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# Solving Linear Equations Using Matrix

Solving linear equations using matrix is done by two prominent methods, namely the matrix method and row reduction or the Gaussian elimination method. In this article, we will look at solving linear equations with matrix and related examples. With the study notes provided below, students will develop a clear idea about the topic.

## How to Solve Linear Equations Using Matrix Method?

Let the equations be:

$$\begin{array}{l}{{a}_{1}}x+{{a}_{2}}y+{{a}_{3}}z={{d}_{1}}\\\end{array}$$
$$\begin{array}{l}{{b}_{1}}x+{{b}_{2}}y+{{b}_{3}}z={{d}_{2}}\\\end{array}$$
$$\begin{array}{l}{{c}_{1}}x+{{c}_{2}}y+{{c}_{3}}z={{d}_{3}}\\\end{array}$$

The first method to find the solution to the system of equations is the matrix method. The steps to be followed are given below:

• All the variables in the equations should be written in the appropriate order.
• The variables, their coefficients and constants are to be written on the respective sides.

Solving a system of linear equations by the method of finding the inverse consists of two new matrices, namely

• Matrix A: which represents the variables
• Matrix B: which represents the constants

A system of equations can be solved using matrix multiplication.

We can write the above equations in the matrix form as follows:

$$\begin{array}{l}\left[ \begin{matrix} {{a}_{1}}x+{{a}_{2}}y+{{a}_{3}}z \\ {{b}_{1}}x+{{b}_{2}}y+{{b}_{3}}z \\ {{c}_{1}}x+{{c}_{2}}y+{{c}_{3}}z \\ \end{matrix} \right]=\left[ \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \right]\Rightarrow \left[ \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \right]\\ \Rightarrow AX=B…..(i)\end{array}$$

Where,

$$\begin{array}{l}A=\left[ \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right],\ X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right],\ B=\left[ \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \right]\end{array}$$
.

A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

Multiplying (i) by A-1 we get

$$\begin{array}{l}{{A}^{-1}}AX={{A}^{-1}}B\Rightarrow I.X={{A}^{-1}}B\\\Rightarrow X={{A}^{-1}}B\end{array}$$

The second method to find the solution for the system of equations is row reduction or Gaussian elimination.

1. The augmented matrix for the linear equations is written.
2. Use elementary such that all the elements below the main diagonal are zero. If a zero is obtained on the diagonal, perform the row operation such that a nonzero element is obtained.
3. Back substitution is used to find the solution.

Related Topics:

## Solution to a System of Equations

A set of values of x, y, and z which simultaneously satisfy all the equations, is called a solution to the system of equations.

Consider, x + y + z = 9

2x – y + z = 5

4x + y – z = 7

Here, the set of values – x = 2, y = 3, z = 4, is a solution to the system of linear equations.

Because, 2 + 3 + 4 = 9, 4 – 3 + 4 = 5, 8 + 3 – 4 = 7

### Consistent Equations

If the system of equations has one or more solutions, then it is said to be a consistent system of equations; otherwise, it is an inconsistent system of equations. For example, the system of linear equations x + 3y = 5;  x – y = 1 is consistent because x = 2, y = 1 is a solution to it. However, the system of linear equations x + 3y = 5; 2x + 6y = 8 is inconsistent because there is no set of values of x and y, which may satisfy the two equations simultaneously.

Condition for consistency of a system of linear equation AX = B

(a) If |A| ≠ 0, then the system is consistent and has a unique solution, given by X =

$$\begin{array}{l}{{A}^{-1}}B\end{array}$$

(b) If |A| = 0, and (Adj A) B ≠ 0, then the system is inconsistent.

(c) If |A| = 0, and (Adj A) B = 0, then the system is consistent and has infinitely many solutions.

Note AX = 0 is known as the homogeneous system of linear equations, and here, B = 0. A system of homogeneous equations is always consistent.

The system has a non-trivial solution (non-zero solution), if | A | = 0

Theorem 1: Let AX = B be a system of linear equations, where A is the coefficient matrix. If A is invertible, then the system has a unique solution, given by X = A-1 B

Proof: AX = B; Multiplying both sides by A-1Since A-1 exists

$$\begin{array}{l}\Rightarrow \left| A \right|\ne 0\end{array}$$
$$\begin{array}{l}\Rightarrow \,\,{{A}^{-1}}AX={{A}^{-1}}B\\ \Rightarrow \,\,IX={{A}^{-1}}B\\ \Rightarrow \,\,X={{A}^{-1}}B\end{array}$$

Thus, the system of equations AX = B has a solution given by

$$\begin{array}{l}X = {{A}^{-1}}B\end{array}$$

Uniqueness: If AX = B has two sets of solutions X1 and X2, then AX1 = B and AX2 = B (Each equal to B).

