Properties of Matrices Inverse: Definition, Methods, Solved Examples

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Namrata Das

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The inverse of a matrix is the matrix which on multiplying with the original matrix provides the multiplicative identity. It is preferably determined using elementary transformations and inverse matrix formulas. If A is a non-singular square matrix, there will be n x n matrix A-1, which is called the inverse of a matrix A in a way that it satisfies the property: AA-1 = A-1A = I, where I is the Identity matrix. If two matrices are inverse of each other, then their product is equal to the identity matrix.

Read more: Transpose of a Matrix 

Key Terms: Matrices, Inverse of a matrix, multiplicative identity, determinant, identity matrix, non-singular square matrix, square matrix, addition, subtraction, multiplication


What is the Inverse of a Matrix?

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Matrices, while difficult to use manually, are extremely useful when incorporated into computers and. Matrices are capable of solving complex problems quickly and easily.

There are methods for adding, subtracting, and multiplying matrices, but there is no concept of matrix division. This is where a matrix's inverse comes into play. The closest we can get to matrix division is to multiply by its inverse.

For all you know about integer functions, dividing by a number yields the same result as multiplying by its reciprocal. 20/10 equals 10 x (1/10) = 2. The same is true for matrix inverses – until the matrix in question has an inverse.

The video below explains this:

Matrices Detailed Video Explanation:

Read more: Symmetric and Skew Symmetric Matrices


How to Find the Inverse of Matrix?

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There are three methods for finding inverse of matrices or for calculating invertible matrices.

  • If A is equal to the matrix of the form , then the inverse of A is determined using the formula,
  • The inverse of a matrix can be determined using minors, co-factors, and adjoints of a matrix. The inverse matrix formula is given as,
  • It can also be determined using elementary transformations in a matrix and involves two types, elementary row transformations and elementary column transformations.

Read more: Determination of Order of Matrices


Properties of the Inverse of Matrices 

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Some of the properties of the inverse of non-singular matrices are:

  • (A-1)-1 =A
  • (AB)-1 =A-1B-1
  • (ABC)-1 = C-1B-1A-1
  • (A1 A2….An)-1 =An-1An-1-1……A2-1A1-1
  • (AT)-1 =(A-1)T
  • (kA)-1 = (1/k)A-1
  • AB = In, where A and B are inverse of each other.
  • If A is a square matrix where n>0, then (A-1)n =A – n

Where A – n = (An)-1

Read more: Determinants and Matrices


Things to Remember

  • Invertible matrices are matrices having a unique inverse. Also, a matrix is not invertible if its determinant is zero.
  • Non-singular matrices are square matrices that have their determinant not equal to zero.
  • For non-singular matrices A and B that are inverse of each other,
  • For a matrix multiplied by a constant k, the inverse is given by,
  • For a square matrix A, where

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Sample Questions

Ques. Find the inverse A – 1 of the matrix. \(A = \begin{bmatrix}2 & 1 & 1\\[0.3em]3 & 2 & 1\\[0.3em]2 & 1 & 2\\[0.3em] \end{bmatrix}\)(4 marks)

Ans. We know that

A – 1 = adj (A) / det(A)

A – 1 = adj (A) / |A|

|A| = 2 \(\begin{bmatrix}2 & 1 \\[0.3em]1 & 2 \\[0.3em] \end{bmatrix}\)– 1 \(\begin{bmatrix}3 & 1 \\[0.3em]2 & 2 \\[0.3em] \end{bmatrix}\)+ 1\(\begin{bmatrix}3 & 2 \\[0.3em]2 & 1 \\[0.3em] \end{bmatrix}\)

|A| = 2 (4 – 1) – 1(6 – 2) + 1(3 – 4)

|A| = 2(3) – 1(4) + 1( – 1)

|A| = 6 – 4 – 1

|A| = 1

Now, find Adj(A):

Ac\(\begin{bmatrix}(4 - 1) & -(6-2) & (3-4)\\[0.3em]-(2-1) & (4-2) & -(2-2)\\[0.3em](1-2) & -(2-3) & (4-3)\\[0.3em] \end{bmatrix}\)

\(A_c = \begin{bmatrix}3 & -4 & 1\\[0.3em]-1 & 2 & 0\\[0.3em]-1 & 1 & 1\\[0.3em] \end{bmatrix}\)

Therefore,

A – 1 = adj (A) / |A|

A-1\(\frac{1}{1}\)\( \begin{bmatrix}3 & -1 & -1\\[0.3em]-4 & 2 & 1\\[0.3em]-1 & 0 & 1\\[0.3em] \end{bmatrix}\)

