How to Solve Matrices

How to Solve Matrices
Go back to  'Algebra'

In this mini-lesson, we will explore the world of matrices by finding the answers to questions like how to solve systems of equations with matrices, how to solve matrices, and matrix operations with examples and formulas while discovering interesting facts around them.

The word "matrix" was termed by a 19th century English mathematician, James Sylvester.

His friend, Authur Cayley, developed the algebraic aspect of it in the 1850s. 

Look at the simulation given below.

Try selecting any of the basic operations listed and see how matrices are solved using them.

Lesson Plan

How to Solve Matrices?

Matrix (plural matrices) is a rectangular array of numbers, symbols, or expressions arranged in \(m\) rows and \(n\) columns.

Hence, the order of a matrix is given as \( m \times n \).

A general matrix may look like:

\[\begin{bmatrix}
a_{11}&a_{12}&\cdots &a_{1n} \\
a_{21}&a_{22}&\cdots &a_{2n} \\
\vdots & \vdots & \ddots & \vdots\\
a_{m1}&a_{m2}&\cdots &a_{mn}
\end{bmatrix}\]

For example,

\( \begin{bmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
\end{bmatrix}\)

Each term in the matrix can be in general written as \(a_{ij}\), where "\(i\)" is the row and "\(j\)" is the column.

We can solve matrices by performing operations on them like addition, subtraction, multiplication, and so on.

Let's learn about them in detail.


How to Solve System of Equations Using Matrices?

We have two matrices \(A\) and \(B\) where \(A\) is known as the coefficient matrix and \(B\) is known as the constant matrix.

There is a third matrix \(X\) containing all the variables of the equations; this matrix is known as a variable matrix. 

Matrix \(A\) is of the order \( m \times n \), while \(B\) is the column matrix of the order \( m \times 1 \).

The product of matrix \(A\) and matrix \(X\) results in matrix \(B\); hence, \(X\) is a column matrix as well of the order \( n \times 1\).

The matrices are arranged as:

\( A \cdot X = B\)

Let's understand how to solve a system of equations using matrices with the help of an example.

We have a set of two equations as given below.

The equations are:

 \(x + y = 8 \)

\(2x + 3y = 10\)

  • Arrange all the coefficients, variables, and constants in the matrix in such a way that whenever we find the product of the matrices, the result obtained must result in the equation.

The matrix \(A\) is:

\(A = \begin{bmatrix}
1 & 1\\
2 & 3\\
\end{bmatrix}\) 

The matrix \(X\) is:

\(X = \begin{bmatrix}
x\\
y\\
\end{bmatrix}\) 

The matrix \(B\) is:

\(B = \begin{bmatrix}
8\\
10\\
\end{bmatrix}\) 

To solve the equations, we need to find matrix \(X\).

It can be found by multiplying the inverse of matrix \(A\) with \(B\), which is given as \( X = (A^{-1})B\).

To find the inverse of \(A\), we will need the determinant and adjoint of matrix \(A\).

To find the determinant of matrix \(A\), we will follow the below steps:

\( |A| = \begin{vmatrix}
1 & 1\\
2 & 3\\
\end{vmatrix}\)

Hence, \( |A| = 3 - 2 = 1\)

\(\because\) \(|A| \neq 0\), it is possible to find the inverse of matrix \(A\).

The adjoint of matrix \(A\) is found by finding each element in it. 

Each element in the cofactor matrix is called a minor and found by taking the determinant of the elements leaving the row and column for which the number is to found. The matrix obtained is a cofactor matrix. To find the adjoint of matrix, we take the transpose of the cofactor matrix.

Hence, the adjoint of matrix is given as 

\(Adj A = C^T\)

\(C = \begin{bmatrix}
C_{11} & C_{12}\\
C_{21} & C_{22}\\
\end{bmatrix}\) 

Each term in a cofactor matrix is obtained by \(C_{ij} = (-1)^{i+j} \times M_{ij}\), where \(i\) indicates the row, \(j\) the column and \(M_{ij}\) being the minor of the matrix.

