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Find the inverse of matrix A=\begin{bmatrix}1&-1&1\\ 1&3&-1\\ 2&4&1\end{bmatrix} using adjoint method

Option: 1

\mathrm{A}^{-1}=\begin{bmatrix}-\frac{7}{8}&\frac{5}{8}&-\frac{1}{4}\\ -\frac{3}{8}&-\frac{1}{8}&\frac{1}{4}\\ -\frac{1}{4}&-\frac{3}{4}&-\frac{1}{2}\end{bmatrix}


Option: 2

\mathrm{A}^{-1}=\begin{bmatrix}\frac{7}{8}&\frac{5}{8}&\frac{1}{4}\\ -\frac{3}{8}&-\frac{1}{8}&\frac{1}{4}\\ -\frac{1}{4}&\frac{3}{4}&\frac{1}{2}\end{bmatrix}


Option: 3

\mathrm{A}^{-1}=\begin{bmatrix}\frac{7}{8}&\frac{5}{8}&-\frac{1}{4}\\ -\frac{3}{8}&-\frac{1}{8}&\frac{1}{4}\\ -\frac{1}{4}&-\frac{3}{4}&\frac{1}{2}\end{bmatrix}


Option: 4

\mathrm{A}^{-1}=\begin{bmatrix}\frac{7}{8}&\frac{5}{8}&-\frac{1}{4}\\ -\frac{3}{8}&\frac{1}{8}&\frac{1}{4}\\ -\frac{1}{4}&-\frac{3}{4}&\frac{1}{2}\end{bmatrix}


Answers (1)

best_answer

 

 

Inverse of a Matrix of order 3 using adjoint -

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

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

Steps to Find the Inverse of a matrix

  1. Calculating the Matrix of Minors,

  2. then turn that into the Matrix of Cofactors,

  3. then take the transpose, and

  4. multiply that by 1/Determinant of the given matrix.

    

-

First, find the determinant of A 

\left | A \right |=1\cdot \det \begin{bmatrix}3&-1\\ 4&1\end{bmatrix}-\left(-1\right)\det \begin{bmatrix}1&-1\\ 2&1\end{bmatrix}+1\cdot \det \begin{bmatrix}1&3\\ 2&4\end{bmatrix}

\left | A \right |=1\cdot \:7-\left(-1\right)\cdot \:3+1\cdot \left(-2\right)=8

Now find minor of each element

M_{11}=\det \begin{bmatrix}3&-1\\ 4&1\end{bmatrix}= 7

M_{1 2}=\det \begin{bmatrix}1&-1\\ 2&1\end{bmatrix}= 3

M_{1 3}=\det \begin{bmatrix}1&3\\ 2&4\end{bmatrix}=-2

M_{2 1}=\det \begin{bmatrix}-1&1\\ 4&1\end{bmatrix}= -5

M_{22}=\det \begin{bmatrix}1&1\\ 2&1\end{bmatrix}= -1

M_{2 3}=\det \begin{bmatrix}1&-1\\ 2&4\end{bmatrix}= 6

M_{3 1}=\det \begin{bmatrix}-1&1\\ 3&-1\end{bmatrix}=-2

M_{3 2}=\det \begin{bmatrix}1&1\\ 1&-1\end{bmatrix}= -2

M_{3 3}=\det \begin{bmatrix}1&-1\\ 1&3\end{bmatrix}= 4

\begin{array}{l}{\text { Matrix of Minors } M_{i, j}} \\ {\left[\begin{array}{ccc}{7} & {3} & {-2} \\ {-5} & {-1} & {6} \\ {-2} & {-2} & {4}\end{array}\right]}\end{array}

\begin{array}{l}{\text { Matrix of Cofactors } C_{i, j}=(-1)^{i+j} M_{i, j}} \\\\ {\left[\begin{array}{ccc}{(-1)^{1+1} \cdot 7} & {(-1)^{2+1} \cdot 3} & {(-1)^{1+3}(-2)} \\ {(-1)^{2+1}(-5)} & {(-1)^{2+2}(-1)} & {(-1)^{2+3} \cdot 6} \\ {(-1)^{3+1}(-2)} & {(-1)^{3+2}(-2)} & {(-1)^{3+3} \cdot 4}\end{array} \right]}\end{array}\\\\

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

adjoint of A is 

\begin{bmatrix}7&-3&-2\\ 5&-1&-6\\ -2&2&4\end{bmatrix}^T=\begin{bmatrix}7&5&-2\\ -3&-1&2\\ -2&-6&4\end{bmatrix}

hence,

\mathrm{A}^{-1}=\frac{\text { adj } \mathrm{A}}{|\mathrm{A}|}=\frac{1}{8}\begin{bmatrix}7&5&-2\\ -3&-1&2\\ -2&-6&4\end{bmatrix}

\mathrm{A}^{-1}=\begin{bmatrix}\frac{7}{8}&\frac{5}{8}&-\frac{1}{4}\\ -\frac{3}{8}&-\frac{1}{8}&\frac{1}{4}\\ -\frac{1}{4}&-\frac{3}{4}&\frac{1}{2}\end{bmatrix}

 

Posted by

Ritika Jonwal

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