CBSE Syllabus for Class 8 Maths 2023-24: Students in CBSE Class 8 need to be thorough with their syllabus so that they can prepare for the...

CBSE Syllabus for Class 8 Maths 2024: Download Free PDF

March 1, 2024**Applications of Determinants and Matrices:** There are many applications of matrices and determinants in daily life and various fields of science, ranging from statistical studies of a population to encrypting data in cryptography to remain confidential. When it comes specifically to mathematics, one of the most important applications is to solve a system of \(n\) linear equations in \(n\) variables using Cramer’s Rule and to use the Inverse Matrix Method. Though both use determinant operations, the latter is discussed in detail in this article. Before going into the details, let us understand the basics and applicants of matrices and determinants.

A matrix is an ordered rectangular array of numbers or mathematical expressions in rows and columns that represents a mathematical object or property. A matrix is always enclosed using brackets [ ] or ( ).

An \(m×n\) matrix has \(m\) rows and \(n\) columns.

For example,

\(\left[ {\begin{array}{*{20}{c}} \begin{array}{l} – 1\\ 6 \end{array}&\begin{array}{l} 0\\ – 3 \end{array}&\begin{array}{l} 5\\ – 2 \end{array} \end{array}} \right]\) is a \(2×3\) matrix.

The algebraic operations addition and multiplication are defined for matrix.

The determinant of a square matrix is a scalar value that is a function of the entries in the matrix. The value of the determinant of a matrix is the sum of the product of elements of a row (or a column) with their cofactors. A matrix is invertible if its determinant is non-zero.

The determinant of a \(2\) by \(2\) matrix \(\left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) is given by:

\(\left| {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right|=\left(a_{11} \times a_{22}\right)-\left(a_{21} \times a_{22}\right)\)

When it comes to a \(3\) by \(3\) matrix \(\left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) is given by:

\({\left( { – 1} \right)^{1 + 1}}{a_{11}}\left| {\begin{array}{*{20}{c}} {{a_{22}}}&{{a_{23}}}\\ {{a_{32}}}&{{a_{33}}} \end{array}} \right| + {\left( { – 1} \right)^{1 + 2}}{a_{12}}\left| {\begin{array}{*{20}{c}} {{a_{21}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{33}}} \end{array}} \right| + {\left( { – 1} \right)^{1 + 3}}{a_{13}}\left| {\begin{array}{*{20}{c}} {{a_{21}}}&{{a_{22}}}\\ {{a_{32}}}&{{a_{33}}} \end{array}} \right|\)\(=a_{11}\left[\left(a_{22} \times a_{33}\right)-\left(a_{32} \times a_{23}\right)\right]-a_{12}\left[\left(a_{21} \times a_{33}\right)-\left(a_{31} \times a_{23}\right)\right]\)

\(+a_{13}\left[\left(a_{21} \times a_{32}\right)-\left(a_{31} \times a_{22}\right)\right]\)

1. **Solving System of Linear Equations**

A system of linear equations can be represented in the matrix form as:

\(AX = B\)

Where,

\(A\)** – Coefficient Matrix**

\(X\)** – Variable Matrix**

\(B\)** – Constant Matrix**

Consider a system of linear equations in three variables:

\(a_{1} x+b_{1} y+c_{1} z=d_{1}\)

\(a_{2} x+b_{2} y+c_{2} z=d_{2}\)

\(a_{3} x+b_{3} y+c_{3} z=d_{3}\)

Here,

Coefficient matrix \(A = \left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right]\)

Variable matrix \(X = \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right]\)

Constant matrix \(B = \left[ {\begin{array}{*{20}{c}} {{d_1}}\\ {{d_2}}\\ {{d_3}} \end{array}} \right]\)

Then, the system can be represented as \(\left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{d_1}}\\ {{d_2}}\\ {{d_3}} \end{array}} \right]\)

**Case 1:** A is a non-singular matrix. That is, its determinant is non-zero.So, \(A^{-1}\) exists and multiplying both sides of the equation \(A X=B\) by \(A^{-1}:\)