In this case, the adjoint of the matrix \(A\) is 

\(Adj A = \begin{bmatrix}
3 & -1\\
-2 & 1\\
\end{bmatrix}\)

To find the inverse of \(A\), we divide the adjoint of \(A\) by determinant of \(A\).

Hence, \(A^{-1} = \dfrac{Adj  A}{|A|}\)

Applying the same to our example we get, 

\(A^{-1} = \begin{bmatrix}
3 & -1\\
-2 & 1\\
\end{bmatrix}\)

Now to find the matrix \(X\), we'll multiply \(A^{-1}\) and \(B\). We get,

\[\begin{bmatrix}
   3 & -1 \\
-2 & 1
\end{bmatrix}
%
\begin{bmatrix}
   8 \\
10
\end{bmatrix}
\

\begin{bmatrix}
    14  \\
    -6      
\end{bmatrix} 
\]

Hence, the value of matrix \(X\) is, 

\(X = \begin{bmatrix}
14\\
-6\\
\end{bmatrix}\)

\(\therefore\) The solution of the equation is \(x = 14\) and \(y = -6\)
tips and tricks
Tips and Tricks

 

  • To find the inverse of any (\(2\times2\)), we can use \(A^{-1} = \frac{1}{|A|} \begin{bmatrix}
    d & -b\\
    -c & a\\
    \end{bmatrix}\)

Try using this simulation below to observe how equations are solved using matrices.


What Are Matrix Operations?

The operations that can be performed on matrices are known as matrix operations.

Let's say we have two matrices \(A\) and \(B\),

\(A = \begin{bmatrix}
a & b\\
c & d\\
\end{bmatrix}\)

\(B = \begin{bmatrix}
e & f\\
g & h\\
\end{bmatrix}\)

 

  1. Addition of Matrices

We add the matrices by adding the numbers, symbols, or expressions in the same position.

They are added in the following way:

Addition of Matrices

  1. Subtraction of Matrices

We subtract the matrices by subtracting the numbers, symbols, or expressions in the same position.

They are subtracted in the following way:

Subtraction of Matrices

  1. Multiplying a Matrix by a Constant

On multiplying a constant to a matrix, it gets multiplied in the entire matrix.

It is found as:

Multiplying a matrix by a constant

  1. Negative of a Matrix

The negative of a matrix is found by negating the elements of the matrix.

The negative of a matrix is found in the following way:

Negative of a Matrix

  1. Multiplying a Matrix by a Matrix

We can multiply a matrix by itself in the following way, 

Multiplying a matrix by a matrix

  1. Transposing a Matrix

The transpose of a matrix can be found in the following way,

Transposing a matrix

Take a look at this simulation below, select your desired operation, and see the output obtained:

 

Examples

Example 1: For \(A\) and \(B\), find the output of the matrix operations given below.

\(A = \begin{bmatrix}
1 & 2\\
3 & 4\\
\end{bmatrix}\) 

\(B = \begin{bmatrix}
5 & 7\\
9 & 6\\
\end{bmatrix}\) 

  1. \(A+B\)
  2. \(B-A\)
  3. Product of \(A\) and \(B\)

Solution:

  1. Let us say, \(A+B = C\).

The addition of both the matrices is given as,

\[C=\begin{bmatrix}
   1 & 2 \\
3 & 4
\end{bmatrix}
+
\begin{bmatrix}
  5 & 7 \\
9 & 6
\end{bmatrix}
\

\begin{bmatrix}
    6 & 9  \\
    12 & 10      
\end{bmatrix} 
\]

Hence, 

 \[ C = \begin{bmatrix}
    6 & 9  \\
    12 & 10      
\end{bmatrix} 
\]

\[ C = \begin{bmatrix}
    6 & 9  \\
    12 & 10      
\end{bmatrix} 
\]
  1. Let us say, \(D = B-A\)

The subtraction of both the matrices is given as,

\[D=\begin{bmatrix}
   5 & 7 \\
9 & 6
\end{bmatrix}
-
\begin{bmatrix}
  1 & 2 \\
3 & 4
\end{bmatrix}
\

\begin{bmatrix}
    4 & 5  \\
    6 & 2      
\end{bmatrix} 
\]

