Solving Linear Systems with Matrices

Everything in this chapter converges here: determinants decide whether a system of linear equations has a solution, and the inverse computes it.

Consistent system: at least one solution exists. Inconsistent: no solution exists.

The setup: $AX = B$

The system $$a_1 x + b_1 y + c_1 z = d_1, \qquad a_2 x + b_2 y + c_2 z = d_2, \qquad a_3 x + b_3 y + c_3 z = d_3$$ packs into a single matrix equation: $$\underbrace{\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} x \\ y \\ z \end{pmatrix}}_{X} = \underbrace{\begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix}}_{B}$$

$A$ holds the coefficients, $X$ the unknowns, $B$ the constants.

Case I — $A$ nonsingular: the matrix method

If $\vert A \vert \neq 0$, premultiply both sides by $A^{-1}$: $$A^{-1}(AX) = A^{-1}B \;\Rightarrow\; (A^{-1}A)X = A^{-1}B \;\Rightarrow\; \boxed{\;X = A^{-1}B\;}$$

Since the inverse is unique, this solution is unique — the system is consistent with exactly one answer. Watch the order: it is $A^{-1}B$, never $BA^{-1}$ (the orders would not even match).

Case II — $A$ singular: test with $(\operatorname{adj} A)B$

If $\vert A \vert = 0$ the inverse does not exist, and the tie-breaker is the product $(\operatorname{adj} A)B$:

1. $(\operatorname{adj} A)B \neq O$: no solution — the system is inconsistent.
2. $(\operatorname{adj} A)B = O$: the system may be consistent (with infinitely many solutions) or inconsistent — this case does not settle by itself.

(NCERT restricts worked systems to the unique-solution case; the table is what the examiner expects you to quote.)

Solved Examples

Example 1 — A $2 \times 2$ system by the matrix method

Solve $2x + 5y = 1$ and $3x + 2y = 7$.

Step 1 — pack into $AX = B$: $A = \begin{pmatrix} 2 & 5 \\ 3 & 2 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \end{pmatrix}$, $B = \begin{pmatrix} 1 \\ 7 \end{pmatrix}$.

Step 2 — check $\vert A \vert$: $4 - 15 = -11 \neq 0$ — unique solution guaranteed.

Step 3 — invert and multiply: $$A^{-1} = -\frac{1}{11}\begin{pmatrix} 2 & -5 \\ -3 & 2 \end{pmatrix}, \qquad X = A^{-1}B = -\frac{1}{11}\begin{pmatrix} 2 - 35 \\ -3 + 14 \end{pmatrix} = -\frac{1}{11}\begin{pmatrix} -33 \\ 11 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$$

Answer: $x = 3$, $y = -1$. Check in the second equation: $9 - 2 = 7$. ✓

Example 2 — A $3 \times 3$ system by the matrix method

Solve $3x - 2y + 3z = 8$, $\;2x + y - z = 1$, $\;4x - 3y + 2z = 4$.

Step 1 — pack: $A = \begin{pmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{pmatrix}$, $B = \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix}$.

Step 2 — determinant: expanding along $R_1$: $3(2 - 3) + 2(4 + 4) + 3(-6 - 4) = -3 + 16 - 30 = -17 \neq 0$.

Step 3 — adjoint and inverse: computing the nine cofactors and transposing: $$\operatorname{adj} A = \begin{pmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{pmatrix}, \qquad A^{-1} = -\frac{1}{17}\begin{pmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{pmatrix}$$

Step 4 — multiply: $$X = A^{-1}B = -\frac{1}{17}\begin{pmatrix} -8 - 5 - 4 \\ -64 - 6 + 36 \\ -80 + 1 + 28 \end{pmatrix} = -\frac{1}{17}\begin{pmatrix} -17 \\ -34 \\ -51 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$

Answer: $x = 1$, $y = 2$, $z = 3$. Check the first equation: $3 - 4 + 9 = 8$. ✓

Example 3 — From words to a system

The sum of three numbers is $6$. If the third number is multiplied by $3$ and the second is added, the result is $11$. Adding the first and third numbers gives double the second. Find the numbers by the matrix method.

Step 1 — translate: with numbers $x, y, z$: $$x + y + z = 6, \qquad y + 3z = 11, \qquad x + z = 2y \;\Rightarrow\; x - 2y + z = 0$$

Step 2 — pack and test: $A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{pmatrix}$; expanding, $\vert A \vert = 1(1 + 6) - 1(0 - 3) + 1(0 - 1) = 7 + 3 - 1 = 9 \neq 0$.

Step 3 — solve $X = A^{-1}B$ with $B = (6, 11, 0)^{\prime}$: the computation gives $X = (1, 2, 3)^{\prime}$.

Answer: the numbers are $1$, $2$ and $3$. Check the second condition: $2 + 3(3) = 11$. ✓

Example 4 — Detecting an inconsistent system

Examine the consistency of $x + 3y = 5$ and $2x + 6y = 8$.

Step 1 — determinant: $A = \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}$, $\vert A \vert = 6 - 6 = 0$ — the matrix method stops here; move to the adjoint test.

Step 2 — compute $(\operatorname{adj} A)B$: $$(\operatorname{adj} A)B = \begin{pmatrix} 6 & -3 \\ -2 & 1 \end{pmatrix}\begin{pmatrix} 5 \\ 8 \end{pmatrix} = \begin{pmatrix} 30 - 24 \\ -10 + 8 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \end{pmatrix} \neq O$$

Answer: inconsistent — no solution. Geometrically the two equations are parallel lines: doubling the first gives $2x + 6y = 10$, which contradicts $2x + 6y = 8$.