A-1\( \begin{bmatrix}3 & -1 & -1\\[0.3em]-4 & 2 & 1\\[0.3em]-1 & 0 & 1\\[0.3em] \end{bmatrix}\)

Ques. Find the Inverse of the Following Matrix. \(\begin{bmatrix}5 & 6 \\[0.3em]-1 & 2 \\[0.3em] \end{bmatrix}\)(2 marks)

Ans.  Inverse of the Following Matrix

Inverse of Matrix

Ques. Find out the inverse of the following matrix: \( \begin{bmatrix}1 & -1 & 2\\[0.3em]4 & 0 & 6\\[0.3em]0 & 1 & -1\\[0.3em] \end{bmatrix}\)(4 marks)

Ans: Let A = \( \begin{bmatrix}1 & -1 & 2\\[0.3em]4 & 0 & 6\\[0.3em]0 & 1 & -1\\[0.3em] \end{bmatrix}\)

be the given matrix.

A – 1 = adj (A)/ |A|

To find out the adj(A), first we have to find out cofactor(A).

a11 = – 6, a12 = 4, a13 = 4

a21 = 1, a22 = – 1, a23 = – 1

a13 = – 6, a32 = 2, a33 = 4 So, cofactor(A) = \( \begin{bmatrix}-6 & 4 & 4\\[0.3em]1 & -1 & 1\\[0.3em]-6 & 2 & 4\\[0.3em] \end{bmatrix}\)

[cofactor(A)]T\( \begin{bmatrix}-6 & 4 & 4\\[0.3em]1 & -1 & 1\\[0.3em]-6 & 2 & 4\\[0.3em] \end{bmatrix}\)T

adj(A) = \( \begin{bmatrix}-6 & 1 & -6\\[0.3em]4 & -1 & 2\\[0.3em]4 & -1 & 4\\[0.3em] \end{bmatrix}\)

Then |A| = 1(0-6) + 1(-4-0) + 2(4-0) = -6-4+8 = -2

A -1 = \(\frac{adj(A)}{|A|}\) = \(\frac{ \begin{bmatrix}-6 & 1 & -6\\[0.3em]4 & -1 & 2\\[0.3em]4 & -1 & 4\\[0.3em] \end{bmatrix}}{-2}\)

A -1 = \( \begin{bmatrix}3 & -\frac{1}{2} & 3\\[0.3em]-2 & \frac{1}{2} & -1\\[0.3em]-2 & \frac{1}{2} & -2\\[0.3em] \end{bmatrix}\)

Ques. If  \(\begin{bmatrix}x - y & z \\[0.3em]2x - y & w \\[0.3em] \end{bmatrix} =\begin{bmatrix}-1 & 4 \\[0.3em]0 & 5 \\[0.3em] \end{bmatrix} \), then Find the value of x + y? (2 marks)

Ans. 

\(\begin{bmatrix}a+4 & 3b \\[0.3em]8 & -6 \\[0.3em] \end{bmatrix} =\begin{bmatrix}2a+2 & b+2 \\[0.3em]8 & a-8b \\[0.3em] \end{bmatrix} \)

We know, two matrices are equal if their corresponding elements are equal. 

Therefore, a + 4 = 2a + 2 … (i)

3b = b + 2 … (ii)

And -6 = a - 8b … (iii)

On solving (i), (ii) and (iii) we get

a - 2b = 2 - 2 (1) = 2 - 2 = 0


Previous Year Questions

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CBSE CLASS XII Related Questions

  • 1.
    Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).


      • 2.

        Find:
        Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

          • \(\begin{bmatrix} \frac{1}{3} & 3 \\ 0 & \frac{4}{3} \end{bmatrix}\)
          • \(\begin{bmatrix} \frac{1}{3} & 0 \\ 3 & \frac{4}{3} \end{bmatrix}\)
          • \(\begin{bmatrix} \frac{4}{3} & 3 \\ 1 & 0 \end{bmatrix}\)
          • \(\begin{bmatrix} \frac{4}{3} & 0 1 & \frac{3}{3} \end{bmatrix}\)

        • 3.
          Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).


            • 4.

              At a birthday party, children are being served orange juice in conical cups, as shown in the figure. 


              Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
              On the basis of the above information, answer the following questions :


                • 5.

                  Evaluate:
                  \[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]


                    • 6.
                      Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

                        CBSE CLASS XII Previous Year Papers

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