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Find the inverse of the matrix A=\begin{bmatrix}1&2&0\\ 3&2&5\\ 1&2&3\end{bmatrix}

Option: 1

A^{-1}=\frac{1}{3}\begin{bmatrix}\frac{1}{3}&\frac{1}{2}&-\frac{5}{6}\\ \frac{1}{3}&\frac{1}{4}&\frac{5}{12}\\ \frac{1}{3}&0&\frac{1}{3}\end{bmatrix}


Option: 2

A^{-1}=\begin{bmatrix}\frac{1}{3}&\frac{1}{2}&-\frac{5}{6}\\ \frac{1}{3}&\frac{1}{4}&\frac{5}{12}\\ \frac{1}{3}&0&\frac{1}{3}\end{bmatrix}


Option: 3

A^{-1}=\frac{1}{3}\begin{bmatrix}\frac{1}{3}&\frac{1}{2}&-\frac{5}{6}\\ \frac{1}{3}&-\frac{1}{4}&\frac{5}{12}\\ -\frac{1}{3}&0&\frac{1}{3}\end{bmatrix}


Option: 4

A^{-1}=\begin{bmatrix}\frac{1}{3}&\frac{1}{2}&-\frac{5}{6}\\ \frac{1}{3}&-\frac{1}{4}&\frac{5}{12}\\ -\frac{1}{3}&0&\frac{1}{3}\end{bmatrix}


Answers (1)

best_answer

 

 

Using elementary operations to compute the inverse of matrix of order 3 -

Algorithm for Finding the lnverse of a Non singular 3 x 3 Matrix by Elementary Row Transformations

  1. Introduce unity at the intersection of first row and first column either by interchanging two rows or by adding a constant multiple of elements of some other row to first row.
  2. After introducing unity at (1,1) place introduce zeros at all other places in first column.
  3. Introduce unity at the intersection of 2nd row and 2nd column with the help of 2nd and 3rd row.
  4. Introduce zeros at all other places in the second column except at the intersection of 2nd row and 2nd column.
  5. Introduce unity at the intersection of 3rd row and third column.
  6. Finally introduce zeros at all other places in the third column except at the intersection of third row and third column.

 

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Use AA-1=I

\\\begin{bmatrix}1&2&0\\ 3&2&5\\ 1&2&3\end{bmatrix}=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}A\\\\\:R_1\:\leftrightarrow \:R_2\\\\\begin{bmatrix}3&2&5\\ 1&2&0\\ 1&2&3\end{bmatrix}=\begin{bmatrix}0&1&0\\ 1&0&0\\ 0&0&1\end{bmatrix}A

R_2\:\rightarrow \:R_2-\frac{1}{3}\cdot \:R_1\\\\\begin{bmatrix}3&2&5\\ 0&\frac{4}{3}&-\frac{5}{3}\\ 1&2&3\end{bmatrix}=\begin{bmatrix}0&1&0\\ 1&-\frac{1}{3}&0\\ 0&0&1\end{bmatrix}A

R_3\:\rightarrow \:R_3-\frac{1}{3}\cdot \:R_1\\\\\begin{bmatrix}3&2&5\\ 0&\frac{4}{3}&-\frac{5}{3}\\ 0&\frac{4}{3}&\frac{4}{3}\end{bmatrix}=\begin{bmatrix}0&1&0\\ 1&-\frac{1}{3}&0\\ 0&-\frac{1}{3}&1\end{bmatrix}A

R_3\:\rightarrow \:R_3-1\cdot \:R_2\\\\\begin{bmatrix}3&2&5\\ 0&\frac{4}{3}&-\frac{5}{3}\\ 0&0&3\end{bmatrix}=\begin{bmatrix}0&1&0\\ 1&-\frac{1}{3}&0\\ -1&0&1\end{bmatrix}A

R_3\:\rightarrow \frac{1}{3}\cdot \:R_3\\\\\begin{bmatrix}3&2&5\\ 0&\frac{4}{3}&-\frac{5}{3}\\ 0&0&1\end{bmatrix}=\begin{bmatrix}0&1&0\\ 1&-\frac{1}{3}&0\\ -\frac{1}{3}&0&\frac{1}{3}\end{bmatrix}A

R_2\:\rightarrow \:R_2+\frac{5}{3}\cdot \:R_3\\\\\begin{bmatrix}3&2&5\\ 0&\frac{4}{3}&0\\ 0&0&1\end{bmatrix}=\begin{bmatrix}0&1&0\\ \frac{4}{9}&-\frac{1}{3}&\frac{5}{9}\\ -\frac{1}{3}&0&\frac{1}{3}\end{bmatrix}A

R_1\:\rightarrow \:R_1-5\cdot \:R_3\\\\\begin{bmatrix}3&2&0\\ 0&\frac{4}{3}&0\\ 0&0&1\end{bmatrix}=\begin{bmatrix}\frac{5}{3}&1&-\frac{5}{3}\\ \frac{4}{9}&-\frac{1}{3}&\frac{5}{9}\\ -\frac{1}{3}&0&\frac{1}{3}\end{bmatrix}A

R_2\:\rightarrow \frac{3}{4}\cdot \:R_2\\\\\begin{bmatrix}3&2&0\\ 0&1&0\\ 0&0&1\end{bmatrix}=\begin{bmatrix}\frac{5}{3}&1&-\frac{5}{3}\\ \frac{1}{3}&-\frac{1}{4}&\frac{5}{12}\\ -\frac{1}{3}&0&\frac{1}{3}\end{bmatrix}A

R_1\:\rightarrow \:R_1-2\cdot \:R_2\\\\\begin{bmatrix}3&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}=\begin{bmatrix}1&\frac{3}{2}&-\frac{5}{2}\\ \frac{1}{3}&-\frac{1}{4}&\frac{5}{12}\\ -\frac{1}{3}&0&\frac{1}{3}\end{bmatrix}A

\:R_1\:\rightarrow \frac{1}{3}\cdot \:R_1\\\\\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}=\begin{bmatrix}\frac{1}{3}&\frac{1}{2}&-\frac{5}{6}\\ \frac{1}{3}&-\frac{1}{4}&\frac{5}{12}\\ -\frac{1}{3}&0&\frac{1}{3}\end{bmatrix}A

hence, A^{-1}=\begin{bmatrix}\frac{1}{3}&\frac{1}{2}&-\frac{5}{6}\\ \frac{1}{3}&-\frac{1}{4}&\frac{5}{12}\\ -\frac{1}{3}&0&\frac{1}{3}\end{bmatrix}

option D is correct.

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chirag

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