Inverse matrices
Objective. Find inverses and connect them to unique solvability.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
An inverse matrix A⁻¹ satisfies AA⁻¹=A⁻¹A=I, and exists precisely when A is square and det(A)≠0, in which case the system Ax=b has the unique solution x=A⁻¹b. For a 2×2 matrix A=[[a,b],[c,d]], the inverse is (1/det(A))[[d,−b],[−c,a]] where det(A)=ad−bc. For A=[[2,1],[1,1]], det(A)=2−1=1, so A⁻¹=[[1,−1],[−1,2]]. Checking AA⁻¹=I is good practice to confirm an inverse computation, and a zero determinant always signals that no inverse exists and the associated system either has no solution or infinitely many.
Worked example
Find the inverse matrix of A=[[2,1],[1,1]].
- 1Compute the determinant: det(A)=(2)(1)−(1)(1)=1.
- 2Since det(A)=1≠0, the inverse exists.
- 3Apply the 2×2 inverse formula: swap diagonal entries and negate off-diagonal entries to get [[1,−1],[−1,2]].
- 4Divide by det(A)=1 (no change) and check: A·A⁻¹=[[2−1,−2+2],[1−1,−1+2]]=[[1,0],[0,1]]=I.
Answer: A⁻¹ = [[1, −1], [−1, 2]]
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Start practisingCommon mistake. Students compute (1/det(A))[[a,b],[c,d]] without swapping a and d and negating b and c, giving an incorrect inverse; the formula requires swapping the diagonal entries and negating the off-diagonal entries.
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