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Use matrix inversion to solve the system of equations. $$\left\\{\begin{array}{r}3 x-6 y+2 z=-6 \\\x+2 y+3 z=-1 \\\y-z=5\end{array}\right.$$

Short Answer

Expert verified
Solving the given system of equations using matrix inversion yields \( x = 1/3, y = 1, z = 6 \)

Step by step solution

01

Express the System in Matrix Form

The system of equations can be expressed in matrix form \( AX = B \), where A is the coefficient matrix, X is the matrix of unknowns (x, y, z) and B is the constant matrix. So the system of equations can be expressed as follows: \[ A = \begin{pmatrix} 3 & -6 & 2 \ 1 & 2 & 3 \ 0 & 1 & -1\ \end{pmatrix}, X = \begin{pmatrix} x \ y \ z\ \end{pmatrix}, B = \begin{pmatrix} -6 \ -1 \ 5\ \end{pmatrix} \]
02

Calculate Inverse of the Coefficient Matrix

This involves several substeps including calculating minors, cofactors, adjoint and determinant. Use algorithms or a calculator to compute the inverse \( A^{-1} \) of coefficient matrix A. After computations, \[ A^{-1} = \begin{pmatrix} 1/3 & -1/3 & -2/3 \ 0 & 1 & -1 \ 1/3 & -1/3 & -1/3\ \end{pmatrix} \]
03

Multiply Inverse of A with Matrix B

To find the matrix of unknowns X, we multiply the inverse of A with matrix B. \( X = A^{-1}B \). After performing the multiplication, we have\[ X = \begin{pmatrix} 1/3 \ 1 \ 6 \ \end{pmatrix} \]
04

Express the Solution

This gives us a solution for the system of equations as the following set of values for the unknowns: \( x = 1/3, y = 1, z = 6 \)

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Key Concepts

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

Coefficient Matrix
A coefficient matrix is an organized way to present the coefficients of variables in a system of linear equations. It is essentially a matrix that plays a crucial role in solving linear systems using matrices. If you have a system of equations, every equation is typically organized in such a way that the coefficients of each variable are aligned correspondingly. This alignment forms what we call a coefficient matrix.

The coefficient matrix helps in transferring the system of equations into a matrix format which is more manageable for computations.
  • For example, from the given system:
    \[\begin{array}{r}3x - 6y + 2z = -6 \x + 2y + 3z = -1 \y - z = 5 \\end{array} \,\]The coefficients of the variables \( x, y, z \) form the coefficient matrix \( A \).
  • The matrix \( A \) is:\[A = \begin{pmatrix} 3 & -6 & 2 \ 1 & 2 & 3 \ 0 & 1 & -1 \end{pmatrix}\]
The coefficient matrix is vital for determining if the system has a unique solution, infinite solutions, or no solution by further calculating its determinant and, if possible, its inverse.
Inverse Matrix
An inverse matrix is a matrix that, when multiplied with the original matrix, yields the identity matrix. This concept is vital in resolving linear systems using matrix equations such as \( AX = B \).

To find the solution using matrix inversion, we need the inverse of the coefficient matrix, denoted as \( A^{-1} \). Calculating the inverse involves several steps, including the use of minors, cofactors, the adjoint, and the determinant.
  • The inverse matrix \( A^{-1} \) is used to solve the equation \( AX = B \) by transforming it to \( X = A^{-1}B \).
  • In our example, the calculated inverse is:
    \[A^{-1} = \begin{pmatrix} 1/3 & -1/3 & -2/3 \ 0 & 1 & -1 \ 1/3 & -1/3 & -1/3\end{pmatrix}\]
To verify if the matrix has an inverse, determining its determinant is essential. If the determinant is non-zero, then the inverse exists, confirming the system of equations is solvable using this approach.
System of Equations
A system of equations comprises multiple equations that you solve simultaneously. The goal is to find a common solution to all the equations—values for the variables that satisfy each equation when substituted back.

Solving a system of linear equations through matrix inversion involves transforming the system into a matrix equation form, specifically \( AX = B \), and then solving for the matrix \( X \), which contains the values of the unknowns.
  • For our given system, the process of solving starts by expressing the system in matrix form and identifying the coefficient matrix \( A \) and the constant matrix \( B \).
  • Next, we compute the inverse of \( A \) and use it to find \( X \) in \( X = A^{-1}B \).
  • Ultimately, this process yields the solution to the system:
    \[x = \frac{1}{3}, \, y = 1, \, z = 6\]
Systems of equations appear in many fields, especially in problems involving multiple constraints and objectives, making understanding matrix operations and solutions crucial.

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Most popular questions from this chapter

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. Electrical Engineering An electrical circuit consists of three resistors connected in series. The formula for the total resistance \(R\) is given by \(R=R_{1}+R_{2}+R_{3},\) where \(R_{1}, R_{2},\) and \(R_{3}\) are the resistances of the individual resistors. In a circuit with two resistors \(A\) and \(B\) connected in series, the total resistance is 60 ohms. The total resistance when \(B\) and \(C\) are connected in series is 100 ohms. The sum of the resistances of \(B\) and \(C\) is 2.5 times the resistance of \(A\). Find the resistances of \(A, B\), and \(C\).

The sum of the squares of two positive integers is \(85 .\) If the squares of the integers differ by 13 find the integers.

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?

Consider the following system of equations.$$\left\\{\begin{array}{r}x+y=3 \\\\-x+y=1 \\\2 x+y=4\end{array}\right.$$ Use Gauss-Jordan elimination to find the solution, if it exists. Interpret your answer in terms of the graphs of the given equations.

If \(A=\left[\begin{array}{rrr}1 & 0 & 1 \\ 0 & 0 & 1 \\ 2 & -1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{rrr}0 & 3 & -1 \\ -1 & 2 & 0 \\\ 0 & 0 & 1\end{array}\right],\) for what values of \(a\) and \(b\) does \(A B=\left[\begin{array}{rrr}0 & 2 a+2 b+1 & 0 \\ 3 a+4 b & 0 & 1 \\ 1 & 4 & -2\end{array}\right] ?\)

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