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Solve using Cramer's rule. $$\begin{aligned}&3 x+2 y-z=4\\\&3 x-2 y+z=5\\\&4 x-5 y-z=-1\end{aligned}$$

Short Answer

Expert verified
x \approx 0.2241, y \approx 0.6724, z \approx 1.7069

Step by step solution

01

- Write the System of Equations in Matrix Form

First, represent the system of equations in matrix form \(AX = B\). The coefficient matrix \(A\), variable matrix \(X\), and constant matrix \(B\) are:\[A = \begin{pmatrix} 3 & 2 & -1 \ 3 & -2 & 1 \ 4 & -5 & -1 \end{pmatrix}\], \[X = \begin{pmatrix} x \ y \ z \end{pmatrix}\], and \[B = \begin{pmatrix} 4 \ 5 \ -1 \end{pmatrix}\].
02

- Calculate the Determinant of Matrix A

Find the determinant of matrix \(A\):\[\text{det}(A) = \begin{vmatrix} 3 & 2 & -1 \ 3 & -2 & 1 \ 4 & -5 & -1 \end{vmatrix}\]. Using the rule of Sarrus or cofactor expansion, we compute:\[\text{det}(A) = 3(-2(-1) - 1(-5)) - 2(3(-1) - 4(1)) + (-1)(3(-5) - (-2)(4))\]. Simplifying each term gives:\[\text{det}(A) = 3(2 + 5) - 2(-3 - 4) + (-1)(-15 - 8)\].\[\text{det}(A) = 3(7) - 2(-7) - (-23)\].\[\text{det}(A) = 21 + 14 + 23 = 58\].
03

- Calculate the Determinant of Matrix Aâ‚“

Replace the first column of matrix \(A\) with matrix \(B\), forming matrix \(Aâ‚“\) and find its determinant \( \text{det}(Aâ‚“) \):\[Aâ‚“ = \begin{pmatrix} 4 & 2 & -1 \ 5 & -2 & 1 \ -1 & -5 & -1 \end{pmatrix}\].\[\text{det}(Aâ‚“) = \begin{vmatrix} 4 & 2 & -1 \ 5 & -2 & 1 \ -1 & -5 & -1 \end{vmatrix}\]. Simplifying, we get:\[\text{det}(Aâ‚“) = 4(-2(-1) - 1(-5)) - 2(5(-1) - (-1)(1)) + (-1)(5(-5) - (-2)(-1))\].\[\text{det}(Aâ‚“) = 4(2 + 5) - 2(-5 - 1) - (25 + 2)\].\[\text{det}(Aâ‚“) = 4(7) - 2(-6) - 27\].\[\text{det}(Aâ‚“) = 28 + 12 - 27 = 13\].
04

- Calculate the Determinant of Matrix Aáµ§

Replace the second column of matrix \(A\) with matrix \(B\), forming matrix \(Aáµ§\) and find its determinant \( \text{det}(Aáµ§) \):\[Aáµ§ = \begin{pmatrix} 3 & 4 & -1 \ 3 & 5 & 1 \ 4 & -1 & -1 \end{pmatrix}\].\[\text{det}(Aáµ§) = \begin{vmatrix} 3 & 4 & -1 \ 3 & 5 & 1 \ 4 & -1 & -1 \end{vmatrix}\]. Simplifying, we get:\[\text{det}(Aáµ§) = 3(5(-1) - 1(-1)) - 4(3(-1) - 4(1)) + (-1)(3(-1) - 5(4))\].\[\text{det}(Aáµ§) = 3(-5 + 1) - 4(-3 - 4) - (-3 - 20)\].\[\text{det}(Aáµ§) = 3(-4) - 4(-7) + 23\].\[\text{det}(Aáµ§) = -12 + 28 + 23 = 39\].
05

