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Inverse Matrix - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Inverse of a Matrix, Properties of Inverse of a Matrix - Part 3 are considered the most difficult concepts.

  • Properties of Inverse of a Matrix - Part 1 are considered the most asked concepts.

  • 22 Questions around this concept.

Solve by difficulty

 Let A be any 3×3 invertible matrix.  Then which one of the following is not always true ?

Let A be a square matrix all of whose entries are integers. Then which one of the following is true?

If A is an 3 x 3 non - singular matrix such that AA' = A'A   and  B = A-1  A' , then BB' equals :

Concepts Covered - 5

Inverse of a Matrix

Inverse of a Matrix

A non-singular square matrix A is said to be invertible if there exists a non-singular square matrix B such that

AB = I = BA

and the matrix B is called the inverse of matrix A. Clearly, B should also have the same order as A.

Hence, \\\mathrm{A^{-1} = B \Leftrightarrow AB = \mathbb{I}_n = BA}

 

Formula for A-1

We know

\\\\ \\\mathrm{A(adj A)=|A|\mathbb{I}_n} \\\text{Multiplying both sides by A}^{-1} \\\mathrm{\Rightarrow A^{-1}A (adj A)=A^{-1}\mathbb{I}_n|A|} \\\mathrm{\Rightarrow\mathbb{I}_n(adj A)=A^{-1}|A|\mathbb{I}_n} \,\,\,\,\quad (As \,\,A^{-1}.A= I)\\\mathrm{A^{-1}=\frac{adj A}{\left | A \right |}}

 

Inverse of a 2 x 2 matrix

\\\mathrm{Let\;A\;is\;a\;square\;matrix\;of\;order\;2}\\\mathrm{A=\begin{bmatrix} a &b \\ c & d \end{bmatrix}}\\\mathrm{Then,}\\\mathrm{A^{-1}=\begin{bmatrix} a &b \\ c & d \end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d &-b \\- c & a \end{bmatrix}}

Inverse of a Matrix of order 3 using adjoint

To compute the inverse of matrix A of order 3, first, check whether the matrix is singular or non-singular.

If the matrix is singular, then its inverse does not exist.

If the matrix in non-singular, then the following ar ethe steps to find the Inverse

We\,\,use\,\,the\,\,formula\,\,A^{-1}= \frac{1}{|A|}.adj(A)

  1. Calculating the Matrix of Minors,

  2. then turn that into the Matrix of Cofactors,

  3. then take the transpose (These 3 steps give us the adjoint of matrix A)

  4. multiply that by 1/|A|.

 

For example, 

Let compute the inverse of matrix A,

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

First, find the determinant of A

\\\mathrm{\left |A \right |=\begin{vmatrix} 1 &1 &2 \\ 1& 2 & 3\\ 3 &1 &1 \end{vmatrix}=1\cdot(2\times1-3\times1)-1\cdot(1\times1-3\times3)+2\cdot(1\times1-3\times2)} \\\mathrm{\left | A \right |=-3\neq0}\\\mathrm{\therefore A^{-1}\;exists}

Now, find the minor of each element

\\\mathrm{M_{11}=\begin{vmatrix} 2&3 \\1 &1 \end{vmatrix}=2\times 1-3\times1=-1}\\\\\mathrm{M_{12}=\begin{vmatrix} 1&3 \\3 &1 \end{vmatrix}=1\times 1-3\times3=-8}\\\\\mathrm{M_{13}=\begin{vmatrix} 1&2 \\3 &1 \end{vmatrix}=1\times 1-2\times3=-5}

And here is the calculation for the whole matrix:

Minor matrix 

M=\begin{bmatrix}2\times \:1-3\times 1&1\times 1-3\times 3&1\times 1-2\times 3\\ \:1\times 1-2\times 1&1\times \:1-2\times 3&1\times 1-3\times \:1\\ \:1\times \:3-2\times 2&1\times 3-2\times \:1&1\times 2\:-1\times \:1\end{bmatrix}=\begin{bmatrix}-1&-8&-5\\ -1&-5&-2\\ -1&1&1\end{bmatrix}

 

Now Cofactor of the given matrix

We need to change the sign of alternate cells, like this \begin{bmatrix}+&-&+\\ \:-&+&-\\ \:+&-&+\end{bmatrix}

 

So, Cofactor matrix C = \begin{bmatrix}+\left(-1\right)&-\left(-8\right)&+\left(-5\right)\\ \:-\left(-1\right)&+\left(-5\right)&-\left(-2\right)\\ \:+\left(-1\right)&-\left(1\right)&+\left(1\right)\end{bmatrix}=\begin{bmatrix}-1&8&-5\\ \:1&-5&2\\ \:-1&-1&1\end{bmatrix}

Now to find the adj A, take the transpose of matrix C

Adj\:A=C'=\begin{bmatrix}-1&1&-1\\ \:\:8&-5&-1\\ \:\:-5&2&1\end{bmatrix}

\\\mathrm{Hence, A^{-1}=\frac{adj\;A}{|A|}}\\\mathrm{A^{-1}=\frac{1}{-3}\begin{bmatrix}-1&1&-1\\ \:\:8&-5&-1\\ \:\:-5&2&1\end{bmatrix} =\begin{bmatrix}-\frac{1}{-3}&\frac{1}{-3}&-\frac{1}{-3}\\ \:\:\frac{8}{-3}&-\frac{5}{-3}&-\frac{1}{-3}\\ \:\:-\frac{5}{-3}&\frac{2}{-3}&\frac{1}{-3}\end{bmatrix}}\\\mathrm{A^{-1}=\begin{bmatrix}\;\frac{1}{3}&-\frac{1}{3}&\frac{1}{3}\\ \:\:-\frac{8}{3}&\frac{5}{3}&\frac{1}{3}\\ \:\:\frac{5}{3}&-\frac{2}{3}&-\frac{1}{3}\end{bmatrix}}

