Problem 23 - Elementary and Intermediate Algebra

Question

Solve using Cramer's rule. $$ \begin{aligned} &r-2 s+3 t=6\\\ &\begin{aligned} 2 r-s-t &=-3 \\ r+s+t &=6 \end{aligned} \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( r = 0.8 \), \( s = 2.8 \), and \( t = 3 \).

Step 1: Write the system in matrix form

First, represent the system of equations in the form of a matrix equation, which is given by \[ AX = B \]. The coefficient matrix \( A \), the variable vector \( X \) and the constant vector \( B \) are: \[ A = \begin{pmatrix} 1 & -2 & 3 \ 2 & -1 & -1 \ 1 & 1 & 1 \end{pmatrix} \], \[ X = \begin{pmatrix} r \ s \ t \end{pmatrix} \], \[ B = \begin{pmatrix} 6 \ -3 \ 6 \end{pmatrix} \].

Step 2: Find the determinant of matrix A

Calculate the determinant of the matrix \( A \): \[ \text{det}(A) = \begin{vmatrix} 1 & -2 & 3 \ 2 & -1 & -1 \ 1 & 1 & 1 \end{vmatrix} \]. Use the rule for 3x3 determinants to solve: \[ \text{det}(A) = 1(-1 \times 1 - (-1 \times 1)) - (-2)(2 \times 1 - (-1 \times 1)) + 3(2 \times 1 - (-1 \times 1)) \]. Calculate each term to find \[ \text{det}(A) = 1(0) + 2(3) + 3(3) = 15. \].

Step 3: Find the determinant of matrix A_r

Replace the first column of the matrix \( A \) with the vector \( B \) to form matrix \( A_r \): \[ A_r = \begin{pmatrix} 6 & -2 & 3 \ -3 & -1 & -1 \ 6 & 1 & 1 \end{pmatrix} \]. Calculate the determinant: \[ \text{det}(A_r) = 6(-1 \times 1 - (-1 \times 1)) - (-2)(-3 \times 1 - (-1 \times 6)) + 3(-3 \times 1 - (-1 \times 6)) \]. Simplify to find \[ \text{det}(A_r) = 6 \times 0 - (-2)(-3 + 6) + 3(-3 + 6) = 12 \].

Step 4: Find the determinant of matrix A_s

Replace the second column of the matrix \( A \) with the vector \( B \) to form matrix \( A_s \): \[ A_s = \begin{pmatrix} 1 & 6 & 3 \ 2 & -3 & -1 \ 1 & 6 & 1 \end{pmatrix} \]. Calculate the determinant: \[ \text{det}(A_s) = 1(-3 \times 1 - 6 \times -1) - 6(2 \times 1 - 1 \times 1) + 3(2 \times 6 - (-3) \times 1) \]. Simplify to find \[ \text{det}(A_s) = 1(3) - 6(1) + 3(15) = 42 \].

Step 5: Find the determinant of matrix A_t

Replace the third column of the matrix \( A \) with the vector \( B \) to form matrix \( A_t \): \[ A_t = \begin{pmatrix} 1 & -2 & 6 \ 2 & -1 & -3 \ 1 & 1 & 6 \end{pmatrix} \]. Calculate the determinant: \[ \text{det}(A_t) = 1(-1 \times 6 - 1 \times -3) - (-2)(2 \times 6 - 1 \times 1) + 6(2 \times 1 - (-1) \times 1) \]. Simplify to find \[ \text{det}(A_t) = 1(-3) - (-2)(11) + 6(3) = 45 \].

Step 6: Calculate the variables

Use Cramer's rule to find the variables: \[ r = \frac{\text{det}(A_r)}{\text{det}(A)} = \frac{12}{15} = 0.8 \], \[ s = \frac{\text{det}(A_s)}{\text{det}(A)} = \frac{42}{15} = 2.8 \], and \[ t = \frac{\text{det}(A_t)}{\text{det}(A)} = \frac{45}{15} = 3 \].

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from a square matrix. It is a key concept in linear algebra, especially when solving systems of equations, finding inverses, and understanding matrix properties.
To find the determinant of a 3x3 matrix \(\begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix}\), follow the formula:
\[\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\]
For the coefficient matrix \(\begin{pmatrix} 1 & -2 & 3 \ 2 & -1 & -1 \ 1 & 1 & 1 \end{pmatrix}\), we compute:
\[\text{det}(A) = 1(-1 \times 1 - (-1 \times 1)) - (-2)(2 \times 1 - (-1 \times 1)) + 3(2 \times 1 - (-1 \times 1)) = 15\]
A determinant helps us in applications like checking matrix invertibility and solving systems of equations using Cramer's Rule.

Linear Algebra

Linear algebra is a branch of mathematics that studies vectors, vector spaces, linear transformations, and systems of linear equations.
It involves understanding matrices, determinants, and vector operations. Key topics include:

• Vectors and vector spaces
• Matrices and their properties
• Systems of linear equations
• Eigenvalues and eigenvectors

Linear algebra is used in various fields like computer science, engineering, physics, and economics.
Matrices form the backbone of linear algebra, representing systems of equations compactly.
By working with matrices and their determinants, we simplify solving complex problems, showing the power and utility of linear algebra.

System of Equations

A system of equations is a collection of two or more equations with a common set of variables.
They can be solved using various methods like substitution, elimination, and matrix operations.
When represented in matrix form \(\begin{pmatrix} A & B & C \ D & E & F \ G & H & I \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} J \ K \ L \end{pmatrix}\), we focus on solving for \(\begin{pmatrix} x \ y \ z \end{pmatrix}\).
In the provided example, using Cramer's Rule, we substitute columns of the coefficient matrix with the constants to form new matrices and calculate their determinants:

• Matrix \(A_r = \begin{pmatrix} 6 & -2 & 3 \ -3 & -1 & -1 \ 6 & 1 & 1 \end{pmatrix} \)
• Matrix \(A_s = \begin{pmatrix} 1 & 6 & 3 \ 2 & -3 & -1 \ 1 & 6 & 1 \end{pmatrix} \)
• Matrix \(A_t = \begin{pmatrix} 1 & -2 & 6 \ 2 & -1 & -3 \ 1 & 1 & 6 \end{pmatrix} \)

With determinants computed, variables are found by dividing determinant values:
\[ r = \frac{\text{det}(A_r)}{\text{det}(A)},\underline{\phantom{xxx}} s = \frac{\text{det}(A_s)}{\text{det}(A)},\underline{\phantom{xxx}} t = \frac{\text{det}(A_t)}{\text{det}(A)} \](For the example: \ r = \frac{12}{15} = 0.8,\underline{\phantom{xxx}} s = \frac{42}{15} = 2.8,\underline{\phantom{xxx}} t = 3).
Thus, we solve the system effectively, showcasing a powerful method in linear algebra.