Solving a System of Linear Equations (Matrix Method) & Consistency

MATSystems of Linear Equations

A system of linear equations can be written in matrix form $AX = B$ , where $A$ is the coefficient matrix, $X$ the column of unknowns and $B$ the column of constants. The matrix method solves it as $X = A^{-1}B$ whenever $A$ is non-singular, and the value of $|A|$ together with the product $(\text{adj}\,A)B$ decides whether the system is consistent (a unique or infinitely many solutions) or inconsistent.

This is the central application of determinants in the ISC syllabus and frequently carries a word problem reducible to three equations in three unknowns.

Matrix form of a system

AX = B \quad\Rightarrow\quad X = A^{-1}B = \dfrac{1}{|A|}(\text{adj}\,A)\,B

A

is the coefficient matrix,

X

the variable column,

B

the constant column; valid only when

|A| \ne 0

Three-variable coefficient matrix

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

Rows are formed from the coefficients of

x,y,z

in each equation;

d_1,d_2,d_3

are the right-hand constants.

Inverse via adjoint

A^{-1} = \dfrac{1}{|A|}\,\text{adj}\,A,\qquad |A| \ne 0

\text{adj}\,A

is the transpose of the cofactor matrix; required to compute

X = A^{-1}B

Consistency test

\begin{cases} |A| \ne 0 & \text{unique solution} \\ |A| = 0,\ (\text{adj}\,A)B \ne O & \text{no solution (inconsistent)} \\ |A| = 0,\ (\text{adj}\,A)B = O & \text{infinitely many solutions} \end{cases}

O

is the null column matrix; this classifies the system after computing

|A|

Step 1: write the equations in the standard order $x, y, z$ (insert a $0$ coefficient for any missing variable) and form $A$ , $X$ , $B$ so that $AX = B$ .
Step 2: evaluate $|A|$ . If $|A| \ne 0$ the system is consistent with a unique solution $X = A^{-1}B$ .
Step 3: find $\text{adj}\,A$ (transpose of the cofactor matrix), then $A^{-1} = \dfrac{1}{|A|}\text{adj}\,A$ , and finally multiply $A^{-1}B$ to read off $x, y, z$ .
When $|A| = 0$ the matrix is singular: $A^{-1}$ does not exist, so compute $(\text{adj}\,A)\,B$ to decide between no solution and infinitely many.
A homogeneous system $AX = O$ always has the trivial solution $x=y=z=0$ ; it has non-trivial solutions only if $|A| = 0$ .
For word problems, define the unknowns clearly, translate each condition into one linear equation, then apply the matrix method; the inverse must pre-multiply, so the order $X = A^{-1}B$ is essential.
Always verify the final answer by substituting $x, y, z$ back into the original equations.

Where the marks go

Computing $B A^{-1}$ instead of $A^{-1}B$ ; the inverse must pre-multiply, giving $X = A^{-1}B$ .
Forgetting to insert a $0$ coefficient for a missing variable (e.g. $2x - 3y = 1$ has $z$ -coefficient $0$ ), which corrupts the matrix $A$ .
Declaring a system inconsistent the moment $|A| = 0$ without checking $(\text{adj}\,A)B$ ; $|A|=0$ can still allow infinitely many solutions.
Sign errors in cofactors while building $\text{adj}\,A$ , or dividing by $|A|$ before taking the transpose.

How the board asks it

Numericalmatrix equation $AX = B$ solved as $X = A^{-1}B$
Using matrices, solve the following system of equations: $2x - 3y + 5z = 11$ , $3x + 2y - 4z = -5$ , $x + y - 2z = -3$ .
Numericalinverse via adjoint, then $X = A^{-1}B$
If $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$ , find $A^{-1}$ and hence solve the system $x - y + z = 4$ , $2x + y - 3z = 0$ , $x + y + z = 2$ .
Applicationword problem reducible to three equations in three unknowns
The cost of $4\,\text{kg}$ onion, $3\,\text{kg}$ wheat and $2\,\text{kg}$ rice is $60$ rupees; the cost of $2\,\text{kg}$ onion, $4\,\text{kg}$ wheat and $6\,\text{kg}$ rice is $90$ rupees; and the cost of $6\,\text{kg}$ onion, $2\,\text{kg}$ wheat and $3\,\text{kg}$ rice is $70$ rupees. Using the matrix method, find the cost per kg of each item.
Identify / classifyconsistency test using $|A|$ and $(\text{adj}\,A)B$
Examine the consistency of the system $x + 2y + z = 3$ , $2x + 3y + 2z = 5$ , $3x + 5y + 3z = 8$ , and state whether it has a unique solution, infinitely many solutions or no solution.
Numericalhomogeneous system has a non-trivial solution only if $|A| = 0$
Find the value of $\lambda$ for which the homogeneous system $\lambda x + 2y + z = 0$ , $2x + y - z = 0$ , $x - y + 2z = 0$ has a non-trivial solution.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.