How a Matrix Inverse Solves a Whole System of Equations at Once

A set of linear equations can be written as a single matrix equation $AX=B$. If $A$ has an inverse, the solution is simply $X=A^{-1}B$ — and determinants tell us when that inverse exists.

This part covers the adjoint, singular matrices and the inverse, the consistency of linear systems, and the matrix method.

**The adjoint of a square matrix is the transpose of its matrix of cofactors, and multiplying a matrix by its adjoint in either order gives $|A|$ times the identity matrix.

Construction.** Replace every entry by its cofactor, then take the transpose:

$$\text{adj }A=[A_{ij}{]}^{\prime }$$

Shortcut for order 2. For $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, swap the diagonal entries and change the signs of the other two:

$$\text{adj }A=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$$

Worked example — verify. Let $A=\left[\begin{array}{cc}2& 3\\ 1& 4\end{array}\right]$, so $|A|=8-3=5$ and $\text{adj }A=\left[\begin{array}{cc}4& -3\\ -1& 2\end{array}\right]$.

$$A(\text{adj }A)=\left[\begin{array}{cc}8-3& -6+6\\ 4-4& -3+8\end{array}\right]=\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]=5I$$

An everyday example. A cook who has scaled up every ingredient can recover the original recipe by reversing the steps and then dividing — the adjoint handles the reversal, and $|A|$ the scaling.

The substance. For order 3, forgetting the final transpose leaves the cofactor matrix, which is not the adjoint.

**A square matrix is singular if its determinant is zero and non-singular otherwise, and only a non-singular matrix has an inverse, given by $A^{-1}=\frac{1}{|A|}\text{adj }A$.

Classification:

- Singular** — $|A|=0$; no inverse
- Non-singular — $|A|\ne 0$; invertible, with $A{A}^{-1}=A^{-1}A=I$

Worked example — an inverse. For $A=\left[\begin{array}{cc}2& 3\\ 1& 4\end{array}\right]$ with $|A|=5$:

$$A^{-1}=\frac{1}{5}\left[\begin{array}{cc}4& -3\\ -1& 2\end{array}\right]$$

Worked example — a singular matrix. $\left|\begin{array}{cc}2& 4\\ 3& 6\end{array}\right|=12-12=0$, so $\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]$ has no inverse — its second row is $1.5$ times its first.

An everyday example. Converting a recipe from 4 servings to 10 and back again works because the scaling can be undone; a step that throws information away, like a singular matrix, cannot be reversed.

The substance. **Check $|A|\ne 0$ before computing an inverse** — the adjoint exists even when the inverse does not.

**For $AX=B$: if $|A|\ne 0$ the system is consistent with a unique solution; if $|A|=0$, compute $(\text{adj }A)B$ — a non-zero result means no solution, and a zero result means infinitely many solutions or none.

The test:**

- $|A|\ne 0$ — unique solution
- $|A|=0$ and $(\text{adj }A)B\ne O$ — no solution
- $|A|=0$ and $(\text{adj }A)B=O$ — infinitely many solutions or none; check further

Worked example — inconsistent. $x+2y=3$ and $2x+4y=7$.

$$|A|=\left|\begin{array}{cc}1& 2\\ 2& 4\end{array}\right|=0(\text{adj }A)B=\left[\begin{array}{cc}4& -2\\ -2& 1\end{array}\right]\left[\begin{array}{c}3\\ 7\end{array}\right]=\left[\begin{array}{c}-2\\ 1\end{array}\right]\ne O$$

The system is inconsistent: the lines are parallel.

Worked example — infinitely many. For $x+2y=3$ and $2x+4y=6$, $(\text{adj }A)B=\left[\begin{array}{c}12-12\\ -6+6\end{array}\right]=O$, and both equations describe the same line.

An everyday example. If one shop says 1 pen and 2 pencils cost ₹30, and another says 2 pens and 4 pencils cost ₹70, no single set of prices fits both — the system is inconsistent.

The substance. **$|A|=0$ alone never decides between no solution and infinitely many** — the extra check is essential.

**Write the system as $AX=B$, check that $|A|\ne 0$, find $A^{-1}$ using the adjoint, and compute $X=A^{-1}B$ to read off the unique solution.

Worked example — two variables.** Solve $2x+3y=13$ and $x+4y=14$.

$$A=\left[\begin{array}{cc}2& 3\\ 1& 4\end{array}\right],B=\left[\begin{array}{c}13\\ 14\end{array}\right],|A|=5$$

$$X=A^{-1}B=\frac{1}{5}\left[\begin{array}{cc}4& -3\\ -1& 2\end{array}\right]\left[\begin{array}{c}13\\ 14\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}52-42\\ -13+28\end{array}\right]=\left[\begin{array}{c}2\\ 3\end{array}\right]$$

So $x=2$ and $y=3$. Check: $2(2)+3(3)=13$ and $2+4(3)=14$.

Three variables. The same steps work for a $3\times 3$ system, using all nine cofactors.

An everyday example. A shopkeeper who knows the total cost of two different mixes of rice and dal can find the price per kilogram of each by solving exactly this kind of system.

The substance. **The method needs $|A|\ne 0$** — when the determinant is zero, use the consistency test instead.