- Calculate the Determinant of Matrix A_z

Replace the third column of matrix \(A\) with matrix \(B\), forming matrix \(A_z\) and find its determinant \( \text{det}(A_z) \):\[A_z = \begin{pmatrix} 3 & 2 & 4 \ 3 & -2 & 5 \ 4 & -5 & -1 \end{pmatrix}\].\[\text{det}(A_z) = \begin{vmatrix} 3 & 2 & 4 \ 3 & -2 & 5 \ 4 & -5 & -1 \end{vmatrix}\]. Simplifying, we get:\[\text{det}(A_z) = 3(-2(-1) - 5(-5)) - 2(3(-1) - 4(5)) + 4(3(-5) - (-2)(4))\].\[\text{det}(A_z) = 3(2 + 25) - 2(-3 - 20) + 4(-15 + 8)\].\[\text{det}(A_z) = 3(27) - 2(-23) - 4(7)\].\[\text{det}(A_z) = 81 + 46 - 28 = 99\].
06

- Apply Cramer's Rule

Using Cramer's rule, the solutions for \(x\), \(y\), and \(z\) are:\[ x = \frac{\text{det}(Aâ‚“)}{\text{det}(A)} = \frac{13}{58} \], \[ y = \frac{\text{det}(Aáµ§)}{\text{det}(A)} = \frac{39}{58} \], \[ z = \frac{\text{det}(A_z)}{\text{det}(A)} = \frac{99}{58} \]. Simplifying these, we get: \[ x \approx 0.2241 \], \[ y \approx 0.6724 \], \[ z \approx 1.7069 \].

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Key Concepts

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

System of Linear Equations
To solve the given problem using Cramer's Rule, we first need to understand what a system of linear equations is. A system of linear equations is a set of equations that you deal with all at once where each equation is linear. Linear equations are those that graph as straight lines and can be written in the form: \[Ax + By + Cz = D\]In our example, we had three equations: \[3x + 2y - z = 4\], \[3x - 2y + z = 5\],\[4x - 5y - z = -1\].Each equation involves three unknowns: \(x\), \(y\), and \(z\). The goal is to find the values of these unknowns that satisfy all the given equations simultaneously.

Using various methods like substitution, elimination, or matrix operations such as Cramer's Rule, we can solve these systems. Each method has its own strengths and is chosen based on the problem's complexity and the form of equations.
Determinants
A determinant is a special number that can be calculated from a square matrix. It is crucial in many areas of linear algebra, including solving systems of linear equations using Cramer's Rule.

Given a 3x3 matrix:

\[A = \begin{pmatrix} 3 & 2 & -1 \ 3 & -2 & 1 \ 4 & -5 & -1 \end{pmatrix}\]


We can find its determinant using cofactor expansion:

\(\text{det}(A) = 3(-2(-1) - 1(-5)) - 2(3(-1) - 4(1)) + (-1)(3(-5) - (-2)(4))\)

When simplifying, we calculate:
\[\text{det}(A) = 3(7) - 2(-7) - (-23) = 21 + 14 + 23 = 58\]


Determinants have different properties and can indicate if a matrix is invertible (non-zero determinant).

In Cramer's rule, determinants are used to find the unique solution to a system of linear equations by finding values of different modified matrices.
Matrix Algebra
Matrix Algebra is a branch of mathematics that deals with matrices and the operations on them. The system of linear equations can be represented in matrix form, simplifying many operations.

In our example, we represented our system of linear equations as:
Formula matrix \(A\):
\[A = \begin{pmatrix} 3 & 2 & -1 \ 3 & -2 & 1 \ 4 & -5 & -1 \end{pmatrix}\]
Variable matrix \(X\):
\[X = \begin{pmatrix} x \ y \ z \end{pmatrix}\]
Constant matrix \(B\):
\[B = \begin{pmatrix} 4 \ 5 \ -1 \end{pmatrix}\]

In Cramer's rule, we use the determinant values of these matrices. First, we calculate the determinant of the coefficient matrix \(A\). Then, we replace each column in \(A\) with matrix \(B\) one-by-one to create matrices \(A_x\), \(A_y\), and \(A_z\). We find the determinants of these new matrices as well.

Using the formula:
\[x = \frac{\text{det}(A_x)}{\text{det}(A)}\]
\[y = \frac{\text{det}(A_y)}{\text{det}(A)}\]
\[z = \frac{\text{det}(A_z)}{\text{det}(A)}\]
We solve for the unknowns. This formula emphasizes the importance of matrix operations in algebra and how they can simplify complex linear problems.

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