/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 The inverse of the matrix $$A=... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 36

The inverse of the matrix $$A=\left(\begin{array}{rrr}1 & -1 & 1 \\\2 & -3 & 2 \\\\-4 & 6 & 1\end{array}\right) \quad \text { is } \quad A^{-1}=\left(\begin{array}{rrr}3 & -\frac{7}{5} & -\frac{1}{5} \\\2 & -1 & 0 \\\0 & \frac{2}{5} & \frac{1}{5}\end{array}\right)$$ Use this fact to solve the following system. $$\left\\{\begin{aligned}x-y+z &=5 \\\2 x-3 y+2 z &=-15 \\\\-4 x+6 y+z &=25\end{aligned}\right.$$

Short Answer

Expert verified
The solution is \( x = 31 \), \( y = 25 \), \( z = -1 \).

Step by step solution

01

Understand the Problem

We need to solve the system of linear equations by using the inverse of the matrix method. The given system is: \( \begin{aligned} x-y+z &=5 \ 2x-3y+2z &=-15 \ -4x+6y+z &=25 \end{aligned} \)
02

Write the system of equations in matrix form

The system of equations can be expressed in matrix form as \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the column matrix of variables, and \( B \) is the column matrix of constants. \( A = \begin{pmatrix} 1 & -1 & 1 \ 2 & -3 & 2 \ -4 & 6 & 1 \end{pmatrix} \) and \( B = \begin{pmatrix} 5 \ -15 \ 25 \end{pmatrix} \).
03

Multiply by the inverse matrix

To solve for \( X \), multiply both sides of the equation \( AX = B \) by \( A^{-1} \), the inverse of \( A \). Thus, \( A^{-1}AX = A^{-1}B \), which simplifies to \( X = A^{-1}B \).
04

Calculate \( X \) using given \( A^{-1} \)

Substitute the given inverse matrix \( A^{-1} = \begin{pmatrix} 3 & -\frac{7}{5} & -\frac{1}{5} \ 2 & -1 & 0 \ 0 & \frac{2}{5} & \frac{1}{5} \end{pmatrix} \) and \( B = \begin{pmatrix} 5 \ -15 \ 25 \end{pmatrix} \) into the equation to find \( X \).
05

Compute the matrix product \( A^{-1}B \)

Perform the matrix multiplication: \[\begin{pmatrix} 3 & -\frac{7}{5} & -\frac{1}{5} \ 2 & -1 & 0 \ 0 & \frac{2}{5} & \frac{1}{5} \end{pmatrix} \begin{pmatrix} 5 \ -15 \ 25 \end{pmatrix} = \begin{pmatrix} x \ y \ z \end{pmatrix}\]. Compute each entry of the resulting matrix: 1. First row: \((3)(5) + \left(-\frac{7}{5}\right)(-15) + \left(-\frac{1}{5}\right)(25)\) 2. Second row: \((2)(5) + (-1)(-15) + (0)(25)\) 3. Third row: \((0)(5) + \left(\frac{2}{5}\right)(-15) + \left(\frac{1}{5}\right)(25)\).
06

Simplify the computations

Calculate each component:- First row: \(15 + 21 - 5 = 31\)- Second row: \(10 + 15 + 0 = 25\)- Third row: \(0 - 6 + 5 = -1\).Thus, \( X = \begin{pmatrix} 31 \ 25 \ -1 \end{pmatrix} \).
07

