Problem 15
Question
perform each matrix row operation and write the new matrix. $$ \left[\begin{array}{rrr|r} 1 & -3 & 2 & 0 \\ 3 & 1 & -1 & 7 \\ 2 & -2 & 1 & 3 \end{array}\right] $$ $$ -3 R_{1}+R_{2} $$
Step-by-Step Solution
Verified Answer
The new matrix after performing the row operation \(-3R_{1}+R_{2}\) is \[\begin{array}{rrr|r} 1 & -3 & 2 & 0 \ 0 & 10 & -7 & 7 \ 2 & -2 & 1 & 3 \end{array}\]
1Step 1: Identify the Rows in the Matrix
The matrix is presented with three rows: \(R_1 = [1,-3,2,0]\), \(R_2 = [3,1,-1,7]\) and \(R_3 = [2,-2,1,3]\). The row operation is described as \(-3 R_{1}+R_{2}\), indicating that the first row will be multiplied by -3 and the result will be added to the second row.
2Step 2: Perform the Row Operation
Multiply each number in the first row by -3, giving us \([-3,9,-6,0]\). These calculated values will be added to the corresponding values in the second row. The new second row \(R_2'\) will be \([0,10,-7,7]\).
3Step 3: Write the New Matrix
Substitute the original second row with the newly calculated row. The new matrix after the row operation will be:\[\begin{array}{rrr|r} 1 & -3 & 2 & 0 \ 0 & 10 & -7 & 7 \ 2 & -2 & 1 & 3 \end{array}\]
Key Concepts
Elementary Row OperationsAugmented MatrixLinear SystemsMatrix Manipulation
Elementary Row Operations
Elementary row operations are essential manipulations in linear algebra that allow us to simplify and solve matrix-related problems. They help us to perform operations without changing the solutions of the associated linear systems. These operations include:
- Row Swapping: Interchanging two rows in a matrix.
- Row Multiplication: Multiplying all elements of a row by a non-zero constant.
- Row Addition: Adding or subtracting the elements of one row from another.
Augmented Matrix
An augmented matrix is a powerful tool used to represent a system of linear equations in matrix form. This type of matrix includes both the coefficients of the variables and the constants from the equations' right-hand side, separated by a vertical line. The layout helps in visually organizing equations to apply row operations conveniently. For instance, in the exercise given, the matrix\[\begin{array}{ccc|c}1 & -3 & 2 & 0 \3 & 1 & -1 & 7 \2 & -2 & 1 & 3\end{array}\]represents a system of three equations with three unknowns. The vertical bar separates the coefficients on the left from the constants on the right. Working with augmented matrices allows you to efficiently apply elementary row operations and solve for the unknowns by transforming the matrix to a simpler form.
Linear Systems
Linear systems comprise multiple linear equations that are solved simultaneously to find values for their variables. The number of solutions to a linear system can vary:
- Unique Solution: A single, distinct set of values for the variables.
- Infinite Solutions: An endless number of possibilities for the variable values.
- No Solution: No set of values exists that will satisfy all equations simultaneously.
Matrix Manipulation
Matrix manipulation involves a set of methods and techniques used to change the form or elements of a matrix. The tools of matrix manipulation, including elementary row operations, are useful for solving linear systems and other matrix problems. The goal can be
- Row Echelon Form: Simplifying the matrix to make finding solutions easier.
- Determinants and Inverses: In more advanced applications, determining matrix properties or finding matrix inverses requires manipulation.
Other exercises in this chapter
Problem 15
Use Cramer's Rule to solve each system. $$\left\\{\begin{array}{l}4 x-5 y=17 \\\2 x+3 y=3\end{array}\right.$$
View solution Problem 15
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$\left\\{\begin{array}{l}2 x+y-z=2 \\\3 x+3 y-2 z
View solution Problem 16
Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$A=\left[\begin{array}{rrr}6 & -3 & 5 \\\6 & 0 & -2 \\\\-4 & 2 & -1\end{array}\ri
View solution Problem 16
In Exercises \(13-18,\) use the fact that if \(A=\left[\begin{array}{ll}a & b \\\ c & d\end{array}\right]\), then \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{
View solution