Systems of linear equations

Three linear equations in three unknowns describe three flat planes in space, and "solving the system" means finding the single point $(x,y,z)$ that lies on all three at once — if such a point exists at all. ISC Class 12 packages this search into a compact matrix recipe called Martin's Rule: write the equations as one matrix equation $AX=B$ , and if the coefficient matrix $A$ is invertible, the unknowns fall out in a single stroke as $X=A^{-1}B$ . This article rebuilds that rule from the algebra of matrix inverses you already know, works a full three-variable example cofactor by cofactor, and shows — through a second example — how to read the case $|A|=0$ correctly, since a vanishing determinant can mean either no solution or infinitely many, and the difference matters on the exam.

The core insight

Writing a linear system as $AX=B$ turns "solve for $x,y,z$ " into "multiply by $A^{-1}$ ," so a unique solution exists exactly when $|A|\neq0$ , giving $X=A^{-1}B=\dfrac{\operatorname{Adj}A}{|A|}B$ — Martin's Rule.

The core idea

A system of three linear equations in three unknowns,

a_1x+b_1y+c_1z=d_1,\qquad a_2x+b_2y+c_2z=d_2,\qquad a_3x+b_3y+c_3z=d_3,

packs its coefficients, unknowns and constants into three matrices, so that the whole system becomes one matrix equation $AX=B$ , where

A=\begin{pmatrix}a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3\end{pmatrix},\qquad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\qquad B=\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}.

where: $A$ is the coefficient matrix built from the numbers multiplying $x$ , $y$ and $z$ ; $X$ is the column of unknowns; $B$ is the column of right-hand-side constants. Each row of $A$ is really the coefficients of one plane in three-dimensional space, and a solution $(x,y,z)$ is a point lying on all three planes simultaneously. Geometrically there are only three outcomes: the three planes meet in exactly one point (unique solution), they meet along a common line or coincide (infinitely many solutions), or they fail to share any common point at all (no solution). Algebra decides which case holds through a single number, $|A|$ , the determinant of the coefficient matrix.

\text{unique solution} \iff |A|\neq0.

where: $|A|$ is the determinant of the coefficient matrix; when it is non-zero, $A$ is invertible and exactly one point satisfies all three equations. This is worth pausing on, because it is the whole reason the chapter builds systems of equations directly on top of matrix inverses rather than teaching elimination or substitution as the primary tool. Once $A^{-1}$ (or, equivalently, $\operatorname{Adj}A/|A|$ ) has been worked out for a given coefficient matrix, any right-hand side $B$ for that same $A$ is solved instantly by one matrix multiplication — useful when several word problems share identical left-hand coefficients but different totals. The determinant test also mirrors the single-variable case a student already trusts: solving $ax=b$ for a number $a$ requires $a\neq0$ before writing $x=b/a$ ; solving $AX=B$ for a matrix $A$ requires $|A|\neq0$ before writing $X=A^{-1}B$ . The syllabus calls the three-outcome classification "consistency by examples" precisely because, unlike the clean if-and-only-if test for a unique solution, distinguishing "infinitely many" from "no solution" when $|A|=0$ is not handed to you by a single formula — it has to be read off the equations themselves, which the worked example below demonstrates.

The main formula — Martin's Rule

Derivation ADeriving Martin's Rule from A inverse

The goal is to show, starting only from the matrix equation $AX=B$ and the definition of a matrix inverse, exactly how the solution $X=A^{-1}B$ arises, and why substituting the adjoint formula for $A^{-1}$ turns this into the concrete working rule $X=\dfrac{\operatorname{Adj}A}{|A|}B$ that ISC calls Martin's Rule, so the formula is never applied as a memorised black box.

Step 1. Start from the packed system $AX=B$ , valid whenever the coefficient matrix $A$ is non-singular, meaning $|A|\neq0$ . Because $|A|\neq0$ , the inverse $A^{-1}$ exists.

Step 2. Multiply both sides of $AX=B$ on the left by $A^{-1}$ : $A^{-1}(AX)=A^{-1}B$ . Matrix multiplication is associative, so the left side regroups as $(A^{-1}A)X$ .

Step 3. Use the defining property of an inverse, $A^{-1}A=I$ , and the identity matrix's property $IX=X$ , to collapse the left side: $(A^{-1}A)X = IX = X$ . This gives the clean result $X=A^{-1}B$ — the unknowns are recovered by one matrix multiplication once $A^{-1}$ is known.

Step 4. Substitute the standard formula for an inverse in terms of the adjoint, $A^{-1}=\dfrac{\operatorname{Adj}A}{|A|}$ , into $X=A^{-1}B$ to reach the working rule $X=\dfrac{\operatorname{Adj}A}{|A|}B$ . Every entry of $\operatorname{Adj}A$ is a cofactor of $A$ , so this step converts an abstract inverse into an entirely computable object: nine cofactors, one determinant, one matrix product.

Worked example

Solve using Martin's Rule: $x+y+z=6,\ x-y+z=2,\ 2x+y-z=1$ .

