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Solving Systems of Equations Using Matrices. $$\left\\{\begin{aligned} x+y-z &=0 \\ 3 x+2 y-z &=-1 \\\\-2 x+y-2 z &=-1 \end{aligned}\right.$$

Short Answer

Expert verified
The solution of the system of equations is \(x = -0.4\), \(y = 0.6\), \(z = -0.2\).

Step by step solution

01

Create Augmented Matrix

First, rewrite the system of equations as an augmented matrix: \[ \begin{bmatrix} 1 & 1 & -1 & 0\ 3 & 2 & -1 & -1\ -2 & 1 & -2 & -1 \end{bmatrix} \]. The coefficients of x, y, z form the three columns on the left of the matrix, and the constants in the system form the column on the right.
02

Row Operations

Next, perform row operations to transform the matrix into row-echelon form. Subtract three times the first row from the second, and add two times the first row to the third. Get: \[ \begin{bmatrix} 1 & 1 & -1 & 0\ 0 & -1 & 2 & -1\ 0 & 3 & -4 & -1 \end{bmatrix} \]. Subtracting three times the second row from the third row gives: \[ \begin{bmatrix} 1 & 1 & -1 & 0\ 0 & -1 & 2 & -1\ 0 & 0 & -10 & 2 \end{bmatrix} \].
03

Express Variables

Finally, express the variables from the obtained triangular matrix: the third equation is -10z = 2, which gives z = -0.2. Substitute z back into the second equation for y: -y + 2*(-0.2) = -1, leading to y = 0.6. Substitute y and z back into the first equation for x: x + 0.6 - (-0.2) = 0, which gives x = -0.4.

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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 linear algebra, especially when dealing with systems of equations. It represents both the coefficients and the constants of a system of equations in a compact, matrix format. Think of it like a snapshot of the entire system all in one place. For example, the equations \[\begin{equation}\left\begin{aligned} x+y-z &=0 \ 3x+2y-z &=-1 \ -2x+y-2z &=-1\end{aligned}\right.\end{equation}\]are neatly compiled into an augmented matrix by listing the coefficients and the equations' constants in columns, leading to:\[\begin{equation}\begin{bmatrix} 1 & 1 & -1 &|& 0\ 3 & 2 & -1 &|& -1\ -2 & 1 & -2 &|& -1\end{bmatrix}\end{equation}\]This visual representation not only streamlines the information but also sets the stage for the use of row operations to solve the system.
Row Operations
In the context of matrices, row operations are maneuvers that manipulate the rows of a matrix to simplify its structure. There are three basic row operations:
  • Swapping two rows,
  • Multiplying a row by a non-zero constant,
  • Adding or subtracting a multiple of one row from another row.

These operations are pivotal when solving systems of equations because they help us to isolate the variables. In our example, to convert the augmented matrix into a simpler form, we can perform the following operations: Subtract three times the first row from the second to eliminate the x coefficient in row 2, and add two times the first row to the third to eliminate the x coefficient in row 3. This results in:\[\begin{equation}\begin{bmatrix} 1 & 1 & -1 &|& 0\ 0 & -1 & 2 &|& -1\ 0 & 3 & -4 &|& -1\end{bmatrix}\end{equation}\]Then subtract three times the now simplified second row from the third to target the y coefficient. Such systematic manipulation, using row operations, drives us towards clarity from complexity.
Row-Echelon Form
Row-echelon form is kind of like a ladder within a matrix that helps us climb to the solution of a system of equations. It is a specific arrangement where each row has a leading entry (that is, the first non-zero number from the left in a row) to the right of the leading entry of the row above it. If the bottom rows are all zeros, it's even better because they can effectively be ignored. The ultimate aim is to have a triangular shape of non-zero entries, which makes back-substitution straightforward.Our matrix, after row operations, looks like this:\[\begin{equation}\begin{bmatrix} 1 & 1 & -1 &|& 0\ 0 & -1 & 2 &|& -1\ 0 & 0 & -10 &|& 2\end{bmatrix}\end{equation}\]This final form showcases the triangular ladder we're looking for—the row-echelon form. From here, we can decode the values of z, y, and x successively, which are hiding within the simplified matrix. This triumphant move from a system of equations to an easier-to-understand row-echelon form is the linchpin of solving these sorts of algebraic puzzles.

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Most popular questions from this chapter

A supply function for widgets is modeled by \(P(q)=a q+b,\) where \(q\) is the number of widgets supplied and \(P(q)\) is the total price of \(q\) widgets, in dollars. It is known that 200 widgets can be supplied for \(\$ 40\) and 100 widgets can be supplied for \(\$ 25 .\) Use a system of linear equations to find the constants \(a\) and \(b\) in the expression for the supply function.

Find the decoding matrix for each encoding matrix. $$\left[\begin{array}{rrrr}1 & 1 & 4 & 1 \\\2 & -3 & 4 & 1 \\\3 & -4 & 6 & 2 \\\\-1 & 0 & -2 & -1\end{array}\right]$$

Consider the following system of equations. $$\left\\{\begin{aligned} x^{2}+y^{2} &=r^{2} \\ (x-h)^{2}+y^{2} &=r^{2} \end{aligned}\right.$$ Let \(r\) be a (fixed) positive number. For what value(s) of \(h\) does this system have (a) exactly one real solution? (b) exactly two real solutions? (c) infinitely many real solutions? (d) no real solution? (Hint: Visualize the graphs of the two equations.)

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. A grocery store carries two brands of diapers. For a certain week, the number of boxes of Brand \(A\) diapers sold was 4 more than the number of boxes of Brand B diapers sold. Brand A diapers cost \(\$ 10\) per box and Brand B diapers cost \(\$ 12\) per box. If the total revenue generated that week from the sale of diapers was \(\$ 172,\) how many of each brand did the store sell?

Answer the question pertaining to the matrices. $$A=\left[\begin{array}{ll}a & b \\\c & d \\\e & f\end{array}\right] \text { and } B=\left[\begin{array}{lll}g & h & i \\\j & k & l\end{array}\right]$$ Let \(Q=B A,\) and find \(q_{11}\) and \(q_{22}\) without performing the entire multiplication of matrix \(B\) by matrix \(A\).

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