Matrix inverse and equations solver
Find the inverse of a matrix by the adjoint method - the determinant, minors, cofactors and adjoint shown - or solve simultaneous linear equations by the matrix method, X = A⁻¹B, as ISC Mathematics asks.
Answer
- |A|
- 1
A⁻¹ =
| -24 | 18 | 5 |
| 20 | -15 | -4 |
| -5 | 4 | 1 |
Determinant
Expanding along the first row: |A| = (1)(-24) - (2)(-20) + (3)(-5)
= (-24) + (40) + (-15) = 1
Minors and cofactors
Minors =
| -24 | -20 | -5 |
| -18 | -15 | -4 |
| 5 | 4 | 1 |
Cofactors =
| -24 | 20 | -5 |
| 18 | -15 | 4 |
| 5 | -4 | 1 |
Cofactor Cᵢⱼ = (-1)ⁱ⁺ʲ × minor Mᵢⱼ.
Adjoint and inverse
- adj A is the transpose of the cofactor matrix:adj A =
-24 18 5 20 -15 -4 -5 4 1 - A⁻¹ = (1/|A|) adj A = (1) adj A:A⁻¹ =
-24 18 5 20 -15 -4 -5 4 1 - Check: A·A⁻¹ = I ✓
Transpose
Aᵀ =
| 1 | 0 | 5 |
| 2 | 1 | 6 |
| 3 | 4 | 0 |
A is neither symmetric nor skew-symmetric.
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