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Use Gaussian Elimination to put the given matrix into reduced row echelon form. $$\left[\begin{array}{lll}-1 & 1 & 4 \\ -2 & 1 & 1\end{array}\right]$$

Short Answer

Expert verified
The final reduced row echelon form is \( \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 3 \end{bmatrix}.\)

Step by step solution

01

Write the Matrix

Write the matrix from the problem: \[\begin{bmatrix} -1 & 1 & 4 \ -2 & 1 & 1 \end{bmatrix}\] This matrix has two rows and three columns.
02

Create a Zero Below the First Pivot

Perform a row operation to eliminate the value below the pivot in the first column. We can add twice the first row to the second row to achieve this: - \(R_2 = R_2 + 2R_1\)New matrix:\[\begin{bmatrix} -1 & 1 & 4 \ 0 & 3 & 9 \end{bmatrix}\]
03

Scale the Second Row to Make the Pivot 1

Scale the second row to make the second pivot equal to 1 by dividing the entire row by 3:- \(R_2 = \frac{1}{3}R_2\)New matrix:\[\begin{bmatrix} -1 & 1 & 4 \ 0 & 1 & 3 \end{bmatrix}\]
04

Eliminate Above the Second Pivot

To create a zero above the second pivot, modify the first row by subtracting the second row:- \(R_1 = R_1 - R_2\)New matrix:\[\begin{bmatrix} -1 & 0 & 1 \ 0 & 1 & 3 \end{bmatrix}\]
05

Scale First Row to Make the Pivot 1

Finally, scale the first row to make the first pivot equal to 1 by multiplying the entire row by -1:- \(R_1 = -1 \times R_1\)Final matrix:\[\begin{bmatrix} 1 & 0 & -1 \ 0 & 1 & 3 \end{bmatrix}\] This is the reduced row-echelon form of the matrix.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reduced Row Echelon Form
When we talk about matrices, the reduced row echelon form (RREF) is a special form that simplifies solving systems of linear equations. In RREF, each leading entry (also known as a pivot) is 1, and is the only non-zero entry in its column. Moreover, each pivot is to the right of the pivot in the row above it, and rows consisting entirely of zeros are at the bottom.
This form is incredibly useful because it allows us to easily read off the solutions to a system of equations. If a matrix is in RREF, the solutions can often be immediately extracted by inspection. This is especially useful in larger systems where row operations might lead to confusion. Understanding RREF is a fundamental skill for solving linear systems efficiently.
Matrix Operations
Matrix operations are the calculations you perform to transform and manipulate matrices. There are several types of operations:
  • Addition and Subtraction: These are performed element-wise between matrices of the same dimensions.
  • Scalar Multiplication: Each element of a matrix is multiplied by a scalar value.
  • Matrix Multiplication: This involves taking the dot product of the rows of the first matrix with the columns of the second matrix.
In the context of Gaussian elimination, these operations are crucial for transforming a matrix into its reduced row echelon form. When you develop a good understanding of matrix operations, you're able to manipulate and solve complex linear algebra problems with ease. Each operation follows a specific set of rules which keeps these manipulations structured and predictable.
Row Operations
Row operations are the tools we use to transform matrices, particularly when performing Gaussian elimination. There are three primary types of row operations:
  • Row Switching: Swap the positions of two rows.
  • Row Multiplying: Multiply all elements of a row by a non-zero scalar.
  • Row Addition: Add a multiple of one row to another row to produce a new row.
These operations are fundamental because they allow for the alteration of a matrix's form without changing the solutions of the corresponding system of equations. Each operation is reversible, ensuring that the integrity of the system is maintained. In practice, these operations are performed in a specific sequence to systematically solve for the row-echelon and reduced row-echelon forms of a matrix.

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Most popular questions from this chapter

The general exponential function has the form \(f(x)=a e^{b x},\) where \(a\) and \(b\) are constants and \(e\) is Euler's constant \((\approx\) 2.718). We want to find the equation of the exponential function that goes through the points (1,2) and (2,4) . (a) Show why we cannot simply subsitute in values for \(x\) and \(y\) in \(y=a e^{b x}\) and solve using the techniques we used for polynomials. (b) Show how the equality \(y=a e^{b x}\) leads us to the linear equation \(\ln y=\ln a+b x\) (c) Use the techniques we developed to solve for the unknowns In \(a\) and \(b\). (d) Knowing In \(a,\) find \(a\); find the exponential function \(f(x)=a e^{b x}\) that goes through the points (1,2) and (2,4)

State whether or not the given matrices are in reduced row echelon form. If it is not, state why. (a) \(\left[\begin{array}{llll}2 & 0 & 0 & 2 \\ 0 & 2 & 0 & 2 \\ 0 & 0 & 2 & 2\end{array}\right]\) (b) \(\left[\begin{array}{llll}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]\) (c) \(\left[\begin{array}{cccc}0 & 0 & 1 & -5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]\) (d) \(\left[\begin{array}{llllll}1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0\end{array}\right]\)

Find the solution to the given linear system. If the system has infinite solutions, give 2 particular solutions. $$ \begin{aligned} x_{1}+x_{2} &=3 \\ 2 x_{1}+x_{2} &=4 \end{aligned} $$

Find the solution to the given linear system. If the system has infinite solutions, give 2 particular solutions. $$ \begin{aligned} -x_{1}-x_{2}+x_{3}+x_{4} &=0 \\ -2 x_{1}-2 x_{2}+x_{3} &=-1 \end{aligned} $$

Use Gaussian Elimination to put the given matrix into reduced row echelon form. $$\left[\begin{array}{ccc}4 & 5 & -6 \\ -12 & -15 & 18\end{array}\right]$$

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