Matrix multiplication is not commutative AB ≠ BA in general.
Aᵀ is formed by turning rows into columns; (AB)ᵀ = BᵀAᵀ.
The identity matrix I has 1s on the diagonal and 0s elsewhere, and AI = IA = A.
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Inverse of a matrix
Inverse of a matrix
When an inverse exists
A square matrix A has an inverse A−1 only if det A ≠ 0; such a matrix is called non-singular.
If det A = 0 the matrix is singular and has no inverse.
Definition AA−1 = A−1A = I.
Comparison aspect
Main idea
Keywords
For A = [[a, b], [c, d]] with det A ≠ 0
A−1 = (1/det A) × [[d
−b]
[−c
a]]
•
The matrix [[d
−b]
[−c
a]] is the adjugate (transpose of the cofactor matrix)
matrixadjugatetransposecofactor
Example
A = [[3
4]]
det A = 2 → A−1 = ½ [[4
−5]
[−2
3]] = [[2
−2.5]
[−1
1.5]]
Check
A−1 = [[3·2 + 5·(−1)
3·(−2.5) + 5·1.5]
[2·2 + 4·(−1)
2·(−2.5) + 4·1.5]] = [[1
1]] = I
(A−1)−1 = A · (AB)−1 = B−1A−1 · (Aᵀ)−1 = (A−1)ᵀ · det(A−1) = 1/det A.
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Solving systems of linear equations
Solving systems of linear equations
The inverse-matrix method
Write the system as AX = B, then X = A−1B, provided det A ≠ 0.
Example 2x + 3y = 7 and x − 2y = −7.A = [[2, 3], [1, −2]], det A = −4 − 3 = −7, A−1 = (−1/7)[[−2, −3], [−1, 2]].
X = A−1B = (−1/7)[[−2·7 + (−3)(−7)], [−1·7 + 2(−7)]] = (−1/7)[[7], [−21]] = [[−1], [3]], so x = −1 and y = 3.
For AX = B with det A ≠ 0 x = Dx/D, y = Dy/D, z = Dz/D, where D = det A and Dx is D with the x-column replaced by B.
Same system D = −7, Dx = |[[7, 3], [−7, −2]]| = −14 + 21 = 7 → x = 7/(−7) = −1.
Dy = |[[2, 7], [1, −7]]| = −14 − 7 = −21 → y = (−21)/(−7) = 3.
Both methods give the same answer; Cramer's rule is faster for a single unknown, the inverse method is better when several right-hand sides share the same A.
det A
System AX = B
≠ 0
Unique solution
= 0 and consistent
Infinitely many solutions
= 0 and inconsistent
No solution
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