Q. 4.77 - Intermediate Algebra

Question

Perform the indicated operations on the augmented matrix: $\left[\begin{array}{rrrr}5& -2& -2& -2\\ 4& -1& -4& 4\\ -2& 3& 0& -1\end{array}\right]$

$\left(a\right)$ Interchange rows $1$ and $3$.

$\left(b\right)$ Multiply row $3$ by $3$.

$\left(c\right)$ Multiply row $3$ by $2$ and add to row $2$. 

Step-by-Step Solution

Answer

Part $\left(a\right)$ Interchange rows $1$ and $3$ :-

$\left[\begin{array}{rrrr}-2& 3& 0& -1\\ 4& -1& -4& 4\\ 5& -2& -2& -2\end{array}\right]$

Part $\left(b\right)$ Multiply row $3$ by $3$ :-

$\left[\begin{array}{rrrr}5& -2& -2& -2\\ 4& -1& -4& 4\\ -6& 9& 0& -3\end{array}\right]$

Part $\left(c\right)$ Multiply row $3$ by $2$ and add to row $2$ :-

$\left[\begin{array}{rrrr}5& -2& -2& -2\\ 0& 5& -4& 2\\ -2& 3& 0& -1\end{array}\right]$

Part (a) Step 1. Interchange the rows 1 and 3

Given augmented matrix is $\left[\begin{array}{rrrr}5& -2& -2& -2\\ 4& -1& -4& 4\\ -2& 3& 0& -1\end{array}\right]$

We interchange rows $1$ and $3$

$\left[\begin{array}{rrrr}5& -2& -2& -2\\ 4& -1& -4& 4\\ -2& 3& 0& -1\end{array}\right] \stackrel{R_1\leftrightarrow R_3}{\to } \left[\begin{array}{rrrr}-2& 3& 0& -1\\ 4& -1& -4& 4\\ 5& -2& -2& -2\end{array}\right]$

Part (b) Step 1. Multiply row 3 by 3 .

Given augmented matrix is $\left[\begin{array}{rrrr}5& -2& -2& -2\\ 4& -1& -4& 4\\ -2& 3& 0& -1\end{array}\right]$

We multiply row $3$ by $3$

$\left[\begin{array}{rrrr}5& -2& -2& -2\\ 4& -1& -4& 4\\ -2& 3& 0& -1\end{array}\right] \stackrel{3R_3}{\to } \left[\begin{array}{rrrr}5& -2& -2& -2\\ 4& -1& -4& 4\\ -6& 9& 0& -3\end{array}\right]$

Part (c) Step 1. Multiply row 3 by 2 and add to row 2 .

Given augmented matrix is $\left[\begin{array}{rrrr}5& -2& -2& -2\\ 4& -1& -4& 4\\ -2& 3& 0& -1\end{array}\right]$

We multiply row $3$ by $2$ and add to row $2$

$\left[\begin{array}{rrrr}5& -2& -2& -2\\ 4& -1& -4& 4\\ -2& 3& 0& -1\end{array}\right]\stackrel{2R_3+R_2}{\to }\left[\begin{array}{rrrr}5& -2& -2& -2\\ (2*(-2\left)\right)+4& (2*3)-1& (2*0)-4& (2*(-1)+4\\ -2& 3& 0& -1\end{array}\right]$

$\left[\begin{array}{rrrr}5& -2& -2& -2\\ 4& -1& -4& 4\\ -2& 3& 0& -1\end{array}\right]\stackrel{2R_3+R_2}{\to }\left[\begin{array}{rrrr}5& -2& -2& -2\\ 0& 5& -4& 2\\ -2& 3& 0& -1\end{array}\right]$