Step 1. Form $A$ , $X$ and $B$ from the coefficients: $A=\begin{pmatrix}1&1&1\\1&-1&1\\2&1&-1\end{pmatrix}$ , $B=\begin{pmatrix}6\\2\\1\end{pmatrix}$ .

Step 2. Evaluate $|A|$ by expanding along the first row: $|A|=1(1-1)-1(-1-2)+1(1+2)=0+3+3=6$ . Since $|A|=6\neq0$ , a unique solution exists and Martin's Rule applies.

Step 3. Compute the nine cofactors and assemble the adjoint (transpose of the cofactor matrix): $C_{11}=0,\ C_{12}=3,\ C_{13}=3,\ C_{21}=2,\ C_{22}=-3,\ C_{23}=1,\ C_{31}=2,\ C_{32}=0,\ C_{33}=-2$ , giving $\operatorname{Adj}A=\begin{pmatrix}0&2&2\\3&-3&0\\3&1&-2\end{pmatrix}$ .

Step 4. Compute $X=\dfrac{1}{6}(\operatorname{Adj}A)B$ : $(\operatorname{Adj}A)B=\begin{pmatrix}0+4+2\\18-6+0\\18+2-2\end{pmatrix}=\begin{pmatrix}6\\12\\18\end{pmatrix}$ , so $X=\dfrac16\begin{pmatrix}6\\12\\18\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}$ . Substituting back confirms $1+2+3=6$ , $1-2+3=2$ and $2(1)+2-3=1$ , all satisfied. Final answer: $x=1,\ y=2,\ z=3$ .

Competitive stretch

Test $2x+3y=5,\ 4x+6y=10$ for consistency. Here $|A|=2(6)-3(4)=0$ , so Martin's Rule cannot be applied directly. But the second equation is exactly twice the first, so both describe the same line: the system is consistent with infinitely many solutions. Had the second equation instead read $4x+6y=9$ — not a clean multiple of the first — the two lines would be parallel and distinct, giving no solution, still with $|A|=0$ . This side-by-side check, comparing equations once $|A|=0$ is found, is exactly the "consistency by examples" reasoning the syllabus expects, since a zero determinant alone never tells you which of the two outcomes you're in.

Common traps

Applying $X=A^{-1}B$ when $|A|=0$ — the inverse simply does not exist there, so the formula cannot fire; the equations must be compared directly instead.
Writing $X=BA^{-1}$ instead of $X=A^{-1}B$ — matrix multiplication is not commutative, and the orders are not interchangeable.
Treating " $|A|=0$ " as automatically "no solution" — it can equally mean infinitely many solutions; only comparing the actual equations (are they multiples of one another, or contradictory?) settles which.
Skipping the back-substitution check — a single sign slip in one cofactor silently propagates into a wrong $x$ , $y$ or $z$ , and it is caught only by plugging the answer back into all three original equations.
Forgetting to transpose the cofactor matrix before calling it the adjoint — the cofactor matrix and $\operatorname{Adj}A$ look similar but are not interchangeable; only the transposed version satisfies $A\cdot\operatorname{Adj}A=|A|\,I$ , which is what makes the whole rule work.

Ask the tutor: if you'd like another Martin's Rule system worked cofactor by cofactor, or help spotting whether a given $|A|=0$ system is consistent or inconsistent, ask the AI tutor about "systems of linear equations."

Why does |A| equal zero sometimes give no solution and sometimes give infinitely many?

Because $|A|=0$ only tells you that the three planes (or two lines) fail to meet in a single unique point — it says nothing about whether they still share some points. If one equation is a genuine scalar multiple of another, the planes coincide or overlap along a line, giving infinitely many shared points. If the equations instead contradict each other (same left-hand pattern, different right-hand constant, or parallel non-coincident planes), no point satisfies both. You resolve the ambiguity by inspecting the equations themselves once $|A|=0$ is found, not from the determinant alone.

Why do we bother transposing the cofactor matrix to get the adjoint, instead of just using the cofactor matrix directly?

The formula $A^{-1}=\operatorname{Adj}A/|A|$ is built so that $A\cdot\operatorname{Adj}A=|A|\,I$ exactly, and that identity only holds after the transpose. Each entry of the plain cofactor matrix pairs a row of $A$ with the "wrong" row's cofactors when multiplied directly; transposing lines up every row of $A$ with the matching column of cofactors, which is precisely what makes the off-diagonal terms of the product collapse to zero and the diagonal terms collapse to $|A|$ .

Can Martin's Rule be used for a system with two equations in two unknowns, or only three in three, and is there a shortcut around computing all nine cofactors for a 3 by 3 system?

Both sizes work — the derivation never used the size of the matrix, only that $A$ is square and $|A|\neq0$ , so a $2\times2$ coefficient matrix follows the identical route with a $2\times2$ adjoint built from four cofactors. For $3\times3$ systems, nine cofactors is genuinely the expected cost within this syllabus; there is no shortcut around the method itself, though expanding $|A|$ along the row or column with the most zeros cuts arithmetic, and the same nine minors feed both $|A|$ and $\operatorname{Adj}A$ , so nothing is wasted. Cross-checking with the final substitution step remains the safety net either way.