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Multiple matrix equations with the same matrix can be solved simultaneously. (a) Solve both systems by performing elimination on just one augmented matrix. $$\left(\begin{array}{rrr}2 & -1 & -1 \\\\-1 & 1 & 1 \\ 1 & -1 & 0\end{array}\right)\left(\begin{array}{l}x \\\y \\\z \end{array}\right)=\left(\begin{array}{l}0 \\\1 \\ 0\end{array}\right),\left(\begin{array}{rrr}2 & -1 & -1 \\\\-1 & 1 & 1 \\\1 & -1 & 0\end{array}\right)\left(\begin{array}{l}a \\\b \\\c\end{array}\right)=\left(\begin{array}{l}2 \\\1 \\\1\end{array}\right)$$ (b) Give an interpretation of the columns of \(M^{-1}\) in \((M \mid I) \sim\left(I \mid M^{-1}\right)\) in terms of solutions to certain systems of linear equations.

Short Answer

Expert verified
The solutions are \(x = 0, y = 0, z = 0\) and \(a = 2, b = 2, c = 4\).

Step by step solution

01

Write the Augmented Matrix

First, create a single augmented matrix from the given systems of equations. Combine both right-hand sides of the equations into an extended matrix.\[\left(\begin{array}{ccc|cc}2 & -1 & -1 & 0 & 2 \-1 & 1 & 1 & 1 & 1 \1 & -1 & 0 & 0 & 1 \\end{array}\right)\]
02

Row Reduce to Row Echelon Form

Use row operations to simplify the matrix to row echelon form. Start with the first row and eliminate elements below the leading coefficient in the first column.1. Add the first row to the third row:\[\left(\begin{array}{ccc|cc}2 & -1 & -1 & 0 & 2 \-1 & 1 & 1 & 1 & 1 \3 & -2 & -1 & 0 & 3 \\end{array}\right)\]2. To make the pivot in the first column equal to 1, multiply the first row by \(\frac{1}{2}\):\[\left(\begin{array}{ccc|cc}1 & -\frac{1}{2} & -\frac{1}{2} & 0 & 1 \-1 & 1 & 1 & 1 & 1 \3 & -2 & -1 & 0 & 3 \\end{array}\right)\]
03

Eliminate First Column Below Pivot

3. Add the first row to the second row and subtract 3 times the first row from the third row to eliminate the elements below the pivot:\[\left(\begin{array}{ccc|cc}1 & -\frac{1}{2} & -\frac{1}{2} & 0 & 1 \0 & \frac{1}{2} & \frac{1}{2} & 1 & 2 \0 & -\frac{1}{2} & \frac{1}{2} & 0 & 0 \\end{array}\right)\]
04

Simplify Second and Third Rows

4. Multiply the second and third rows by 2 to simplify:\[\left(\begin{array}{ccc|cc}1 & -\frac{1}{2} & -\frac{1}{2} & 0 & 1 \0 & 1 & 1 & 2 & 4 \0 & -1 & 1 & 0 & 0 \\end{array}\right)\]
05

Eliminate the Second Column in Third Row

5. Add the second row to the third row to eliminate the -1 in the second column of the third row:\[\left(\begin{array}{ccc|cc}1 & -\frac{1}{2} & -\frac{1}{2} & 0 & 1 \0 & 1 & 1 & 2 & 4 \0 & 0 & 2 & 2 & 4 \\end{array}\right)\]
06

Simplify the Third Row

6. Divide the third row by 2:\[\left(\begin{array}{ccc|cc}1 & -\frac{1}{2} & -\frac{1}{2} & 0 & 1 \0 & 1 & 1 & 2 & 4 \0 & 0 & 1 & 1 & 2 \\end{array}\right)\]
07

Back Substitution

Utilize back substitution to solve for each variable:\(z = -1\) from the first column, \(b + c = 2\), and \(b + z = 4\) solving reveal:The solutions are:\(x = 0, y = 0, z = 0, a = 2, b = 2, c = 4\).

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Key Concepts

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

Augmented Matrix
In the context of solving systems of linear equations, an augmented matrix is an extremely helpful tool. It combines the coefficients and constants of the system of equations into a single matrix form. This matrix makes the process of solving equations more systematic and allows for the use of matrix operations like row reduction.

For example, if you have the system of equations:
\[\begin{align*}2x - y - z &= 0, \-x + y + z &= 1, \x - y &= 0\end{align*}\]
it would be represented by the augmented matrix:
\[\begin{bmatrix}2 & -1 & -1 & | & 0 \-1 & 1 & 1 & | & 1 \1 & -1 & 0 & | & 0\end{bmatrix}\]
The vertical bar separates the coefficient matrix from the constants, aligning them together for easier handling during row reduction steps.
Row Reduction
Row reduction, also known as Gaussian elimination, is a series of operations performed on the augmented matrix to simplify it into a row-echelon form, and subsequently to a reduced row-echelon form if needed. Here are the essential steps involved in row reduction:

  • Swap rows – switches places between two rows.
  • Multiply a row by a nonzero scalar – changes the scale of one whole row.
  • Add or subtract a multiple of one row to or from another row – effectively combines equations to eliminate variables.

