Let Abe an invertible \(n \times n\) matrix, and let B be an \(n \times p\) matrix. Show that the equation \(AX = B\) has a unique solution \({A^{ - 1}}B\).

Theorem 5states that Ais an invertible \(n \times n\) matrix, then for each b in \({\mathbb{R}^n}\), the equation \(Ax = b\) has a unique solution \(x = {A^{ - 1}}b\).

It is given that \(B\) is a \(n \times p\) matrix (it is arbitrary). The matrix \({A^{ - 1}}B\) satisfies the equation \(AX = B\)because Ais invertible. Hence,

\(\begin{aligned}{c}A\left( {{A^{ - 1}}B} \right) = \left( {A{A^{ - 1}}} \right)B\\ = IB\\ = B\end{aligned}\)

Consider \(X\) to be any solution of the equation \(AX = B\)to demonstrate that the solution is unique.

Multiply each side of the equation \(AX = B\) by \({A^{ - 1}}\) to show that \(X\) must be \({A^{ - 1}}B\):

\(\begin{aligned}{c}{A^{ - 1}}\left( {AX} \right) = {A^{ - 1}}B\\\left( {{A^{ - 1}}A} \right)X = {A^{ - 1}}B\\IX = {A^{ - 1}}B\\X = {A^{ - 1}}B\end{aligned}\)

Hence, it is proved that the equation \(AX = B\) has a unique solution \({A^{ - 1}}B\).