⇒  AX1 = AX2

By cancellation law, A is invertible.

⇒  X1 = X2

Hence, the given system AX = B has a unique solution.

Note: A homogeneous system of equations is always consistent.

### Problems on Solving Linear Equations Using Matrix Method

Illustration: Let A =

$$\begin{array}{l}\left[ \begin{matrix} x+y & y \\ 2x & x-y \\ \end{matrix} \right],\ B=\left[ \begin{matrix} 2 \\ -1 \\ \end{matrix} \right],\ C = \left[ \begin{matrix} 3 \\ 2 \\ \end{matrix} \right]\end{array}$$
. If AB = C, then find the matrix A2.

Solution:

By solving AB = C, we get the values of x and y. Then by substituting these values in A, we obtain A2

Here

$$\begin{array}{l}\left[ \begin{matrix} x+y & y \\ 2x & x-y \\ \end{matrix} \right]\left[ \begin{matrix} 2 \\ -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 \\ 2 \\ \end{matrix} \right]\\ \Rightarrow \left[ \begin{matrix} 2\left( x+y \right)-y \\ 2x.2-\left( x-y \right) \\ \end{matrix} \right]=\left[ \begin{matrix} 3 \\ 2 \\ \end{matrix} \right]\\ \Rightarrow 2\left( x+y \right)-y=3 \:\:and \:\:4x-\left( x-y \right)=2\end{array}$$
$$\begin{array}{l}\Rightarrow \,\,2x+y=3\;\;\;\;and\;\;\;\;3x+y=2\end{array}$$
Subtracting the two equations, we get, x = -1 So, y = 5 .

$$\begin{array}{l}\therefore A=\left[ \begin{matrix} -1+5 & 5 \\ 2\left( -1 \right) & -1-5 \\ \end{matrix} \right]=\left[ \begin{matrix} 4 & 5 \\ -2 & -6 \\ \end{matrix} \right]\end{array}$$
$$\begin{array}{l}\therefore {{A}^{2}}=\left[ \begin{matrix} 4 & 5 \\ -2 & -6 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 5 \\ -2 & -6 \\ \end{matrix} \right]\end{array}$$
$$\begin{array}{l}=\left[ \begin{matrix} 4\times 4+5\left( -2 \right) & 4\times 5+5\left( -6 \right) \\ -2\times 4+\left( -6 \right)\left( -2 \right) & -2\times 5+\left( -6 \right)\left( -6 \right) \\ \end{matrix} \right]=\left[ \begin{matrix} 6 & -10 \\ 4 & 26 \\ \end{matrix} \right]\end{array}$$

Illustration: Solve the following equations by matrix inversion

2x+y+2z=0

2x-y+z=10

x+3y-z=5

Solution:

The given equation can be written in a matrix form as AX = D, and then by obtaining A-1 and multiplying it on both sides, we can solve the given problem.

$$\begin{array}{l}\left[ \begin{matrix} 2 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\ 10 \\ 5 \\ \end{matrix} \right]\end{array}$$

AX = D where A

$$\begin{array}{l}=\left[ \begin{matrix} 2 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{matrix} \right],X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right],D=\left[ \begin{matrix} 0 \\ 10 \\ 5 \\ \end{matrix} \right]\end{array}$$
$$\begin{array}{l}\Rightarrow \,\,{{A}^{-1}}\left( AX \right)={{A}^{-1}}D\\ \left( {{A}^{-1}}A \right)X={{A}^{-1}}D\\ \Rightarrow IX={{A}^{-1}}D\\ \Rightarrow X={{A}^{-1}}D…..(i)\end{array}$$

Now

$$\begin{array}{l}{{A}^{-1}}=\frac{adj\,\,A}{\left| A \right|}; \;\;\;\;\;\;\left| A \right|=\left| \begin{matrix} 2 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{matrix} \right|=2\left( 1-3 \right)-1\left( -2-1 \right)+2\left( 6+1 \right)=13\end{array}$$