Hence, 

\[ D = \begin{bmatrix}
    4 & 5  \\
    6 & 2      
\end{bmatrix} 
\]

 \[ D = \begin{bmatrix}
    4 & 5  \\
    6 & 2      
\end{bmatrix} 
\]
  1. Let us say, product of \(A\) and \(B\) gives \(F\),

The product of matrices \(A\) and \(B\) is, 

\[F=\begin{bmatrix}
   1 & 2 \\
3 & 4
\end{bmatrix}
%
\begin{bmatrix}
  5 & 7 \\
9 & 6
\end{bmatrix}
\

\begin{bmatrix}
    23 & 19  \\
    51 & 45      
\end{bmatrix} 
\]

Hence, 

\[ F = \begin{bmatrix}
    23 & 19  \\
    51 & 45      
\end{bmatrix} 
\]

 \[ F = \begin{bmatrix}
    23 & 19  \\
    51 & 45      
\end{bmatrix} 
\]

Example 2: Find: 

  1. Negative of the given matrix
  2. Output when matrix is multiplied by 3
  3. Transpose of the given matrix

\[ A = \begin{bmatrix}
    -1 & 5  \\
    4 & -2      
\end{bmatrix} 
\]

Solution:

  1. The negative of the matrix \(A\) is given as,

\[A = -\begin{bmatrix}
    -1 & 5  \\
    4 & -2      
\end{bmatrix} 
\]

The negative of the matrix \(A\) is, 

\[A = \begin{bmatrix}
    1 & -5  \\
    -4 & 2      
\end{bmatrix} 
\]

 \[A = \begin{bmatrix}
    1 & -5  \\
    -4 & 2      
\end{bmatrix} 
\]
  1. The output when matrix is multiplied by 3 is given as,

\[B = 3\begin{bmatrix}
    -1 & 5  \\
     4 & -2      
\end{bmatrix} 
\]

The value of the matrix \(A\) when it is multiplied by 3 is, 

\[A = \begin{bmatrix}
    -3 & 15  \\
      12 & -6      
\end{bmatrix} 
\]

 \[A = \begin{bmatrix}
    -3 & 15  \\
      12 & -6      
\end{bmatrix} 
\]
  1. The transpose of given matrix is,

\[A^T = \begin{bmatrix}
    -1 & 5  \\
     4 & -2      
\end{bmatrix}^T 
\]

The transpose of \(A\) is, 

\[A^T = \begin{bmatrix}
    -1 & 4  \\
      5 & -2      
\end{bmatrix} 
\]

\[A^T = \begin{bmatrix}
    -1 & 4  \\
      5 & -2      
\end{bmatrix} 
\]
 
important notes to remember
Important Notes
  1. Addition, subtraction, and multiplication can be applied to matrices. It is not possible to divide two matrices. 
  2. Matrices can either have whole numbers, fractions or decimals.

Solved Examples

Example 1

 

 

Julia has two matrices \(A\) and \(B\) as below. Help her find the values of "\(a\)" and "\(b\)" if \(C\) is the addition of both the matrices.

\(A = \begin{bmatrix}
1 & 3\\
a & 6\\
\end{bmatrix}\) 

\(B = \begin{bmatrix}
2 & b\\
4 & 7\\
\end{bmatrix}\) 

\(C = \begin{bmatrix}
3 & 10\\
9 & 13 \\
\end{bmatrix}\) 

Solution

Julia knows matrix \(C\) is the sum of matrices \(A\) and \(B\).

\[C=\begin{bmatrix}
   1 & 3 \\
a & 6
\end{bmatrix}
+
\begin{bmatrix}
  2 & b \\
4 & 7
\end{bmatrix}
\

\begin{bmatrix}
    3 & 10  \\
    9 & 13      
\end{bmatrix} 
\]

Hence, the conclusion derived from them is, 

\(b + 3 = 10 \implies b = 7\)

\(a + 4 = 9 \implies a = 5\)

\(\therefore\) Julia found the values of a and b are 5 and 7 respectively.
Example 2

 

 

Oliver's teacher asked him to solve the systems of equations given below with matrices.

