Q.Solve the following system of linear equations by matrix inversion method:

Step 1. Part (i): set up $AX=B$. $2x+5y=-2,\ x+2y=-3 \Rightarrow A=\begin{pmatrix}2&5\\1&2\end{pmatrix},\ X=\begin{pmatrix}x\\y\end{pmatrix},\ B=\begin{pmatrix}-2\\-3\end{pmatrix}$.

$|A|=2(2)-5(1)=4-5=-1 \neq 0$, so $A$ is invertible. For $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, $\operatorname{adj}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}=\begin{pmatrix}2&-5\\-1&2\end{pmatrix}$, so $A^{-1}=\dfrac{1}{-1}\begin{pmatrix}2&-5\\-1&2\end{pmatrix}=\begin{pmatrix}-2&5\\1&-2\end{pmatrix}$.

Step 2. Part (i): compute $X=A^{-1}B$.

So $x=-11,\ y=4$.

Step 3. Part (ii): set up and invert. $2x-y=8,\ 3x+2y=-2 \Rightarrow A=\begin{pmatrix}2&-1\\3&2\end{pmatrix},\ B=\begin{pmatrix}8\\-2\end{pmatrix}$.

$|A|=2(2)-(-1)(3)=4+3=7$. $\operatorname{adj}A=\begin{pmatrix}2&1\\-3&2\end{pmatrix}$, so $A^{-1}=\dfrac17\begin{pmatrix}2&1\\-3&2\end{pmatrix}$.

Step 4. Part (ii): compute $X=A^{-1}B$.

So $x=2,\ y=-4$.

Step 5. Part (iii): set up $AX=B$. $2x+3y-z=9,\ x+y+z=9,\ 3x-y-z=-1 \Rightarrow A=\begin{pmatrix}2&3&-1\\1&1&1\\3&-1&-1\end{pmatrix},\ B=\begin{pmatrix}9\\9\\-1\end{pmatrix}$.

Step 6. Part (iii): compute $|A|$ (expand along row 1).

Step 7. Part (iii): compute the cofactors and assemble $\operatorname{adj}A$.

$C_{11}=0,\ C_{12}=4,\ C_{13}=-4;\quad C_{21}=4,\ C_{22}=1,\ C_{23}=11;\quad C_{31}=4,\ C_{32}=-3,\ C_{33}=-1$.

Transposing the cofactor matrix, $\operatorname{adj}A=\begin{pmatrix}0&4&4\\4&1&-3\\-4&11&-1\end{pmatrix}$, so $A^{-1}=\dfrac{1}{16}\begin{pmatrix}0&4&4\\4&1&-3\\-4&11&-1\end{pmatrix}$.

Step 8. Part (iii): compute $X=A^{-1}B$.

So $x=2,\ y=3,\ z=4$.

Step 9. Part (iv): set up $AX=B$. $x+y+z-2=0,\ 6x-4y+5z-31=0,\ 5x+2y+2z=13$ rewrite as $x+y+z=2,\ 6x-4y+5z=31,\ 5x+2y+2z=13$, so $A=\begin{pmatrix}1&1&1\\6&-4&5\\5&2&2\end{pmatrix},\ B=\begin{pmatrix}2\\31\\13\end{pmatrix}$.

Step 10. Part (iv): compute $|A|$ (expand along row 1).

Step 11. Part (iv): compute the cofactors and assemble $\operatorname{adj}A$.

$C_{11}=-18,\ C_{12}=13,\ C_{13}=32;\quad C_{21}=0,\ C_{22}=-3,\ C_{23}=3;\quad C_{31}=9,\ C_{32}=1,\ C_{33}=-10$.

Transposing, $\operatorname{adj}A=\begin{pmatrix}-18&0&9\\13&-3&1\\32&3&-10\end{pmatrix}$, so $A^{-1}=\dfrac1{27}\begin{pmatrix}-18&0&9\\13&-3&1\\32&3&-10\end{pmatrix}$.

Step 12. Part (iv): compute $X=A^{-1}B$.

So $x=3,\ y=-2,\ z=1$.

• Using the raw cofactor matrix as $\operatorname{adj}A$ instead of transposing it
• Multiplying in the wrong order, computing $BA^{-1}$ instead of $A^{-1}B$
• Sign error in the $2\times2$ adjugate shortcut $\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$