Interpret the result

The solution \( X = \begin{pmatrix} 31 \ 25 \ -1 \end{pmatrix} \) implies \( x = 31 \), \( y = 25 \), \( z = -1 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Matrix Multiplication
Matrix multiplication is more than just multiplying numbers. It involves a sequence of operations to combine matrices. When multiplying two matrices, their inner dimensions must match. If you have a matrix A of size (m x n) and a matrix B of size (n x p), you can multiply them to get a new matrix C of size (m x p). Each element in the resulting matrix C is calculated as the dot product of the corresponding row of A and the column of B.
  • Matrix Elements: The elements of matrix C, denoted as C[i][j], are calculated by summing the products of the elements of the ith row of A and the jth column of B.
  • Example Calculation: For C[i][j], the formula is: \[ C[i][j] = A[i][1] \times B[1][j] + A[i][2] \times B[2][j] + ... + A[i][n] \times B[n][j] \]
This method is essential when solving matrix equations like \( AX = B \). By multiplying A by its inverse, you can isolate and solve for the matrix X.
System of Linear Equations
Systems of linear equations involve multiple equations with multiple variables but appear linear when graphed. In our current exercise, we have three equations with three unknowns, \( x \, y\, \text{and}\, z \). To solve these systems efficiently, they can be represented in a matrix form such as \( AX = B \).
  • Matrix Form Representation: Each equation corresponds to a row in matrix A, a row in matrix B, and a variable in column matrix X.
  • Purpose: This representation allows the application of matrix operations to solve the equations systematically.
Using the inverse matrix method, you reverse the transformations applied to variables by their coefficients (matrix A) to find the solution \( X \). This approach gives a structured and efficient way to arrive at the solution using linear algebra.
Matrix Inverse Calculation
Calculating the inverse of a matrix is essential in solving systems of equations using the inverse matrix method. An inverse of a matrix A is another matrix, denoted as \( A^{-1} \), such that \( A \times A^{-1} = I \), where I is the identity matrix. Only square matrices (same number of rows and columns) can have inverses, and not all square matrices are invertible.
  • Finding the Inverse: To calculate the inverse manually, methods like the Gauss-Jordan elimination or the adjoint method can be used. These involve systematic row reduction or determinant and cofactors calculation.
  • Role in Problem Solving: Once you have the inverse matrix, it can be multiplied by matrix B to find the solution matrix X in the matrix equation \( AX = B \). This is because multiplying both sides of the equation by \( A^{-1} \) rearranges it to \( X = A^{-1}B \).
In the example exercise, we used the given inverse \( A^{-1} \) to solve the system, demonstrating the power of the inverse matrix method in simplifying and solving complex systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The matrices \(A, B, C, D, E, F,\) and \(G\) are defined as follows: $$ A=\left(\begin{array}{rr} 2 & 3 \\ -1 & 4 \end{array}\right) \quad B=\left(\begin{array}{rr} 1 & -1 \\ 3 & 0 \end{array}\right) \quad C=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) $$ $$\begin{aligned} &D=\left(\begin{array}{rrr} -1 & 2 & 3 \\ 4 & 0 & 5 \end{array}\right) \quad E=\left(\begin{array}{rr} 2 & 1 \\ 8 & -1 \\ 6 & 5 \end{array}\right)\\\ &F=\left(\begin{array}{rr} 5 & -1 \\ -4 & 0 \\ 2 & 3 \end{array}\right) \quad G=\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array}\right) \end{aligned}$$ In each exercise, carry out the indicated matrix operations if they are defined. If an operation is not defined, say so. $$A^{2} A$$

Use Cramers rule to solve those systems for which \(D \neq 0 .\) In cases where \(D=0,\) use Gaussian elimination or matrix methods. $$\left\\{\begin{array}{l} 3 x+4 y+2 z=1 \\ 4 x+6 y+2 z=7 \\ 2 x+3 y+z=11 \end{array}\right.$$

Let \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)\) and \(B=\left(\begin{array}{ll}5 & 6 \\ 7 & 8\end{array}\right) .\) Let \(A^{2}\) and \(B^{2}\) denote the matrix products \(A A\) and \(B B\), respectively. Compute each of the following. (a) \((A+B)(A+B)\) (b) \(A^{2}+2 A B+B^{2}\) (c) \(A^{2}+A B+B A+B^{2}\)

Graph the systems of linear inequalities. In each case specify the vertices. Is the region convex? Is the region bounded? $$\left\\{\begin{array}{l} x \geq 0 \\ y \geq 0 \\ 3 x-y+1 \geq 0 \\ 0 \leq x-y+3 \\ y \leq 5 \\ x \leq \frac{1}{3}(17-y) \\ x \leq \frac{1}{2}(y+8) \end{array}\right.$$

So the input (3,5) yields an output of \(\sqrt{2} .\) We define the do- main for this function just as we did in Chapter 3: The domain is the set of all inputs that yield real-number outputs. For instance, the ordered pair (1,4) is not in the domain of the function we have been discussing, because (as you should check for yourself\() f(1,4)=\sqrt{-1},\) which is not a real number. We can determine the domain of the function in equation ( 1 ) by requiring that the quantity under the radical sign be non negative. Thus we require that \(2 x-y+1 \geq 0\) and, consequently, \(y \leq 2 x+1\) (Check this.) The following figure shows the graph of this inequality; the domain of our function is the set of ordered pairs making up the graph. In Exercises follow a similar procedure and sketch the domain of the given function. (Graph cant copy) $$h(x, y)=\sqrt{x}+\sqrt{y}$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.