 \(2x + 4y = 5 \)

\(x + 3y = 7\)

Solution

Oliver knows that the set of equations are:

 \(2x + 4y = 5 \)

\(x - 3y = 7\)

Arrange all the coefficients, variables, and constants in the matrix in such a way that whenever the product of the matrices is found, the result obtained must result in the equation.

The matrix \(A\) is:

\(A = \begin{bmatrix}
2 & 4\\
1 & -3\\
\end{bmatrix}\) 

The matrix \(X\) is:

\(X = \begin{bmatrix}
x\\
y\\
\end{bmatrix}\) 

The matrix \(B\) is:

\(B = \begin{bmatrix}
5\\
7\\
\end{bmatrix}\) 

To solve the equations, it is required to find the matrix \(X\). Hence, it can be found by multiplying inverse of matrix A with B, which is given as, \( X = (A^{-1})B\).

To find the inverse of A, we will need the determinant and adjoint of matrix A.

To find the determinant of matrix A, we will follow the below steps:

\( |A| = \begin{vmatrix}
2 & 4\\
1 & -3\\
\end{vmatrix}\)

Hence, \( |A| = 6 - 4 = 2\)

\(\because\) \(|A| \neq 0\), it is possible to find the inverse of matrix A.

The adjoint of matrix A is found by finding each element in it. 

Each element in the cofactor matrix is called a minor and found by taking the determinant of the elements leaving the row and column for which the number is to found. The matrix obtained is a cofactor matrix. To find the adjoint of matrix, we take the transpose of the cofactor matrix.

Hence, the adjoint of matrix is given as 

\(Adj A = C^T\)

\(C = \begin{bmatrix}
C_{11} & C_{12}\\
C_{21} & C_{22}\\
\end{bmatrix}\) 

Each term in a cofactor matrix is obtained by \(C_{ij} = (-1)^{i+j} \times M_{ij}\)

In this case, the adjoint of the matrix A is 

\(Adj A = \begin{bmatrix}
-3 & -4\\
-1 & 2\\
\end{bmatrix}\)

To find the inverse of A, we divide the adjoint of A by determinant of A.

Hence, \(A^{-1} = \dfrac{Adj  A}{|A|}\)

Applying the same to our example we get, 

\(A^{-1} = \dfrac{1}{-10}\begin{bmatrix}
-3 & -4\\
-1 & 2\\
\end{bmatrix}\)

Now to find the matrix X, we'll multiply \(A^{-1}\) and B. We get,

\[\begin{bmatrix}
   \frac{3}{10} & \frac{4}{10} \\
\frac{1}{10} & \frac{-2}{10}
\end{bmatrix}
%
\begin{bmatrix}
   5 \\
7
\end{bmatrix}
\

\begin{bmatrix}
    \frac{43}{10}  \\
    \frac{-9}{10}     
\end{bmatrix} 
\]

Hence, the value of matrix \(X\) is,

\(X = \begin{bmatrix}
\frac{43}{10}\\
\frac{-9}{10}\\
\end{bmatrix}\)

\(\therefore\) The solution of equations is \(x = \frac{43}{10}\) and \(y = \frac{-9}{10}\)

Interactive Questions

Here are a few activities for you to practice.

Select/Type your answer and click the "Check Answer" button to see the result.

 
 
 
 


 


Let's Summarize

We hope you enjoyed learning about how to solve matrices with the practice questions and simulations. Now, you will have a better understanding of how to solve systems of equations with matrices and matrix operations.

The mini-lesson targeted the fascinating concept of matrices. The math journey around matrices starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

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Frequently Asked Questions (FAQs)

1. How to solve matrices?

We can solve matrices by performing operations on them like addition, subtraction, multiplication, and so on.

2. How to solve systems of equations with matrices?

To solve the system of equations with matrices, we will follow the steps given below.

  • Arrange the elements of equations in matrices and find the coefficient matrix, variable matrix, and constant matrix.
  • Write the equations in \(AX = B\) form.
  • Take the inverse of \(A\) by finding the adjoint and determinant of \(A\).
  • Multiply the inverse of \(A\) to matrix \(B\), thereby finding the value of variable matrix \(X\).

3. What are matrix operations?

The operations that can be performed on the matrices are known as matrix operations.

  
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