Solve a matrix equation

Subtraction removes a known addend

For same-order matrices, X+A=B gives X=B−A. For invertible A, AX=B gives X=A⁻¹B, whereas XA=B gives X=BA⁻¹. Multiply on the side that puts the inverse next to its matching factor

If XA=B and A is invertible, which expression gives X?

Work a left-factor equation

For the matrices A and B and the stated inverse A⁻¹, solve AX=B by left multiplication

Given matrix

\(\displaystyle X=\begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix}\)

For this solution X, find x₁₂

The same data on the other side gives a different answer

For the displayed A and B, compare AX=B with YA=B. Left multiplication gives X=A⁻¹B; right multiplication gives Y=BA⁻¹. The results differ, so the side of A matters

Given matrix

\(\displaystyle Y=\begin{bmatrix} 5 & -5 \\ 2 & -1 \end{bmatrix}\)

For this Y, find y₂₁

Remove addition before cancelling a factor

For AX+C=B with invertible A, first obtain AX=B−C and then X=A⁻¹(B−C). Expand as A⁻¹B−A⁻¹C if needed; do not drop the second inverse factor

Prove the solution is unique, ungraded

For invertible A, prove AX=B has the unique solution A⁻¹B. Distinguish showing a candidate works from showing no other candidate works. Independent proof is ungraded

Model: existence and uniqueness separately

Existence: A(A⁻¹B)=(AA⁻¹)B=IB=B by associativity. Uniqueness: if AY=B then multiply on the left by A⁻¹ to obtain Y=A⁻¹B. Hence every possible solution equals the candidate. The same argument on the right proves XA=B has unique solution BA⁻¹