Solving by the matrix method

If A is non-singular, AX = B gives A⁻¹(AX) = A⁻¹B, so (A⁻¹A)X = IX = X = A⁻¹B. Because the inverse is unique, so is this solution.

Multiply by A⁻¹ on the left. BA⁻¹ is a different product and usually not even defined when B is a column.

Worked example: 2x + 5y = 1 and 3x + 2y = 7 have |A| = 4 − 15 = −11, A⁻¹ = −(1/11)[[2, −5], [−3, 2]], and X = A⁻¹B gives x = 3, y = −1.

Worked example: 3x − 2y + 3z = 8, then 2x + y − z = 1, and 4x − 3y + 2z = 4 have |A| = −17. The cofactors give adj A with rows (−1, −5, −1), (−8, −6, 9), (−10, 1, 7), and X = −(1/17)(adj A)B = (1, 2, 3).

Word problems turn into AX = B first. Three numbers with sum 6, where the second plus three times the third is 11 and the first plus the third is twice the second, give |A| = 9 and the numbers 1, 2 and 3.

If a product such as PQ = I is already known, Q is P⁻¹, and a system with coefficient matrix P is solved by X = QB with no cofactors at all. This way the system x − y + 2z = 1, then 2y − 3z = 1, and 3x − 2y + 4z = 2 gives x = 0, y = 5, z = 3.

Check the answer by substituting it back into every original equation.