How One Matrix Equation Solves Three Equations at Once

Three equations in three unknowns can be written as a single matrix equation, $AX=B$. The determinant of A then tells you at once whether there is exactly one solution, and the inverse of A produces that solution in one step.

This part covers consistency and the number of solutions, solving with the inverse matrix, known in the ICSE syllabus as Martin's Rule, and verifying the solution.

**Writing the system as $AX=B$, it has a unique solution when $|A|\ne 0$; when $|A|=0$, compute $(\text{adj}A)B$ — if it is not the zero matrix there is no solution, and if it is zero the system has infinitely many solutions or none, which must be checked further.

Summary:**

- $|A|\ne 0$ — consistent, with a unique solution
- $|A|=0$ and $(\text{adj}A)B\ne O$ — inconsistent, with no solution
- $|A|=0$ and $(\text{adj}A)B=O$ — infinitely many solutions or none

Worked example (inconsistent). For $x+2y=4$ and $2x+4y=10$, $|A|=4-4=0$ and

$$(\text{adj}A)B=\left[\begin{array}{cc}4& -2\\ -2& 1\end{array}\right]\left[\begin{array}{c}4\\ 10\end{array}\right]=\left[\begin{array}{c}-4\\ 2\end{array}\right]$$

which is not zero, so there is no solution — the lines are parallel.

Worked example (infinitely many). For $x+2y=4$ and $2x+4y=8$, $|A|=0$ and $(\text{adj}A)B=\left[\begin{array}{c}16-16\\ -8+8\end{array}\right]=O$; the second equation is twice the first, so every point on the line is a solution.

An everyday example. If 2 pens and 1 notebook cost ₹50, and 4 pens and 2 notebooks cost ₹100, the second fact adds nothing new, so the separate prices cannot be found.

The substance. A zero determinant never gives a unique solution — it signals parallel or identical lines, or planes that do not meet at a single point.

**When $|A|\ne 0$, the system $AX=B$ has the unique solution $X=A^{-1}B$, found by computing $A^{-1}=\frac{1}{|A|}\text{adj}A$ and multiplying it by the column of constants B.

Steps:**

- Write A, X and B, keeping the variables in the same order in every equation
- Check that $|A|\ne 0$
- Find the adjoint and $A^{-1}$
- Multiply $A^{-1}B$ and read off the unknowns

Worked example. Solve $x+2y=5$, $y+3z=11$ and $2x+z=5$.

Here $A=\left[\begin{array}{ccc}1& 2& 0\\ 0& 1& 3\\ 2& 0& 1\end{array}\right]$ and $B=\left[\begin{array}{c}5\\ 11\\ 5\end{array}\right]$, with $|A|=13$ and

$$A^{-1}=\frac{1}{13}\left[\begin{array}{ccc}1& -2& 6\\ 6& 1& -3\\ -2& 4& 1\end{array}\right]$$

$$X=\frac{1}{13}\left[\begin{array}{c}5-22+30\\ 30+11-15\\ -10+44+5\end{array}\right]=\frac{1}{13}\left[\begin{array}{c}13\\ 26\\ 39\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$$

So $x=1$, $y=2$ and $z=3$.

An everyday example. A canteen working out the prices of tea, samosas and biscuits from three customers' bills sets up exactly this kind of system.

The substance. **The order matters: $X=A^{-1}B$, not $B{A}^{-1}$** — the product $B{A}^{-1}$ does not even exist when B is a $3\times 1$ column.

A solution is verified by substituting the values back into every original equation, or equivalently by checking that the product AX equals B.

Worked example. For $x=1$, $y=2$ and $z=3$ above:

- $x+2y=1+4=5$, as required
- $y+3z=2+9=11$, as required
- $2x+z=2+3=5$, as required

In matrix form, $AX=\left[\begin{array}{c}1+4+0\\ 0+2+9\\ 2+0+3\end{array}\right]=\left[\begin{array}{c}5\\ 11\\ 5\end{array}\right]=B$.

Worked example 2. Solve and verify $3x+4y=18$ and $2x+5y=19$. With $A^{-1}=\frac{1}{7}\left[\begin{array}{cc}5& -4\\ -2& 3\end{array}\right]$,

$$X=\frac{1}{7}\left[\begin{array}{c}90-76\\ -36+57\end{array}\right]=\frac{1}{7}\left[\begin{array}{c}14\\ 21\end{array}\right]=\left[\begin{array}{c}2\\ 3\end{array}\right]$$

Check: $3(2)+4(3)=18$ and $2(2)+5(3)=19$.

An everyday example. A shopkeeper matching a calculated bill against the printed receipt is doing the same kind of check before trusting the answer.

The substance. Verification catches slips in the adjoint — a single wrong cofactor almost always makes at least one equation fail.