Matrix Method: Solve Linear Equations Like a Pro!

Today, we shall discuss the application of matrices and determinants, especially the Matrix Method, for solving the system of linear equations in three variables.

Let’s first understand the meaning of consistency of the system of linear equations.

Consistent System: A system of equations is called consistent if its solution (one or more) exists.

Inconsistent System: A system of equations is called inconsistent if its solution does not exist means it has no common solution.

In this very section, we shall restrict ourselves to the examples of system of linear equations having Unique Solution only (consistent system). (You can also refer to NCERT textbook for Class 12)

Table of Contents show

Consistency Check (for Unique Solution)

Before moving on to solving the system of linear equations by matrix method, firstly, we must check whether the given system is consistent or not.

For that, let’s consider a system of the following type:

$A \cdot X = B$

$\text{Here:}$

$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$

$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$

$\text{And:}$

$B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}$

Key Points:

$\textbf{Case 1: If } \text{det}(A) \neq 0$

A is a non-singular matrix, so its inverse exists.
The system is consistent.
It has a unique solution given by:

$\quad X = A^{-1} \cdot B.$

$\textbf{Case 2: If } \text{det}(A) = 0$

A is a singular matrix, therefore $A^{-1}$ does not exist.
Two possibilities:
1. If Adj (A) . B ≠ 0: No Solution. ( i. e. Incosistent)
2. If Adj (A) . B = 0: May be inconsistent or consistent with infinitely many solutions (in such a case we need to parametrize one variable and solve for others – we are not discussing this case in today’s section).

Solving System of Linear Equations Using the Matrix Method

Let’s understand the concept by taking up various types of questions.

In fact, I have included 8 different problems to solve for x, y and z using Matrix Method.

You can find a complete hand-written solutions of all of them in a single attached pdf at the end. I suggest you to download it for free without even logging into!

Here’s the first one:

Type 1: Standard “AX = B” Type Problem (Check Algorithm)

Question 1: Solve the following system of equations using the matrix method:

$x + 2y + 3z = 6$
$2x - y + z = 2$
$3x + 2y - 2z = 3$

CBSE 2023

Now to solve it, the following steps are required:

Step 1: Find determinant A. Check if it’s equal to zero or some non-zero number.

Step 2: If it’s non-zero (means Non-Singular), continue to find its Inverse Matrix in the next step. (We are not taking up the case here in this section, where determinant of A is zero)

Step 3: Using Adjoint Matrix of A, find its Inverse Matrix, as shown in the image below:

Finding Inverse Matrix Using Adjoint — *Watch it on YouTube*

Step 4: Get the Unique Solution of the given system by using the following:

$\quad X = A^{-1} \cdot B.$

Step 5 (My Suggestion): I strongly recommend you to verify your answer by putting the values so obtained in the given system of linear equations, just to make sure that your answer is correct!

Type 2: CBSE 2024 – An Easy PYQ

Question 2: Find the inverse of the matrix:

$A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}$

And use it to solve the following system of equations:

$x - 2y = 10$
$2x - y - z = 8$
$- 2y + z = 7$

CBSE 2024

This can be easily solved just like Type 1 above. See the solution in the attached pdf.

Type 3: Equations in Terms of Reciprocals

Question 3: Solve the following system of equations using matrices:

$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$

$\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1$

$\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$

where x, y, z ≠ 0

CBSE 2024

In such questions, we should do as follows:

$\text{Let:}$
$u = \frac{1}{x}, \quad v = \frac{1}{y}, \quad w = \frac{1}{z}.$

$\text{Rewrite the system as:}$

$2u + 3v + 10w = 4$
$4u - 6v + 5w = 1$
$6u + 9v - 20w = 2$

Now, solve the system using the matrix method, just as in type 1 or 2 above.

Don’t forget to take the reciprocals of u, v, w at the end to get the values of x, y and z.

Type 4: Finding Inverse Using Product of Two Matrices

Question 4: Compute the product AB, where

$A = \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix}$ and

$B = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}$

Then use it to solve the system, $\quad B^T \cdot X = C$ for a given C.

CBSE 2023

Do check the complete question and solution in the attached PDF. This is an important one. Must give it a try!

Type 5: Solving Word Problem Using Matrix Method

Question 5: Gautam buys 5 pens, 3 bags, and 1 instrument boxes for Rs. 160, Vikram buys 2 pens, 1 bag, and 3 instrument boxes for Rs. 190 and Ankur buys 1 pen, 2 bags, and 4 instrument box for Rs. 250.

Based on the above information, answer the following:

(i) Convert the given situation into the matrix equation $\quad A \cdot X = B.$

$\text{(ii) Find } \text{det}(A).$

$\text{(iii) If } A \text{ is invertible, find } A^{-1}.$

Use $\quad A^{-1}$ to solve the system of equations.

CBSE 2023

Just follow the given statements to first form the linear equations and then proceed accordingly. Don’t forget to verify your answer.

Type 6: By Using Inverse of Transpose of Matrix

Question 6: Given:

$A = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 2 & 2 \\ -3 & 1 & -1 \end{bmatrix}$

Find $\quad A^{-1}$ and hence solve the following system of equations:

$2x + y - 3z = 13$
$3x + 2y + z = 4$
$x + 2y - z = 8$

CBSE 2017

This one is similar to type 4 above, when the given system has Transpose of Matrix A. Give it a shot and verify your answer while checking with a complete solution in the downloadable pdf.

Type 7: NCERT EXAMPLAR: Important Type

Question 7: Determine the product of the matrices:

$A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}$

$B = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}$

and use it to solve the system of equations:

$y + 2z = 7$
$x - y = 3$
$2x + 3y +4z = 17$

NCERT EXEMPLAR

First of all, find the product of the given matrices, it will be of the type kI, where I is an Identity Matrix of order 3 and k is some scalar.

Use the product to find the inverse and solve this challenging question using Matrix Method.

Type 8: Find Inverse Matrix from a Given Equation

Question 8: A non-singular matrix:

$A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}$

satisfies the equation $A^3 + 2A^2 - 15A - 40I = 0$

Use it to find $A^{-1}$ and hence solve the system of equations:

$2x + 3y + 3z = 5$
$x - 2y + z = -4$
$3x - y - 2z = 3$

Maths Better

This is one of the most beautiful questions out of many ones by using Matrix Method.

While solving all these questions, do keep a check on your speed along with accuracy and presentation too.

Conclusion

The matrix method is a simple and effective way to solve systems of linear equations. By understanding when a matrix is invertible and how to handle special cases, you can solve different types of problems easily. Practice these steps, and you’ll find solving linear equations much easier and faster.

Signing-off for the day!

See you with more topics of Maths Class 12.

Drop me an email (mathsbetter@gmail.com) or comment if you have any doubts or want to understand any specific topic (of Class 11 and 12 Maths).

Moreover, you can also watch a good number of videos on PYQs/MCQs here on YouTube (@MathsBetter) based on Matrices and Determinants.

You take care, Bye-Bye!

MASTER MATRIX METHOD (8 Imp. Examples)Download

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