.
Quick Check
Interactive Practice — 5 Questions
For A = [[2, 1], [5, 3]], det(A) =
If det(A) = 0, then:
To solve Ax = b using the inverse, you compute:
A⁻¹ for A = [[3, 1], [5, 2]] is:
The Gauss-Jordan method for finding A⁻¹ starts with:
Common Mistakes
Common Mistakes
Computing x = bA⁻¹ instead of x = A⁻¹b
Matrix multiplication is not commutative. The correct formula is x = A⁻¹b (A⁻¹ on the LEFT). bA⁻¹ may not even be defined.
Forgetting to check det(A) before computing the inverse
Always check det(A) first. If det(A) = 0, the matrix is singular and has no inverse — the method fails.
Applying the 2×2 inverse formula to a 3×3 matrix
The formula A⁻¹ = (1/det)·[[d, −b], [−c, a]] only works for 2×2 matrices. Use Gauss-Jordan for larger matrices.
Not verifying A·A⁻¹ = I after computing the inverse
Always verify by computing A·A⁻¹. If you don't get the identity matrix, there's an arithmetic error.
Math Tips
Math Tips
The 2×2 inverse formula: swap the diagonal entries, negate the off-diagonal entries, divide by the determinant.
The inverse method shines when you need to solve Ax = b for MULTIPLE right-hand sides b. Compute A⁻¹ once, then multiply each b.
For a 2×2 system, the inverse method is often faster than Gaussian elimination. For 3×3 and larger, Gaussian elimination is usually more efficient.
det(A) = 0 means the rows (or columns) of A are linearly dependent — the system either has no solution or infinitely many.
Gauss-Jordan for the inverse: [A|I] → [I|A⁻¹]. The row operations that turn A into I automatically turn I into A⁻¹.