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Applications of Determinants and Matrices

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Estimated time: 8 minutes
CBSE: Class 12

Introduction

Determinants and matrices are useful tools for solving systems of linear equations in two or three variables. This topic explains how a system of equations can be written in matrix form and solved by using the inverse of a matrix when it exists. It also helps in deciding whether a system is consistent or inconsistent.

CBSE: Class 12

Consistent and Inconsistent

Consistent Solution: A system is consistent if it has at least one solution.

Inconsistent Solution: A system is inconsistent if it has no solution.

CBSE: Class 12

Example 1

Solve the system of equations

\[ 2x + 5y = 1 \]
\[ 3x + 2y = 7 \]

Solution: The system of equations can be written in the form

\[ \text{AX} = \text{B} \], where
\[\text{A} = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}, \text{X} = \begin{bmatrix} x \\ y \end{bmatrix} \text{ and } \text{B} = \begin{bmatrix} 1 \\ 7 \end{bmatrix} \]

Now,

\[ |\text{A}| = -11 \neq 0 \], Hence, A is a nonsingular matrix and so has a unique solution.

Note that

\[\text{A}^{-1} = -\frac{1}{11} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix} \]

Therefore

\[ \text{X} = \text{A}^{-1}\text{B} = -\frac{1}{11} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 7 \end{bmatrix} \]

i.e.

\[ \begin{bmatrix} x \\ y \end{bmatrix} = -\frac{1}{11} \begin{bmatrix} -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix} \]

Hence

\[ x = 3, y = -1 \]
CBSE: Class 12

Example 2

The sum of three numbers is 6. If we multiply the third number by 3 and add the second number to it, we get 11. By adding the first and third numbers, we get twice the second number. Represent it algebraically and find the numbers using the matrix method.

Solution: Let the first, second and third numbers be denoted by\[ x, y\] and\[ z \], respectively.

Then, according to the given conditions, we have

\[x + y + z = 6 \]
\[y + 3z = 11 \]
\[ x + z = 2y \text{ or } x - 2y + z = 0 \]

This system can be written as

\[\text{A X} = \text{B} \], where
\[ \text{A} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}, \text{X} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } \text{B} = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix} \]

Here

\[ |\text{A}| = 1(1 + 6) - (0 - 3) + (0 - 1) = 9 \neq 0 \]. Now we find \[ adj \text{ A} \]
\[ \text{A}_{11} = 1 (1 + 6) = 7, \hspace{2cm} \text{A}_{12} = - (0 - 3) = 3, \hspace{2cm} \text{A}_{13} = - 1 \]
\[ \text{A}_{21} = - (1 + 2) = - 3, \hspace{1.5cm} \text{A}_{22} = 0, \hspace{3.2cm} \text{A}_{23} = - (- 2 - 1) = 3 \]
\[ \text{A}_{31} = (3 - 1) = 2, \hspace{2cm} \text{A}_{32} = - (3 - 0) = - 3, \hspace{1.5cm} \text{A}_{33} = (1 - 0) = 1 \]

Hence \[ adj \text{ A} = \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix} \]

Thus \[\text{A}^{-1} = \frac{1}{|\text{A}|} adj (\text{A}) = \frac{1}{9} \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix} \]

Since \[ \text{X} = \text{A}^{-1} \text{B} \]

\[ \text{X} = \frac{1}{9} \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix} \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix} \]

or

\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 42 - 33 + 0 \\ 18 + 0 + 0 \\ -6 + 33 + 0 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \]

Thus \[ x = 1, y = 2, z = 3 \]

Shaalaa.com | Solution of System of Linear Equations by Inversion Method Part 1

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