The adjoint identity $A(\operatorname{adj}A)=|A|I$ from the previous topic gives a direct formula for the inverse, and the inverse in turn solves systems of linear equations.
Singular vs non-singular
A square matrix $A$ is non-singular if $|A|\ne0$ and singular if $|A|=0$. Only non-singular matrices have an inverse, so the very first step in any inverse problem is to compute $|A|$.
The inverse formula
Divide the adjoint identity by $|A|$ (legal precisely when $|A|\ne0$):
$$A^{-1}=\frac{1}{|A|}\,\operatorname{adj}A,\qquad |A|\ne0.$$
If $|A|=0$ the matrix is singular and $A^{-1}$ does not exist. Useful companion facts: $|A^{-1}|=\dfrac{1}{|A|}$, $(A^{-1})^{-1}=A$, $(AB)^{-1}=B^{-1}A^{-1}$, and the inverse, when it exists, is unique.
Matrix method for a system $AX=B$
A system of linear equations can be written in matrix form $AX=B$, where $A$ is the coefficient matrix, $X$ the column of unknowns and $B$ the column of constants. For example,
$$\left.\begin{aligned}a_{1}x+b_{1}y+c_{1}z&=d_{1}\\a_{2}x+b_{2}y+c_{2}z&=d_{2}\\a_{3}x+b_{3}y+c_{3}z&=d_{3}\end{aligned}\right\}\;\Longleftrightarrow\;\underbrace{\begin{bmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{bmatrix}}_{A}\underbrace{\begin{bmatrix}x\\y\\z\end{bmatrix}}_{X}=\underbrace{\begin{bmatrix}d_{1}\\d_{2}\\d_{3}\end{bmatrix}}_{B}.$$
If $|A|\ne0$, multiply both sides on the left by $A^{-1}$ to get the unique solution
$$X=A^{-1}B.$$
This is closely related to Cramer's rule, where each variable equals a ratio of determinants — e.g. $x=\dfrac{|A_{x}|}{|A|}$ with $|A_{x}|$ formed by replacing the $x$-column of $A$ with $B$. Both methods need $|A|\ne0$.
Consistency — what $|A|$ tells you
A system is consistent if it has at least one solution and inconsistent if it has none.
For a homogeneous system $AX=O$: if $|A|\ne0$ the only solution is the trivial $x=y=z=0$; if $|A|=0$ there are infinitely many non-trivial solutions.
Deeper Insight — why $|A|=0$ breaks the whole method: The formula $X=A^{-1}B$ and Cramer's $x=|A_{x}|/|A|$ both divide by $|A|$. When $|A|=0$ that division is illegal, which is the algebra mirroring the geometry: a zero coefficient determinant means the equations are not truly independent — the planes (or lines) either meet in a whole line/plane (infinitely many solutions) or never meet at all (no solution). So $|A|$ is not just a number to compute; its vanishing is the single signal that separates "one clean answer" from "something has gone wrong with independence", and checking $(\operatorname{adj}A)B$ then tells you which of the two failures you face.
Find $A^{-1}$ for $A=\begin{bmatrix}2&3\\1&4\end{bmatrix}$.
Solution$|A|=2\cdot4-3\cdot1=5\ne0$. $\operatorname{adj}A=\begin{bmatrix}4&-3\\-1&2\end{bmatrix}$, so $A^{-1}=\dfrac{1}{5}\begin{bmatrix}4&-3\\-1&2\end{bmatrix}=\begin{bmatrix}\tfrac45&-\tfrac35\\[2pt]-\tfrac15&\tfrac25\end{bmatrix}.$
Answer: $A^{-1}=\begin{bmatrix}\tfrac45&-\tfrac35\\[2pt]-\tfrac15&\tfrac25\end{bmatrix}$
Solve $2x+3y=8,\ x+4y=9$ by the matrix method.
SolutionHere $A=\begin{bmatrix}2&3\\1&4\end{bmatrix}$, $B=\begin{bmatrix}8\\9\end{bmatrix}$. Using $A^{-1}$ from Example 1, $X=A^{-1}B=\dfrac15\begin{bmatrix}4&-3\\-1&2\end{bmatrix}\begin{bmatrix}8\\9\end{bmatrix}=\dfrac15\begin{bmatrix}32-27\\-8+18\end{bmatrix}=\dfrac15\begin{bmatrix}5\\10\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix}.$ So $x=1,\ y=2$.
Answer: $x=1,\ y=2$.
Is $A=\begin{bmatrix}1&2\\2&4\end{bmatrix}$ invertible?
Solution$|A|=1\cdot4-2\cdot2=0$, so $A$ is singular and has no inverse.
Answer: $|A|=0$, so $A$ is singular and has no inverse.
State the nature of the system whose coefficient determinant is non-zero.
SolutionIf $|A|\ne0$ the system $AX=B$ has the unique solution $X=A^{-1}B$ — it is consistent and independent.
Answer: The system has the unique solution $X=A^{-1}B$ (consistent and independent).
Find $A^{-1}$ for $A=\begin{bmatrix}1&2&3\\0&1&4\\5&6&0\end{bmatrix}$.
