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Chapter 2: Problem 13

Use Gauss-Jordan elimination to determine the solution set to the given system. $$\begin{array}{rr} 2 x_{1}-x_{2}-x_{3}= & 2 \\ 4 x_{1}+3 x_{2}-2 x_{3}= & -1 \\ x_{1}+4 x_{2}+x_{3}= & 4 \end{array}$$

Short Answer

Expert verified
The solution set to the given system of linear equations using Gauss-Jordan elimination is \(x_{1} = 1\), \(x_{2} = -1\), and \(x_{3} = 2\).

Step by step solution

01

Create the augmented matrix

We will begin by translating the system of equations into an augmented matrix. The left side of the matrix represents the coefficients of the variables (x1, x2, and x3), and the right side of the matrix represents the constants. $$ \left[\begin{array}{ccc|c} 2 & -1 & -1 & 2 \\ 4 & 3 & -2 & -1 \\ 1 & 4 & 1 & 4 \end{array}\right] $$
02

Row reduction

Perform row operations to obtain a reduced-row echelon form of the matrix. We start with Row1 and make the leading coefficient 1 (if not already). $$ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -\frac{1}{2} & 1 \\ 4 & 3 & -2 & -1 \\ 1 & 4 & 1 & 4 \end{array}\right] $$ Next, we want to make zeros below the leading one in Row1. Subtract Row1 * 4 from Row2 and subtract Row1 from Row3: $$ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -\frac{1}{2} & 1 \\ 0 & 5 & 0 & -5 \\ 0 & \frac{9}{2} & \frac{3}{2} & 3 \end{array}\right] $$ Now, we want the leading coefficient in Row2 to be 1 (by dividing Row2 by 5): $$ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -\frac{1}{2} & 1 \\ 0 & 1 & 0 & -1 \\ 0 & \frac{9}{2} & \frac{3}{2} & 3 \end{array}\right] $$ Make zeros below the leading one in Row2. Subtract Row2 * \(\frac{9}{2}\) from Row3: $$ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -\frac{1}{2} & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 3 & 6 \end{array}\right] $$ Now, divide Row3 by 3 to make the leading coefficient 1: $$ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & -\frac{1}{2} & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right] $$ Lastly, make zeros above the leading ones in rows other than Row1. Add Row3 * \(\frac{1}{2}\) to Row1: $$ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & 0 & \frac{3}{2} \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right] $$
03

Obtain the solution set

Now that the matrix is in a reduced-row echelon form, let us write out the corresponding system of linear equations: $$ \begin{array}{ccc} x_{1} - \frac{1}{2}x_{2} = \frac{3}{2} \\ x_{2} = -1 \\ x_{3} = 2 \end{array} $$ Finally, we can easily calculate the solution for x1 by substituting x2 into the first equation: $$ x_{1} - \frac{1}{2}(-1) = \frac{3}{2} \\ x_{1} + \frac{1}{2} = \frac{3}{2} \\ x_{1} = \frac{3}{2} - \frac{1}{2} \\ x_{1} = 1 $$ Thus, the solution set to the given system of linear equations is \(x_{1} = 1\), \(x_{2} = -1\), and \(x_{3} = 2\).

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Key Concepts

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

Augmented Matrix
An augmented matrix is a powerful tool in the world of linear algebra. It provides a compact representation of a system of linear equations. This matrix format includes the coefficients of the variables as well as the constant terms of the equations.
A typical augmented matrix looks like this:
  • The left-hand side contains the coefficients of the variables.
  • The right-hand side contains the constants from the equations.
For example, the given system of equations is transformed into:\[\begin{bmatrix} 2 & -1 & -1 \ 4 & 3 & -2 \ 1 & 4 & 1 \end{bmatrix}\bigg| \begin{bmatrix} 2 \ -1 \ 4 \end{bmatrix} \]Creating an augmented matrix is the first step in applying methods like Gauss-Jordan elimination to find solutions to systems of linear equations.
Row Operations
Row operations are the backbone of transforming matrices into easier forms. In Gauss-Jordan elimination, these operations help in simplifying the matrix to reach a form that readily provides the solution. Various row operations include:
  • Swapping two rows.
  • Multiplying a row by a non-zero scalar.
  • Adding or subtracting the multiples of one row to another.
For instance, in the solution given, row operations were used to make leading coefficients equal to 1 and to nullify the coefficients below these in each column, making calculations straightforward.
Reduced-Row Echelon Form
The reduced-row echelon form (RREF) of a matrix is a version where each leading entry in the rows is 1, and these 1's are the only non-zero entries in their column. This form provides a clear path to finding the solutions.
In the exercise, the matrix ultimately achieved this reduced-row echelon form:\[\begin{bmatrix} 1 & -\frac{1}{2} & 0 & \frac{3}{2} \0 & 1 & 0 & -1 \0 & 0 & 1 & 2 \end{bmatrix}\]Here鈥檚 why RREF is helpful:
  • Each row corresponds directly to an equation involving the variables.
  • The non-trivial nature of the leading 1's ensures that you can directly read off the solution for each variable.
In this case, x1, x2, and x3 can be determined easily from this form. It simplifies the process of solving the system of equations greatly.

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Most popular questions from this chapter

Integration of matrix functions given in the text was done with definite integrals, but one can naturally compute indefinite integrals of matrix functions as well, by performing indefinite integrals for each element of the matrix function. For each element of the matrix \(\int A(t) d t,\) an arbitrary constant of integration must be included, and the arbitrary constants for different elements should be different. Evaluate the indefinite integral \(\int A(t) d t\) for the given matrix function. You may assume that the constants of all indefinite integrations are zero. $$A(t)=\left[\begin{array}{ll} -5 & \frac{1}{t^{2}+1} \end{array} e^{3 t}\right]$$

Reduce the given matrix to reduced rowechelon form and hence determine the rank of each matrix. $$\left[\begin{array}{cccc} 1 & -2 & 1 & 3 \\ 3 & -6 & 2 & 7 \\ 4 & -8 & 3 & 10 \end{array}\right]$$.

Determine the solution set to the given system. $$\begin{aligned} 2 x_{1}+x_{2}-x_{3}+x_{4} &=0 \\ x_{1}+x_{2}+x_{3}-x_{4} &=0 \\ 3 x_{1}-x_{2}+x_{3}-2 x_{4} &=0 \\ 4 x_{1}+2 x_{2}-x_{3}+x_{4} &=0 \end{aligned}$$

Use the LU factorization of \(A=\left[\begin{array}{rr}2 & -1 \\ -8 & 3\end{array}\right]\) to solve each of the systems \(A \mathbf{x}_{i}=\mathbf{b}_{i}\) if $$\mathbf{b}_{1}=\left[\begin{array}{r} 3 \\\\-1\end{array}\right], \quad \mathbf{b}_{2}=\left[\begin{array}{l}2 \\\7\end{array}\right], \quad \mathbf{b}_{3}=\left[\begin{array}{r}5 \\\\-9\end{array}\right].$$

Let \(A\) and \(B\) be invertible matrices. (a) By computing an appropriate matrix product, verify that \(\left(A^{-1} B\right)^{-1}=B^{-1} A.\) (b) Use properties of the inverse to derive \(\left(A^{-1} B\right)^{-1}=B^{-1} A.\)
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