\(AX=B\)

\(A^{-1}(A X)=A^{-1} B\)

Matrix multiplication is associative. So:

\(\left(A^{-1} A\right) X=A^{-1} B\)

The product of a matrix and its inverse is the identity matrix \(I\)

\(I X=A^{-1} B\)

The product of a matrix and the identity matrix \(I\) is the matrix itself. Then:

\(X=A^{-1} B\)

Since the inverse of a matrix is unique, \(X=A^{-1} B X\) is the unique solution for the system of equations represented by \(AX = B\).

This method of solving a system of equations is known as the Inverse Matrix Method.

**Case 2:** A is a singular matrix. That is, \(A^{-1}\) does not exist.

In such cases, calculate the product of the adjoint of the coefficient matrix and the constant matrix.

(i) If the product is a non-zero matrix, there is no solution, and the system is inconsistent.

(ii) If the product is a zero matrix, the system has infinite solutions.

2. **Equation of a Line**

The equation of a line joining the points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) in determinant form is given as:

\(\left| {\begin{array}{*{20}{c}} x&y&1\\ {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1 \end{array}} \right| = 0\)

3. **Area of a Triangle**

The area of a triangle formed by the vertices \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\), and \(\left(x_{3}, y_{3}\right)\) is given by the formula:

\({\rm{Area}}\,{\rm{of}}\,\Delta = \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

4. **Collinearity of Three Points**

The area of a triangle becoming zero indicates that all the three vertices lie in the same line.

Thus, the condition for collinearity of three points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\) and \(\left(x_{3}, y_{3}\right)\) is:

\(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0\)

** Q.1. Is the system of equations consistent? **\(x+3 y=5\)

Ans:

\(\left[ {\begin{array}{*{20}{c}} 1&3\\ 2&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5\\ 8 \end{array}} \right]\)

Now, find the determinant of the coefficient matrix.

\(\left| {\begin{array}{*{20}{c}} 1&3\\ 2&6 \end{array}} \right|=(1 \times 6)-(2 \times 3)=0\)

So, the coefficient matrix is singular. The system does not have a unique solution.

\({\rm{Adj}}\left[ {\begin{array}{*{20}{c}} 1&3\\ 2&6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 6&{ – 3}\\ { – 2}&1 \end{array}} \right]\)

\(\therefore \left[ {\begin{array}{*{20}{c}} 6&{ – 3}\\ { – 2}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5\\ 8 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\left( {6 \times 5} \right) + \left( { – 3 \times 8} \right)}\\ {\left( { – 2 \times 5} \right) + \left( {1 \times 8} \right)} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} 6\\ { – 2} \end{array}} \right]\)

The product of the adjoint and the coefficient matrices is a non-zero matrix.

Therefore the system has no solution, and it is inconsistent.

** Q.2. Solve the system of linear equations using inverse matrix method.**\(x – y + z = 4\)

\(2x + y – 3z = 0\)

\(x + y+ z = 2\)

Ans:

\(\left[ {\begin{array}{*{20}{c}} 1&{ – 1}&1\\ 2&1&{ – 3}\\ 1&1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4\\ 0\\ 2 \end{array}} \right]\)

Now,

\(|A|=1[(1 \times 1)-(-3 \times 1)]-(-1)[2 \times 1-(-3 \times 1)]+1[(2 \times 1)-(1 \times 1)]\)

\(=4+5+1\)

\(=10\)

Since the determinant is a non-zero number, the coefficient matrix is non-singular. Then, the system has a unique solution.