Properties of Inverse of a Matrix - Part 1

Properties of inverse of a matrix:

1. Inverse of a matrix is unique

        Proof:

        Let A be a square and non-singular matrix and let B and C be two inverses of matrix A

        \\\mathrm{AB= BA =\mathbb{I}_n \;\; (since \;B\; is\; inverse\; of\; A)} \\\mathrm{AC= CA =\mathbb{I}_n \;\; (since \;C\; is\; inverse\; of\; A)} \\\mathrm{now, } \;\;\mathrm{AB=\mathbb{I}_n} \\\mathrm{C(AB)=C\mathbb{I}_n\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left [ Multiplication\;by\;C \right ]} \\\mathrm{(CA)B=C\mathbb{I}_n\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left [ by\;associativity \right ]} \\\mathrm{\mathbb{I}_nB=C\mathbb{I}_n} \Rightarrow B=C

        Hence an invertible matrix has a unique inverse.

 

2.  If A and B are invertible matrices of order n, then AB will also be invertible. and (AB)-1 = B-1A-1.

        Proof : 

        \\\text{A and B are invertible matrices}\text{, so}\;|A|\neq 0\;and\;|B|\neq0\\\text{Hence},\;|A||B|\neq0\Rightarrow |AB|\neq0

        \\\text{now,}\;(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}\;\;\;\;\;\;\;\;\;\;\text{[by\;associative\;law]}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=}A(I_n)A^{-1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{[\because BB^{-1}=I_n]}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=}AA^{-1}=I_n\\\text{also,}\;(B^{-1}A^{-1})(AB)=B^{-1}(A^{-1}A)B\;\;\;\;\;\;\;\;\;\;\text{[by\;associative\;law]}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=}B^{-1}(I_nB)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{[\because A^{-1}A=I_n]}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=}B^{-1}B=I_n\\\text{Thus,}\;\;(AB)(B^{-1}A^{-1})=I_n=(B^{-1}A^{-1})(AB)\\\text{Hence,}\;\;(AB)^{-1}=B^{-1}A^{-1}

Properties of Inverse of a Matrix - Part 2

Properties of Inverse of a Matrix

3. If A is an invertible matrix, then

        (A')-1 = (A-1) '

        Proof: As A is an invertible matrix, so |A| ≠ 0 ⇒ |A' | ≠ 0. Hence, A' is also invertible.

        \\\mathrm{Now, \;\;\;AA^{-1}=\mathbb{I}_n=A^{-1}A} \\\text{Taking transpose of all three sides}\\\mathrm{\Rightarrow (AA^{-1})'=(\mathbb{I}_n)' = (A^{-1}A)'} \\\mathrm{\Rightarrow (A^{-1})'A'=\mathbb{I}=A'(A^{-1})'} \\\mathrm{(A')^{-1} = (A^{-1})'}

4. Let A be an invertible matrix, then, (A-1)-1=A

        Proof:

        Let A be an invertible matrix of order n.

        \\As\,\,A.A^{-1}=I=A^{-1}.A\\\mathrm{\Rightarrow \;\;\;\;\;(A^{-1})^{-1}=A}

5. Let A be an invertible matrix of order n and k is a natural number, then (Ak)-1 = (A-1)= A-k 

        Proof: 

        \\\mathrm{(A^k)^{-1}=(A\times A\times A\times \ldots\times A)^{-1}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;=(A^{-1}\times A^{-1}\times A^{-1}\times \ldots\times A^{-1})}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;=(A^{-1})}^k

Properties of Inverse of a Matrix - Part 3

Properties of Inverse of a Matrix

6. Let A be an invertible matrix of order n, then

          \\\mathrm{|A^{-1}| = \frac{1}{\left | A \right |}}

       Proof: ∵ A is invertible, then |A| ≠ 0.

        \\\mathrm{now, \;\;AA^{-1} =\mathbb{I}_n=A^{-1}A} \\\\\mathrm{\Rightarrow |AA^{-1}|=|\mathbb{I}_n|} \\\\\mathrm{\Rightarrow |A||A^{-1}| = 1} \\\\\mathrm{\Rightarrow|A^{-1}|= \frac{1}{\left | A \right |}}

 

7.  Inverse of a non-singular diagonal matrix is a diagonal matrix

         \\\text{if }A = \begin{bmatrix} a &0 & 0\\ 0 & b &0 \\ 0 &0 & c \end{bmatrix} \text{and}\;\mathrm{|A|\neq 0}

        then

         \mathrm{A^{-1}=\begin{bmatrix} \frac{1}{a} &0 & 0\\ 0& \frac{1}{b} &0 \\ 0& 0 & \frac{1}{c} \end{bmatrix}}

Study it with Videos

Inverse of a Matrix
Inverse of a Matrix of order 3 using adjoint
Properties of Inverse of a Matrix - Part 1

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