MENUMatrix Equations
By Catalin David
AX = B, where A is Invertible
Since matrix multiplication isn't always commutative, we multiply to the left both members of the equations by$ A^{-1}$.
$A^{-1}\cdot|A\cdot X = B$
$A^{-1}\cdot A\cdot X = A^{-1}\cdot B$
$I_{n}\cdot X = A^{-1}\cdot B$
The general form of the solution of the equation is:
$\color{red}{X =A^{-1}\cdot B}$
Example 50
Solve the equation
$\begin{pmatrix}
1 & 3\\
2 & 5
\end{pmatrix}\cdot X
\begin{pmatrix}
3 & 5\\
2 & 1
\end{pmatrix}$
We check if the first matrix is invertible.
$\left|A\right|=5-6=-1\neq 0$, so the matrix is invertible.
We multiply to the left by its inverse.
$\begin{pmatrix}
1 & 3\\
2 & 5\\
\end{pmatrix}^{-1}\cdot
\begin{pmatrix}
1 & 3\\
2 & 5
\end{pmatrix}\cdot
X=
\begin{pmatrix}
1 & 3\\
2 & 5
\end{pmatrix}^{-1}\cdot
\begin{pmatrix}
3 & 5\\
2 & 1
\end{pmatrix}$
$I_{2}\cdot X = \begin{pmatrix} 1 & 3\\ 2 & 5 \end{pmatrix}^{-1}\cdot \begin{pmatrix} 3 & 5\\ 2 & 1 \end{pmatrix}$
$X=\begin{pmatrix} 1 & 3\\ 2 & 5 \end{pmatrix}^{-1}\cdot \begin{pmatrix} 3 & 5\\ 2 & 1 \end{pmatrix}$
$\begin{pmatrix} 1 & 3\\ 2 & 5 \end{pmatrix}^{-1}= \begin{pmatrix} -5 & 3\\ 2 & -1 \end{pmatrix}\rightarrow X= \begin{pmatrix} -5 & 3\\ 2 & -1 \end{pmatrix}\cdot \begin{pmatrix} 3 & 5\\ 2 & 1 \end{pmatrix}= \begin{pmatrix} -9 & -22\\ 4 & 9 \end{pmatrix}$
XA = B, where A is Invertible
Since matrix multiplication isn't always commutative, we multiply to the right both members of the equation by $ A^{-1}.$
$X\cdot A = B |\cdot A^{-1}$
$X\cdot A\cdot A^{-1} = B\cdot A^{-1}$
$X \cdot I_{n} =B\cdot A^{-1}$
The general form of the solution of the equation is:
$\color{red}{X =B\cdot A^{-1}}$
Example 51
Solve the equation
$X \begin{pmatrix}
1 & 3\\
2 & 5\\
\end{pmatrix}=
\begin{pmatrix}
3 & 5\\
2 & 1\\
\end{pmatrix}$
We check if the first matrix is invertible.
$\left|A\right|=5-6=-1\neq 0$, so the matrix is invertible.
We multiply to the right by its inverse.
$X
\begin{pmatrix}
1 & 3\\
2 & 5
\end{pmatrix}\cdot
\begin{pmatrix}
1 & 3\\
2 & 5
\end{pmatrix}^{-1}=
\begin{pmatrix}
3 & 5\\
2 & 1
\end{pmatrix}\cdot
\begin{pmatrix}
1 & 3\\
2 & 5
\end{pmatrix}^{-1}$
$X\cdot I_{2}= \begin{pmatrix} 3 & 5\\ 2 & 1 \end{pmatrix}\cdot \begin{pmatrix} 1 & 3\\ 2 & 5 \end{pmatrix}^{-1}$
$X=\begin{pmatrix} 3 & 5\\ 2 & 1 \end{pmatrix}\cdot \begin{pmatrix} 1 & 3\\ 2 & 5 \end{pmatrix}^{-1}$
$\begin{pmatrix} 1 & 3\\ 2 & 5 \end{pmatrix}^{-1}= \begin{pmatrix} -5 & 3\\ 2 & -1 \end{pmatrix}\rightarrow X= \begin{pmatrix} 3 & 5\\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} -5 & 3\\ 2 & -1 \end{pmatrix}= \begin{pmatrix} -5 & 4\\ -8 & 5 \end{pmatrix}$
