4.3 Solving Systems of Linear Equations

A stationery shop sells notebooks, pens and geometry boxes. One customer pays $\text{₹}\,200$ for three notebooks, two pens and a box; another pays $\text{₹}\,110$ for one of each; a third pays $\text{₹}\,270$ for two notebooks, a pen and three boxes. What does each item cost? That is three linear equations in three unknowns. Elimination works, but it is a different sequence of steps for every system, and it does not tell you in advance whether the equations even have a solution.

Matrices turn every such system into a single equation $AX=B$ that looks just like $ax=b$ for numbers. For numbers, $ax=b$ has the unique solution $x=\tfrac ba$ when $a\ne0$; when $a=0$ there is either no solution ($0x=5$) or every $x$ works ($0x=0$). In this section you will see that the same three outcomes happen for systems, with $|A|$ playing the part of $a$ and $A^{-1}$ (from Section 4.2) playing the part of $\tfrac1a$.

Systems of Equations as $AX=B$; Consistency

The system $$\begin{aligned}a_1x+b_1y+c_1z&=d_1\\a_2x+b_2y+c_2z&=d_2\\a_3x+b_3y+c_3z&=d_3\end{aligned}$$ is exactly the matrix equation $AX=B$ with $$A=\begin{bmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{bmatrix},$$ $$X=\begin{bmatrix}x\\y\\z\end{bmatrix},\qquad B=\begin{bmatrix}d_1\\d_2\\d_3\end{bmatrix},$$ because the row-by-column product $AX$ (Section 3.2) has entries $a_1x+b_1y+c_1z$, and so on. $A$ is the coefficient matrix, $X$ the column of unknowns and $B$ the column of constants. Two equations in two unknowns work in the same way with $2\times2$ and $2\times1$ matrices.

For two equations in $x$ and $y$ each equation is a straight line, and a solution is a point on both lines. Exactly one of three things happens:

• the lines cross at one point — consistent, with a unique solution;
• the lines are parallel and distinct — inconsistent, no solution;
• the lines coincide — consistent, with infinitely many solutions.

The last two happen exactly when the lines have the same slope, i.e. when $\tfrac{a_1}{a_2}=\tfrac{b_1}{b_2}$, i.e. when $a_1b_2-a_2b_1=0$: the determinant of the coefficient matrix is zero. Three equations in $x,y,z$ are three planes in space, and the same three outcomes occur. The rest of the section turns this picture into tests you can apply without drawing anything.

Unique Solutions: the Matrix Method $X=A^{-1}B$

Proof. Since $|A|\ne0$, $A^{-1}$ exists (Theorem 4.6). It is a solution: $A(A^{-1}B)=(AA^{-1})B=IB=B$. It is the only one: if $AX=B$, multiply both sides on the left by $A^{-1}$: $A^{-1}AX=A^{-1}B$, i.e. $IX=X=A^{-1}B$. $\square$

This is the matrix method. It costs one inverse, but once $A^{-1}$ is known, every system with the same left-hand side — the same coefficients, different right-hand sides — is solved by a single multiplication.

When $|A|=0$: No Solution or Infinitely Many

If $A$ is singular there is no $A^{-1}$, so Theorem 4.7 is silent. The adjoint still exists, however, and the identity $(\operatorname{adj}A)A=|A|\,I$ of Theorem 4.4 still holds — with $|A|=0$ it says $(\operatorname{adj}A)A=O$. That is enough to detect many inconsistent systems.

Proof of (a). Suppose, on the contrary, that some $X$ satisfies $AX=B$. Multiply on the left by $\operatorname{adj}A$: $(\operatorname{adj}A)AX=(\operatorname{adj}A)B$. By Theorem 4.4 the left side is $|A|\,IX=0\cdot X=O$. So $(\operatorname{adj}A)B=O$ — contradicting the hypothesis. Hence there is no solution. $\square$

Part (b) cannot be sharpened using the adjoint alone. For example, the system $x+y+z=1$, $\ 2x+2y+2z=2$, $\ 3x+3y+3z=4$ has every $2\times2$ minor equal to $0$, so $\operatorname{adj}A=O$ and $(\operatorname{adj}A)B=O$ — yet it is inconsistent, because the first equation says $x+y+z=1$ and the third says $x+y+z=\tfrac43$. When $(\operatorname{adj}A)B=O$, always finish by solving.

Homogeneous systems. A system $AX=O$ (all constants zero) is always consistent, because $X=O$ — that is, $x=y=z=0$ — is a solution. By Theorem 4.7, if $|A|\ne0$ this is the only solution. Non-zero solutions can exist only when $|A|=0$; Example 6(b) with right-hand sides $(0,0,0)$ instead gives $(t,-2t,t)$.

Exercises