Solveeit Logo
Login

Question

Mathematics Question on Matrices

Using elementary transformations, find the inverse of matrix [−112 123 311].\begin{bmatrix}-1&1&2\\\ 1&2&3\\\ 3&1&1\end{bmatrix}.

A

[1−11 −87−5 5−43]\begin{bmatrix}1&-1&1\\\ -8&7&-5\\\ 5&-4&3\end{bmatrix}

B

[2−11 −67−5 5−43]\begin{bmatrix}2&-1&1\\\ -6&7&-5\\\ 5&-4&3\end{bmatrix}

C

[2−11 −64−5 5−43]\begin{bmatrix}2&-1&1\\\ -6&4&-5\\\ 5&-4&3\end{bmatrix}

D

[1−11 −64−5 5−43]\begin{bmatrix}1&-1&1\\\ -6&4&-5\\\ 5&-4&3\end{bmatrix}

Answer

[1−11 −87−5 5−43]\begin{bmatrix}1&-1&1\\\ -8&7&-5\\\ 5&-4&3\end{bmatrix}

Explanation

Solution

Given, matrix A=[−112 123 311]A =\begin{bmatrix}-1&1&2\\\ 1&2&3\\\ 3&1&1\end{bmatrix}, then A=IAA = IA ⇒[−112 123 311]=[100 010 001]A\Rightarrow\begin{bmatrix}-1&1&2\\\ 1&2&3\\\ 3&1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\end{bmatrix}A Applying R2→R2+R1,R3→R3+3R1R_{2}\rightarrow R_{2}+R_{1}, R3\rightarrow R_{3} + 3R_{1}, we get [−112 035 047]=[100 010 301]A\begin{bmatrix}-1&1&2\\\ 0&3&5\\\ 0&4&7\end{bmatrix}=\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 3&0&1\end{bmatrix}A Applying R1→(−1)R1R_{1}\rightarrow \left(-1\right)R_{1}, we get [1−1−2 035 047]=[−100 110 301]A\begin{bmatrix}1&-1&-2\\\ 0&3&5\\\ 0&4&7\end{bmatrix}=\begin{bmatrix}-1&0&0\\\ 1&1&0\\\ 3&0&1\end{bmatrix}A Applying R2→R2−R3R_{2}\rightarrow R_{2}-R_{3}, we get [1−1−2 0−1−2 047]=[−100 −21−1 301]A\begin{bmatrix}1&-1&-2\\\ 0&-1&-2\\\ 0&4&7\end{bmatrix}=\begin{bmatrix}-1&0&0\\\ -2&1&-1\\\ 3&0&1\end{bmatrix}A Applying R1→R1−R2,andR3→R3+4R2R_{1}\rightarrow R_{1} -R_{2}, and R_{3}\rightarrow R_{3} + 4R_{2 },we get [100 0−1−2 00−1]=[1−11 −21−1 −54−3]A\begin{bmatrix}1&0&0\\\ 0&-1&-2\\\ 0&0&-1\end{bmatrix}=\begin{bmatrix}1&-1&1\\\ -2&1&-1\\\ -5&4&-3\end{bmatrix}A Applying R2→(−1)R2R_{2}\rightarrow\left(-1\right)R_{2}, we get [100 012 00−1]=[1−11 2−11 54−3]A\begin{bmatrix}1&0&0\\\ 0&1&2\\\ 0&0&-1\end{bmatrix}=\begin{bmatrix}1&-1&1\\\ 2&-1&1\\\ 5&4&-3\end{bmatrix}A Applying R2→R2+2R3R_{2}\rightarrow R_{2} +2R_{3}, we get [100 010 00−1]=[1−11 −87−5 5−43]A\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 0&0&-1\end{bmatrix}=\begin{bmatrix}1&-1&1\\\ -8&7&-5\\\ 5&-4&3\end{bmatrix}A A−1=[1−11 −87−5 5−43]A^{-1}=\begin{bmatrix}1&-1&1\\\ -8&7&-5\\\ 5&-4&3\end{bmatrix}