Solveeit Logo
Login

Question

Question: Using elementary row transformation find the inverse of the matrix \(A = \begin{bmatrix} 3 & - 1 & -...

Using elementary row transformation find the inverse of the matrix A=[3−1−220−13−50]A = \begin{bmatrix} 3 & - 1 & - 2 \\ 2 & 0 & - 1 \\ 3 & - 5 & 0 \end{bmatrix}

A

$\begin{bmatrix}

  • 5/8 & 5/4 & 1/8 \
  • 3/8 & 3/4 & - 1/8 \
  • 5/4 & 3/2 & 1/4 \end{bmatrix}$
B

18[5−513−31031]\frac{1}{8}\begin{bmatrix} 5 & - 5 & 1 \\ 3 & - 3 & 1 \\ 0 & 3 & 1 \end{bmatrix}

C

18[55136−110−122]\frac{1}{8}\begin{bmatrix} 5 & 5 & 1 \\ 3 & 6 & - 1 \\ 10 & - 12 & 2 \end{bmatrix}

D

None of these

Answer

$\begin{bmatrix}

  • 5/8 & 5/4 & 1/8 \
  • 3/8 & 3/4 & - 1/8 \
  • 5/4 & 3/2 & 1/4 \end{bmatrix}$
Explanation

Solution

We have A=IA ⇒ [3−1−220−13−50]=[100010001]A\begin{bmatrix} 3 & - 1 & - 2 \\ 2 & 0 & - 1 \\ 3 & - 5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}A

Applying (R1→R1−R2)\left( R _ { 1 } \rightarrow R _ { 1 } - R _ { 2 } \right) [1−1−120−13−50]=[1−10010001]A\begin{bmatrix} 1 & - 1 & - 1 \\ 2 & 0 & - 1 \\ 3 & - 5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}A

Applying R2→R2−2R1R_{2} \rightarrow R_{2} - 2R_{1} and R3→R3−3R1R_{3} \rightarrow R_{3} - 3R_{1},

$\begin{bmatrix} 1 & - 1 & - 1 \ 0 & 2 & - 1 \ 0 & - 2 & 3 \end{bmatrix} = \begin{bmatrix} 1 & - 1 & 0 \

  • 2 & 3 & 0 \
  • 3 & 3 & 1 \end{bmatrix}AApplyingApplyingR_{2} \rightarrow R_{2}/2,\begin{bmatrix} 1 & - 1 & - 1 \ 0 & 1 & 1/2 \ 0 & - 2 & 3 \end{bmatrix} = \begin{bmatrix} 1 & - 1 & 0 \
  • 1 & 3/2 & 0 \
  • 3 & 3 & 1 \end{bmatrix}A$

Applying R1→R1+R2R_{1} \rightarrow R_{1} + R_{2} and R3→R3+2R2R_{3} \rightarrow R_{3} + 2R_{2},

1 & 0 & - 1/2 \\ 0 & 1 & 1/2 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 0 & 1/2 & 0 \\ - 1 & 3/2 & 0 \\ - 5 & 6 & 1 \end{bmatrix}A$$ Applying $R_{3} \rightarrow R_{3}/4$, $\begin{bmatrix} 1 & 0 & - 1/2 \\ 0 & 1 & 1/2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1/2 & 0 \\ - 1 & 3/2 & 0 \\ 5/4 & 6/4 & 1/4 \end{bmatrix}A$Applying $R_{1} \rightarrow R_{1} + \frac{1}{2}R_{3}\text{and }R_{2} \rightarrow R_{2} - \frac{1}{2}R_{3}$,$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} - 5/8 & 5/4 & 1/8 \\ - 3/8 & 3/4 & - 1/8 \\ - 5/4 & 3/2 & 1/4 \end{bmatrix}A$ $A^{- 1} = \begin{bmatrix} - 5/8 & 5/4 & 1/8 \\ - 3/8 & 3/4 & - 1/8 \\ - 5/4 & 3/2 & 1/4 \end{bmatrix}$