These operations are used to progressively eliminate coefficients, turning the matrix into a simpler form where back substitution can be easily carried out to find solutions for the variables. The desired goal is to transform the coefficient part of the augmented matrix into an upper triangular form, where you can isolate variables starting from the bottom row up to the top.
Inverse Matrix
An inverse matrix is a square matrix that, when multiplied with the original matrix, results in an identity matrix. Essentially, it undoes or reverses the effects of the original matrix. In terms of solving systems of linear equations, if you have a matrix equation in the form \(A\mathbf{x} = \mathbf{b}\), you can solve for \(\mathbf{x}\) using the inverse matrix of \(A\) assuming that \(A^{-1}\) exists (i.e., if \(A\) is non-singular).

Mathematically, it’s written as:
\[A^{-1}A\mathbf{x} = A^{-1}\mathbf{b} \]
which simplifies to:
\[\mathbf{x} = A^{-1}\mathbf{b}\]
This usage of the inverse matrix is crucial when interpreting solutions found through row reduction, especially when considering the columns of \(M^{-1}\) as solutions to certain systems of linear equations. Each column of \(M^{-1}\) provides the solutions of the system when the right-hand side vector is a corresponding standard unit vector (e.g., \(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\), etc.).
Systems of Linear Equations
Systems of linear equations are collections of equations that share the same set of variables. Solving a system of linear equations means determining the values of these variables that satisfy all equations simultaneously.

In matrix form, a system of equations can often be represented as \(A\mathbf{x} = \mathbf{b}\), where \(A\) is a matrix composed of coefficients, \(\mathbf{x}\) is a vector of the variables, and \(\mathbf{b}\) is the vector of constants.

There are several methods to solve these systems including substitution, elimination, and matrix-based approaches utilizing augmented matrices and row reduction. These techniques are interconnected, with the choice of method often depending on the specific problem at hand.
For instance, working with augmented matrices and row reduction is particularly useful for dealing with multiple equations and variables at once, streamlining the solution process and providing flexibility in handling more complex systems.

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Most popular questions from this chapter

State whether the following augmented matrices are in RREF and compute their solution sets. $$\begin{array}{l}\left(\begin{array}{lllll|l}1 & 0 & 0 & 0 & 3 & 1 \\ 0 & 1 & 0 & 0 & 1 & 2 \\\0 & 0 & 1 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 & 2 & 0\end{array}\right) \\ \left(\begin{array}{llllll|l}1 & 1 & 0 & 1 & 0 & 1 & 0 \\\0 & 0 & 1 & 2 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 & 3 & 0 \\\0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right),\end{array}$$ \(\left(\begin{array}{lllllll|r}1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 & 0 & 2 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & -2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\end{array}\right) .\)

Write down examples of augmented matrices corresponding to each of the five types of solution sets for systems of equations with three unknowns.

Is \(L U\) factorization of a matrix unique? Justify your answer. If you randomly create a matrix by picking numbers out of the blue, it will probably be difficult to perform elimination or factorization; fractions and large numbers will probably be involved. To invent simple problems it is better to start with a simple answer: (a) Start with any augmented matrix in RREF. Perform EROs to make most of the components non-zero. Write the result on a separate piece of paper and give it to your friend. Ask that friend to find RREF of the augmented matrix you gave them. Make sure they get the same augmented matrix you started with. (b) Create an upper triangular matrix \(U\) and a lower triangular matrix \(L\) with only 1s on the diagonal. Give the result to a friend to factor into \(L U\) form. (c) Do the same with an \(L D U\) factorization.

Another method for solving linear systems is to use row operations to bring the augmented matrix to Row Echelon Form (REF as opposed to RREF). In REF, the pivots are not necessarily set to one, and we only require that all entries left of the pivots are zero, not necessarily entries above a pivot. Provide a counterexample to show that row echelon form is not unique. Once a system is in row echelon form, it can be solved by "back substitution." Write the following row echelon matrix as a system of equations, then solve the system using back-substitution. $$\left(\begin{array}{lll|l}2 & 3 & 1 & 6 \\\0 & 1 & 1 & 2 \\\0 & 0 & 3 & 3\end{array}\right)$$

Solve the following linear system: $$\begin{array}{l}2 x_{1}+5 x_{2}-8 x_{3}+2 x_{4}+2 x_{5}=0 \\ 6 x_{1}+2 x_{2}-10 x_{3}+6 x_{4}+8 x_{5}=6 \\ 3 x_{1}+6 x_{2}+2 x_{3}+3 x_{4}+5 x_{5}=6 \\ 3 x_{1}+1 x_{2}-5 x_{3}+3 x_{4}+4 x_{5}=3 \\ 6 x_{1}+7 x_{2}-3 x_{3}+6 x_{4}+9 x_{5}=9\end{array}$$ Be sure to set your work out carefully with equivalence signs \(\sim\) between each step, labeled by the row operations you performed.

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