To find the \(\text {adj} \,A\) for \(A = \left[ {\begin{array}{*{20}{c}} 1&{ – 1}&1\\ 2&1&{ – 3}\\ 1&1&1 \end{array}} \right]\), the cofactors of each element can be calculated as:

\(A_{11}=(-1)^{1+1}[(1 \times 1)-(1 \times-3)]=4\)

\(A_{12}=(-1)^{1+2}[(2 \times 1)-(1 \times-3)]=-5\)

\(A_{13}=(-1)^{1+3}[(2 \times 1)-(1 \times 1)]=1\)

\(A_{21}=(-1)^{2+1}[(-1 \times 1)-(1 \times 1)]=2\)

\(A_{22}=(-1)^{2+2}[(1 \times 1)-(1 \times 1)]=0\)

\(A_{23}=(-1)^{2+3}[(1 \times 1)-(1 \times-1)]=-2\)

\(A_{31}=(-1)^{3+1}[(-1 \times-3)-(1 \times 1)]=2\)

\(A_{32}=(-1)^{3+2}[(1 \times-3)-(2 \times 1)]=5\)

\(A_{33}=(-1)^{3+3}[(1 \times 1)-(2 \times-1)]=3\)

Then,

Cofactor of \(A = \left[ {\begin{array}{*{20}{c}} 4&{ – 5}&1\\ 2&0&{ – 2}\\ 2&5&3 \end{array}} \right]\)

Now, \({\rm{Adj}}\left( A \right) = {\left[ {\begin{array}{*{20}{c}} 4&{ – 5}&1\\ 2&0&{ – 2}\\ 2&5&3 \end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}} 4&2&2\\ { – 5}&0&5\\ 1&{ – 2}&3 \end{array}} \right]\)

Now, the inverse of the coefficient matrix is:

\(A^{-1}=\frac{1}{|A|} \operatorname{Adj}(A)\)

\(\therefore {A^{ – 1}} = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} 4&2&2\\ { – 5}&0&5\\ 1&{ – 2}&3 \end{array}} \right]\)

Thus, \(X=A^{-1} B\)

\(\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} 4&2&2\\ { – 5}&0&5\\ 1&{ – 2}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4\\ 0\\ 2 \end{array}} \right]\)

\( = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {\left( {4 \times 4} \right) + \left( {2 \times 0} \right) + \left( {2 \times 2} \right)}\\ {\left( { – 5 \times 4} \right) + \left( {0 \times 0} \right) + \left( {5 \times 2} \right)}\\ {\left( {1 \times 4} \right) + \left( { – 2 \times 0} \right) + \left( {3 \times 2} \right)} \end{array}} \right]\)

\( = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {20}\\ { – 10}\\ {10} \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2\\ { – 1}\\ 1 \end{array}} \right]\)

The solutions to the system of equations are \(x = 2 , y = -1, z = 1\).

** Q.3. Use determinants to find the equation of a line joining the points **\((-3, 2)\)

Ans:

\(\left| {\begin{array}{*{20}{c}} x&y&1\\ {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1 \end{array}} \right| = 0\)

Here, \(\left(x_{1}, y_{1}\right)=(-3,2)\) and \(\left(x_{2}, y_{2}\right)=(1,4)\)

Then, \(\left| {\begin{array}{*{20}{c}} x&y&1\\ { – 3}&2&1\\ 1&4&1 \end{array}} \right| = 0\)

\(x(2-4)-y(-3-1)+1(-12-2)=0\)

\(-2x+4y-14 =0\)

\(-x+2y =7\)

** Q.4. Find the area of a triangle formed by the points **\((4, 5), (-1, 2)\)

Ans:

\({\rm{Area}}\,{\rm{of}}\,\Delta = \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

Here.

\(\left(x_{1}, y_{1}\right)=(4,5)\)

\(\left(x_{2}, y_{2}\right)=(-1,2)\)

\(\left(x_{3}, y_{3}\right)=(2,0)\)

\(\therefore {\rm{Area}}\,{\rm{of}}\,\Delta = \left| {\begin{array}{*{20}{c}} 4&5&1\\ { – 1}&2&1\\ 2&0&1 \end{array}} \right|\)

\(=4(2-0)-5(-1-2)+1(0-4)\)

\(=8+15-4\)

\(=19\)

Therefore, the area of the triangle is \(19\) square units.

