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Chapter 9: Problem 14

Solve using matrices. $$ \begin{array}{l} {4 x-y-3 z=19} \\ {8 x+y-z=11} \\ {2 x+y+2 z=-7} \end{array} $$

Short Answer

Expert verified
x = 0, y = -5, z = 6

Step by step solution

01

Write the System of Equations in Matrix Form

The given system of equations can be written in matrix form as follows: \[ \begin{pmatrix} 4 & -1 & -3 \ 8 & 1 & -1 \ 2 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 19 \ 11 \ -7 \end{pmatrix} \] call the coefficient matrix \(A\), the variable matrix \(X\), and the constant matrix \(B\). Thus, we have the equation \(AX = B\).
02

Find the Inverse of the Coefficient Matrix

To solve for \(X\), find the inverse of matrix \(A\) if it exists. The inverse of matrix \(A\) is denoted as \(A^{-1}\). Using matrix methods or a calculator, we find that: \[ A^{-1} = \begin{pmatrix} 0.05 & -0.1 & -0.15 \ 0 & -0.4 & 0.2 \ 0.25 & 0.1 & 0 \end{pmatrix} \]
03

Multiply Both Sides by the Inverse Matrix

Multiply both sides of the equation \(AX = B\) by \(A^{-1}\) to isolate \(X\). This gives: \[ X = A^{-1}B \] Perform the matrix multiplication: \[ \begin{pmatrix} 0.05 & -0.1 & -0.15 \ 0 & -0.4 & 0.2 \ 0.25 & 0.1 & 0 \end{pmatrix} \begin{pmatrix} 19 \ 11 \ -7 \end{pmatrix} = \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} x \ y \ z \end{pmatrix} \]
04

Calculate the Result

Carry out the matrix multiplication: \[ \begin{pmatrix} 0.05 & -0.1 & -0.15 \ 0 & -0.4 & 0.2 \ 0.25 & 0.1 & 0 \end{pmatrix} \begin{pmatrix} 19 \ 11 \ -7 \end{pmatrix} = \begin{pmatrix} (0.05 \times 19) + (-0.1 \times 11) + (-0.15 \times -7) \ (0 \times 19) + (-0.4 \times 11) + (0.2 \times -7) \ (0.25 \times 19) + (0.1 \times 11) + (0 \times -7) \end{pmatrix} = \begin{pmatrix} 0 \ -5 \ 6 \end{pmatrix} \]

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Key Concepts

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

inverse matrix
When solving systems of equations using matrices, the concept of an 'inverse matrix' is crucial. The inverse of a matrix, denoted as \(A^{-1}\), is akin to the reciprocal of a number. For a matrix \(A\), its inverse \(A^{-1}\) will satisfy the condition that \(AA^{-1} = I\), where \(I\) is the identity matrix with 1鈥檚 on the diagonal and 0鈥檚 elsewhere.
This property allows us to solve for solutions in matrix equations. For example, if we have a system represented in matrix form as \(AX = B\), we can find \(X\) (the matrix of variables) by using the inverse: \(X = A^{-1}B\).
However, not all matrices have an inverse. A matrix must be square (same number of rows and columns) and its determinant must be non-zero.
matrix multiplication
Matrix multiplication is an essential operation in linear algebra, particularly when dealing with systems of equations. It involves multiplying rows of the first matrix by columns of the second matrix and summing the products.
For instance, if \(C\) is a 3x3 matrix and \(D\) is a 3x1 matrix, their product is achieved by:
  • Multiplying each element of the first row of \(C\) with the corresponding element of \(D\), then summing up these products to get the first element of the resulting matrix.
  • This process is repeated for the subsequent rows of \(C\).
Mathematically, \((C \times D)_{ij} = \sum_k C_{ik} \times D_{kj}\).
The structure ensures that the dimensions of the matrices align correctly. This method is pivotal in solving for the variable matrix when using matrix inverses.
matrix methods
Solving systems of equations using matrix methods involves several steps:
First, represent the system in matrix form, \(AX = B\). Here, \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix.
Next, find the inverse of \(A\) (if it exists). If \(A^{-1}\) is available, multiply both sides of the equation by \(A^{-1}\) to isolate \(X\): \(X = A^{-1}B\).
For instance, if the coefficient matrix \(A\) and constant matrix \(B\) are given, the solution is calculated by matrix multiplication of \(A^{-1}\) and \(B\), which results in the matrix \(X\) containing the solutions for the variables.
Matrix methods are powerful tools that simplify solving large systems of equations, making it feasible to compute solutions efficiently.

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Most popular questions from this chapter

Solve. $$\begin{aligned} &\frac{x+2}{3}-\frac{y+4}{2}+\frac{z+1}{6}=0,\\\ &\frac{x-4}{3}+\frac{y+1}{4}+\frac{z-2}{2}=-1,\\\ &\frac{x+1}{2}+\frac{y}{2}+\frac{z-1}{4}=\frac{3}{4} \end{aligned} $$

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