The matrix of co-factors of

$$\begin{array}{l}\left| A \right|\;\; is \;\;\left[ \begin{matrix} -2 & 3 & 7 \\ 7 & -4 & -5 \\ 3 & 2 & -4 \\ \end{matrix} \right].\;\;So, \;\;adj\;\; A \;\;\;\;=\left[ \begin{matrix} -2 & 7 & 3 \\ 3 & -4 & 2 \\ 7 & -5 & -4 \\ \end{matrix} \right]\end{array}$$

$$\begin{array}{l}\,\,{{A}^{-1}}=\frac{1}{13}\left[ \begin{matrix} -2 & 7 & 3 \\ 3 & -4 & 2 \\ 7 & -5 & -4 \\ \end{matrix} \right]\end{array}$$

⇒ from (i),

$$\begin{array}{l} X=\frac{1}{13}\left[ \begin{matrix} -2 & 7 & 3 \\ 3 & -4 & 2 \\ 7 & -5 & -4 \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 10 \\ 5 \\ \end{matrix} \right]\end{array}$$
$$\begin{array}{l}=\frac{1}{13}\left[ \begin{matrix} 0+70+15 \\ 0-40+10 \\ 0-50-20 \\ \end{matrix} \right]=\left[ \begin{matrix} 85/13 \\ -30/13 \\ -70/13 \\ \end{matrix} \right]\end{array}$$

∴

$$\begin{array}{l}\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 85/13 \\ -30/13 \\ -70/13 \\ \end{matrix} \right]\Rightarrow \,\,x=\frac{85}{13},y=\frac{-30}{13},z=\frac{-70}{13}\end{array}$$

Illustration: If

$$\begin{array}{l}\left[ \begin{matrix} 2 & 1 \\ 7 & 4 \\ \end{matrix} \right]A\left[ \begin{matrix} -3 & 2 \\ 5 & -3 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right],\end{array}$$

then matrix A equals:

$$\begin{array}{l}(a) \left[ \begin{matrix} 7 & 5 \\ -11 & -8 \\ \end{matrix} \right] \;\;\;\;\;(b) \left[ \begin{matrix} 2 & 1 \\ 5 & 3 \\ \end{matrix} \right] \;\;\;\;\;\; (c)\left[ \begin{matrix} 7 & 1 \\ 34 & 5 \\ \end{matrix} \right] \;\;\;\;\;\; (d) \left[ \begin{matrix} 5 & 3 \\ 13 & 8 \\ \end{matrix} \right]\end{array}$$

Solution:

(a) We know that if XAY = I, then A

$$\begin{array}{l}={{X}^{-1}}{{Y}^{-1}}={{\left( YX \right)}^{-1}}.\end{array}$$

In this case

$$\begin{array}{l}YX=\left[ \begin{matrix} -3 & 2 \\ 5 & -3 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 1 \\ 7 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 8 & 5 \\ -11 & -7 \\ \end{matrix} \right]\end{array}$$

$$\begin{array}{l}A={{\left[ \begin{matrix} 8 & 5 \\ -11 & -7 \\ \end{matrix} \right]}^{-1}}=\left[ \begin{matrix} 7 & 5 \\ -11 & -8 \\ \end{matrix} \right]\end{array}$$

Illustration: The system of equations

$$\begin{array}{l}\left( \begin{matrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \\ \end{matrix} \right)\left( \begin{matrix} x \\ y \\ z \\ \end{matrix} \right)=\left( \begin{matrix} b \\ 3 \\ -1 \\ \end{matrix} \right)\end{array}$$

has no solution if a and b are

$$\begin{array}{l}(a) a=-3,b\ne 1/3\;\;\;\; (b) a=2/3,b\ne 1/3 \;\;\;(c)a\ne 1/4,b=1/3 \;\;\;\;(d) a\ne -3,b\ne 1/3\end{array}$$

Solution:

By applying row operation in the given matrices and comparing them, we can obtain the required result.

(a) The augmented matrix is given by (A|B)

$$\begin{array}{l}=\left( \begin{matrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \\ \end{matrix}\left| \begin{matrix} b \\ 3 \\ -1 \\ \end{matrix} \right. \right)\end{array}$$

Applying

$$\begin{array}{l}{{R}_{1}}\to 2{{R}_{1}}-{{R}_{2}},\end{array}$$
we get
$$\begin{array}{l}\left( A|B \right)\tilde{\ }\left( \begin{matrix} 1 & 4 & -7 \\ 5 & -8 & 9 \\ 2 & 1 & a \\ \end{matrix}\left| \begin{matrix} 2b-3 \\ 3 \\ -1 \\ \end{matrix} \right. \right)\end{array}$$

