Problem 21
Question
In Exercises \(19-24\), perform each matrix row operation and write the new matrix. \(\left[\begin{array}{rrr|r}3 & -12 & 6 & 9 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 & 4\end{array}\right] \quad \frac{1}{3} R_{1}\)
Step-by-Step Solution
Verified Answer
The new matrix after performing the operation \(\frac{1}{3} R_{1}\) is \[\left[\begin{array}{rrr|r}1 & -4 & 2 & 3 \ 1 & -4 & 4 & 0 \ 2 & 0 & 7 & 4\end{array}\right]\].
1Step 1: Identify the operation
The operation to be performed here is \(\frac{1}{3} R_{1}\). This means that each element in row 1 is to be divided by 3. The matrix provided is: \[\left[\begin{array}{rrr|r}3 & -12 & 6 & 9 \ 1 & -4 & 4 & 0 \ 2 & 0 & 7 & 4\end{array}\right]\]
2Step 2: Perform the operation
Perform the operation \(\frac{1}{3} R_{1}\) on the matrix, i.e., divide each element in the first row by 3. This gives us a new matrix as: \[\left[\begin{array}{rrr|r}1 & -4 & 2 & 3 \ 1 & -4 & 4 & 0 \ 2 & 0 & 7 & 4\end{array}\right]\]
Key Concepts
Elementary Row Operations
Elementary Row Operations
In linear algebra, mastering elementary row operations is a critical skill that serves as the foundation for many complex procedures such as solving systems of equations, finding determinants, and determining the rank of a matrix. Elementary row operations consist of three types:
Other exercises in this chapter
Problem 21
find \(A^{-1}\) by forming \([A | I]\) and then using row operations to obtain [ \(I | B],\) where \(A^{-1}=[B]\) Check that \(A A^{-1}=I\) and \(A^{-1} A=I\) $
View solution Problem 21
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$\begin{aligned}2 w-3 x+4 y+z &=7 \\\w-x+3 y-5 z
View solution Problem 22
In Exercises \(17-26,\) let $$ A=\left[\begin{array}{rr} -3 & -7 \\ 2 & -9 \\ 5 & 0 \end{array}\right] \text { and } B=\left[\begin{array}{rr} -5 & -1 \\ 0 & 0
View solution Problem 22
Use Cramer's rule to solve each system or to determine that the system is inconsistent or contains dependent equations. $$ \begin{array}{rr}y= & -4 x+2 \\\2 x=
View solution