Method of Inversion

As the name suggests, this method uses the inverse of the coefficient matrix directly. Consider three linear equations

As explained in section 2.3, these can be expressed as

where $A,X,B$ are of orders $3\times3$, $3\times1$, $3\times1$ respectively.

The derivation. If the system's solution exists, $A$ must be non-singular, so $A^{-1}$ exists (found either by the elementary-transformation method or the adjoint method of section 2.2). Pre-multiplying both sides of $AX=B$ by $A^{-1}$:

which gives the required solution -- every unknown is read off from the single matrix product $A^{-1}B$.

Worked Example -- two equations. Solve $2x+5y=1$ and $3x+2y=7$ by the method of inversion. Writing $AX=B$ with $A=\begin{bmatrix}2&5\\3&2\end{bmatrix}$, $X=\begin{bmatrix}x\\y\end{bmatrix}$, $B=\begin{bmatrix}1\\7\end{bmatrix}$: $|A|=2(2)-3(5)=4-15=-11$, and $\text{adj}\,A=\begin{bmatrix}2&-5\\-3&2\end{bmatrix}$, so $A^{-1}=\dfrac{1}{-11}\begin{bmatrix}2&-5\\-3&2\end{bmatrix}$. Then

By equality of matrices, $x=3$ and $y=-1$.

Worked Example -- three equations. Solve $x-y+z=4$, $2x+y-3z=0$, $x+y+z=2$ by the method of inversion. The matrix equation is $\begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\0\\2\end{bmatrix}$, i.e. $AX=B$. Here $\text{adj}\,A=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$ and $|A|=10$, so $A^{-1}=\dfrac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}$. Then

By equality of matrices, $x=2$, $y=-1$, $z=1$. …