Applying

$$\begin{array}{l}{{R}_{2}}\to {{R}_{2}}-5{{R}_{1}},{{R}_{3}}\to{R}_{3}- 2{{R}_{1}},\end{array}$$
we get
$$\begin{array}{l}\left( A|B \right)\tilde{\ }\left( \begin{matrix} 1 & 4 & -7 \\ 0 & -28 & 44 \\ 0 & -7 & a+14 \\ \end{matrix}\left| \begin{matrix} 2b-3 \\ 18-10b \\ 5-4b \\ \end{matrix} \right. \right)\end{array}$$

The system of equations will have no solution if

$$\begin{array}{l}\frac{-28}{-7}=\frac{44}{a+14}\ne \frac{18-10b}{5-4b}\end{array}$$
$$\begin{array}{l}\Rightarrow \,\,a+14=11\;\;\;and\;\;\;20-16b\ne 18-10b\end{array}$$

$$\begin{array}{l}\Rightarrow \,a=-3 \;\;and \;\;b\ne -1/3 \end{array}$$
.

Illustration: Let A =

$$\begin{array}{l}\left( \begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ \end{matrix} \right).\end{array}$$

If u1 and u2 are column matrices such that

$$\begin{array}{l}A{{u}_{1}}=\left( \begin{matrix} 1 \\ 0 \\ 0 \\ \end{matrix} \right)\;\;\;\;\;and \;\;\;\;\;A{{u}_{2}}=\left( \begin{matrix} 0 \\ 1 \\ 0 \\ \end{matrix} \right), \;\;\;\;then \;\;\;\;\;{{u}_{1}}+{{u}_{2}}\end{array}$$

equals:

$$\begin{array}{l}(a) \left( \begin{matrix} -1 \\ 1 \\ -1 \\ \end{matrix} \right) \;\;\;\;\;\;\; (b) \left( \begin{matrix} -1 \\ -1 \\ 0 \\ \end{matrix} \right)\;\;\;\;\;\; (c)\left( \begin{matrix} 1 \\ -1 \\ -1 \\ \end{matrix} \right) \;\;\;\;\;\; (d) \left( \begin{matrix} -1 \\ 1 \\ 0 \\ \end{matrix} \right)\end{array}$$

Solution:

(c) Adding Au1 and Au2 we get

$$\begin{array}{l}A\left( {{u}_{1}}+{{u}_{2}} \right).\end{array}$$
Then using the invariance method we obtain
$$\begin{array}{l}{{u}_{1}}+{{u}_{2}}.\end{array}$$

$$\begin{array}{l}A\left( {{u}_{1}}+{{u}_{2}} \right)=A{{u}_{1}}+A{{u}_{2}}=\left( \begin{matrix} 1 \\ 0 \\ 0 \\ \end{matrix} \right)+\left( \begin{matrix} 0 \\ 1 \\ 0 \\ \end{matrix} \right)=\left( \begin{matrix} 1 \\ 1 \\ 0 \\ \end{matrix} \right)\end{array}$$

We then solve the above equation for

$$\begin{array}{l}{{u}_{1}}+{{u}_{2}},\end{array}$$
if we consider the augmented matrix (A|B) =
$$\begin{array}{l}\left( \begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ \end{matrix}\left| \begin{matrix} 1 \\ 1 \\ 0 \\ \end{matrix} \right. \right)\end{array}$$

Applying

$$\begin{array}{l}{{R}_{3}}\to {{R}_{3}}-2{{R}_{2}}+{{R}_{1}} \;\;and \;\;\;{{R}_{2}} to {{R}_{2}}-2{{R}_{1}}\end{array}$$
we get;

$$\begin{array}{l}(A|B)\tilde{\ }\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\left| \begin{matrix} 1 \\ -1 \\ -1 \\ \end{matrix} \right. \right)\end{array}$$
$$\begin{array}{l}\Rightarrow {{u}_{1}}+{{u}_{2}}=\left( \begin{matrix} 1 \\ -1 \\ -1 \\ \end{matrix} \right)\end{array}$$

Q1

### What do you mean by a linear equation?

A linear equation is an equation that has one or more variables having degree one.

Q2

### Name two methods to solve linear equations using matrices.

The matrix multiplication method and the Gaussian elimination method are two methods to solve linear equations using matrices.

Q3

### What do you mean by consistent equations?

A consistent system of equations is an equation having one or more solutions.

Q4

### Give the formula used in the matrix multiplication method for solving linear equations.

We use the formula AX = B in the matrix multiplication method for solving linear equations. Here, A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

Test your Knowledge on Solving linear equations