** Q.5. Use determinants to prove that the points **\((1, 2), (-1, 0)\)

Ans:

\(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0\)

Here,

\(\left(x_{1}, y_{1}\right)=(1,2)\)

\(\left(x_{2}, y_{2}\right)=(-1,0)\)

\(\left(x_{3}, y_{3}\right)=(3,4)\)

\(\therefore \left| {\begin{array}{*{20}{c}} 1&2&1\\ { – 1}&0&1\\ 3&4&1 \end{array}} \right| = 1\left( {0 – 4} \right) – 2\left( { – 1 – 3} \right) + 1\left( { – 4 – 0} \right)\)

\(=-4+8-4\)

\(=0\)

Hence, proved.

The article narrates the definitions of matrices and determinants, after which their applications are discussed. It explains one of the most important applications of determinants and matrices as solving the system of linear equations in detail. The classification of a system into consistent or inconsistent is also delineated with the process. The other major applications of determinants and matrices discussed are the equation of a line, given the coordinates of two points on it, the area of a triangle given the coordinates of its vertices, and the collinearity of three points. A few solved examples are also given to reinforce the concepts learned.

** Q.1. What is the application of matrix and determinants?** One of the main applications of determinants and matrices is to solve the system of linear equations using Cramer’s Rule. This allows one to solve the system using the concept of the inverse of a matrix. Another application of matrix and determinant is to form the equation of a line, given two points on the line.

Ans:

The area of a triangle given the coordinates of the vertices can also be found by applying the concept of determinants. Further, this can be applied to check the collinearity of three points because if the area of a triangle is zero, the three vertices lie on the same line.

** Q.2. Define matrix.** Matrix is an ordered rectangular array of numbers or mathematical expressions in rows and columns that represents a mathematical object or property. A matrix is always enclosed using brackets [ ] or ( ). An \(m×n\) matrix has \(m\) rows and \(n\) columns. For example,

Ans:

\(\left[ {\begin{array}{*{20}{c}} 2&{ – 1}\\ \begin{array}{l} 4\\ – 3 \end{array}&\begin{array}{l} 0\\ 9 \end{array} \end{array}} \right]\) is a \(3×2\) matrix.

The algebraic operations addition and multiplication are defined for matrix.

** Q.3. What is meant by determinant?** The determinant of a square matrix is a scalar value that is a function of the entries of the matrix. The value of the determinant of a matrix is obtained by the sum of the product of elements of a row (or a column) with corresponding cofactors.

Ans:

Its value can be used to characterize the matrix in many ways. A matrix is invertible if and only if its determinant is non-zero.

** Q.4. Why do we use determinants?**One of the main intentions to define the determinant of a matrix is to represent the matrix using a singleton number so that some essential characteristics of the matrix can be explored. They are also used to give formulas to solve a system of n linear equations in \(n\) unknowns.

Ans:

** Q.5. What are the important properties of determinants?** The main properties of the determinant of a matrix are:

Ans:

1. The determinant of a matrix and its transpose are equal.

2. If any two rows (or columns) of a matrix are interchanged, the sign of the determinant changes.

3. If any two rows or any two columns of a matrix are identical or proportional, then the determinant of the matrix is zero.

4. If each element of a row or a column is multiplied by a scalar \(k\) then the value of the determinant gets multiplied by \(k\). For, a \(3 \times 3\) matrix \(A\) and \(k\) is a constant, then \(|k \cdot A|=k^{3}|A|\)

5. A system of linear equations with a coefficient matrix that has a non-zero determinant has a unique solution.

6. If elements of a row or a column in a determinant can be expressed as the sum of two or more elements, then the given determinant can be expressed as a sum of two or more determinants.

7. If to each element of a row or a column of a determinant the equimultiples of corresponding elements of other rows or columns are added, then the value of determinant remains the same.

*We hope this detailed article on Applications of Determinants and Matrices will be helpful to you in your preparation. If you have any doubts please reach out to us through the comments section, and we will get back to you as soon as possible.*