Solution$|A|$ (expand along column $1$) $=1\begin{vmatrix}1&4\\6&0\end{vmatrix}-0+5\begin{vmatrix}2&3\\1&4\end{vmatrix}=1(0-24)+5(8-3)=-24+25=1$. The cofactor matrix is $\begin{bmatrix}-24&20&-5\\18&-15&4\\5&-4&1\end{bmatrix}$; transposing gives $\operatorname{adj}A=\begin{bmatrix}-24&18&5\\20&-15&-4\\-5&4&1\end{bmatrix}$. Since $|A|=1$, $A^{-1}=\operatorname{adj}A=\begin{bmatrix}-24&18&5\\20&-15&-4\\-5&4&1\end{bmatrix}.$
Answer: $A^{-1}=\begin{bmatrix}-24&18&5\\20&-15&-4\\-5&4&1\end{bmatrix}$
Solve $x+y+z=6,\ y+3z=11,\ x+z=2y$ (i.e. $x-2y+z=0$) by the matrix method.
SolutionWrite $A=\begin{bmatrix}1&1&1\\0&1&3\\1&-2&1\end{bmatrix}$, $B=\begin{bmatrix}6\\11\\0\end{bmatrix}$. $|A|=1(1+6)-1(0-3)+1(0-1)=7+3-1=9\ne0$. Computing $\operatorname{adj}A$ and using $X=\tfrac{1}{9}(\operatorname{adj}A)B$ gives $X=\begin{bmatrix}1\\2\\3\end{bmatrix}$. So $x=1,\ y=2,\ z=3$. (Check: $1+2+3=6$, $2+9=11$, $1+3=2\cdot2$.)
Answer: $x=1,\ y=2,\ z=3$.
Examine the consistency of $2x+y=3,\ 4x+2y=7$.
Solution$A=\begin{bmatrix}2&1\\4&2\end{bmatrix}$, $|A|=4-4=0$, so the system is not uniquely solvable. $\operatorname{adj}A=\begin{bmatrix}2&-1\\-4&2\end{bmatrix}$ and $(\operatorname{adj}A)B=\begin{bmatrix}2&-1\\-4&2\end{bmatrix}\begin{bmatrix}3\\7\end{bmatrix}=\begin{bmatrix}6-7\\-12+14\end{bmatrix}=\begin{bmatrix}-1\\2\end{bmatrix}\ne O$. Hence the system is inconsistent (no solution).
Answer: The system is inconsistent (no solution).
If $A$ is a $3\times3$ matrix with $|A|=2$, find $|A^{-1}|$ and $|2A^{-1}|$.
Solution$|A^{-1}|=\dfrac{1}{|A|}=\dfrac12$. And $|2A^{-1}|=2^{3}|A^{-1}|=8\cdot\tfrac12=4.$
Answer: $|A^{-1}|=\dfrac12$ and $|2A^{-1}|=4$.
Solve, using $A^{-1}$, the system $x-y+z=4,\ 2x+y-3z=0,\ x+y+z=2$.
Solution$A=\begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}$, $B=\begin{bmatrix}4\\0\\2\end{bmatrix}$. $|A|=1(1+3)-(-1)(2+3)+1(2-1)=4+5+1=10\ne0$. Solving $X=\tfrac{1}{10}(\operatorname{adj}A)B$ gives $x=2,\ y=-1,\ z=1$. (Check: $2-(-1)+1=4$, $4-1-3=0$, $2-1+1=2$.)
Answer: $x=2,\ y=-1,\ z=1$.
Solve the system $5x+2y=4,\ 7x+3y=5$ by the matrix method.
Solution$A=\begin{bmatrix}5&2\\7&3\end{bmatrix}$, $|A|=15-14=1$. $A^{-1}=\dfrac11\begin{bmatrix}3&-2\\-7&5\end{bmatrix}$. $X=A^{-1}B=\begin{bmatrix}3&-2\\-7&5\end{bmatrix}\begin{bmatrix}4\\5\end{bmatrix}=\begin{bmatrix}12-10\\-28+25\end{bmatrix}=\begin{bmatrix}2\\-3\end{bmatrix}.$ So $x=2,\ y=-3$.
Answer: $x=2,\ y=-3$.
Use Cramer-style reasoning to solve $3x+y=7,\ x-2y=0$.
Solution$|A|=\begin{vmatrix}3&1\\1&-2\end{vmatrix}=-6-1=-7$. $|A_{x}|=\begin{vmatrix}7&1\\0&-2\end{vmatrix}=-14$ and $|A_{y}|=\begin{vmatrix}3&7\\1&0\end{vmatrix}=-7$. Then $x=\dfrac{|A_{x}|}{|A|}=\dfrac{-14}{-7}=2$ and $y=\dfrac{|A_{y}|}{|A|}=\dfrac{-7}{-7}=1.$ So $x=2,\ y=1$.
Answer: $x=2,\ y=1$.
Discuss the homogeneous system $x+2y=0,\ 3x+6y=0$.
Solution$A=\begin{bmatrix}1&2\\3&6\end{bmatrix}$, $|A|=6-6=0$. A homogeneous system with $|A|=0$ has infinitely many non-trivial solutions: here every $(x,y)$ with $x=-2y$ works, e.g. $(-2,1),(-4,2),\dots$ as well as the trivial $(0,0)$.
Answer: Infinitely many solutions: every $(x,y)$ with $x=-2y$ (including the trivial $(0,0)$).