Chapter 12: Problem 30
For Problems 27-40, use the method of matrix inverses to solve each system. $$ \left(\begin{array}{rl} 6 x-y & =-14 \\ 3 x+2 y & =-17 \end{array}\right) $$
Short Answer
Expert verified
The solution is \( x = -3 \) and \( y = -4 \).
Step by step solution
01
Write the System in Matrix Form
The given system of equations can be expressed in matrix form as \( AX = B \). Here, \( A \) is the coefficient matrix, \( X \) is the column matrix of variables, and \( B \) is the constant matrix. Thus, the system can be written as: \[A = \begin{pmatrix} 6 & -1 \ 3 & 2 \end{pmatrix}, \, X = \begin{pmatrix} x \ y \end{pmatrix}, \, B = \begin{pmatrix} -14 \ -17 \end{pmatrix}\]
02
Find the Inverse of A
To solve for \( X \), we need to find the inverse of \( A \), denoted \( A^{-1} \). The formula for the inverse of a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is \( \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \). For matrix \( A \), the determinant \( (ad-bc) = (6 \times 2) - (3 \times -1) = 15 \) and its inverse is: \[A^{-1} = \frac{1}{15} \begin{pmatrix} 2 & 1 \ -3 & 6 \end{pmatrix}\]
03
Multiply A-inverse by B to Solve for X
Now compute \( X = A^{-1}B \): \[X = \frac{1}{15} \begin{pmatrix} 2 & 1 \ -3 & 6 \end{pmatrix} \begin{pmatrix} -14 \ -17 \end{pmatrix}\] Perform the matrix multiplication: \[X = \frac{1}{15} \begin{pmatrix} (2 \times -14) + (1 \times -17) \ (-3 \times -14) + (6 \times -17) \end{pmatrix} = \frac{1}{15} \begin{pmatrix} -28 - 17 \ 42 - 102 \end{pmatrix} = \frac{1}{15} \begin{pmatrix} -45 \ -60 \end{pmatrix}\] Finally, divide each element by 15: \[X = \begin{pmatrix} -3 \ -4 \end{pmatrix}\]
04
Verify the Solution
Substitute \( x = -3 \) and \( y = -4 \) back into the original equations to verify: 1. \( 6(-3) - (-4) = -18 + 4 = -14 \), which is correct. 2. \( 3(-3) + 2(-4) = -9 - 8 = -17 \), which is also correct. Both equations are satisfied, confirming the solution is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
System of Equations
A system of equations consists of two or more equations containing common variables. In the context of a 2x2 system, this generally involves two linear equations with two unknowns. These systems can arise in numerous real-life contexts, such as predicting financial outcomes or solving engineering problems.
Here’s how a system of equations can be set up:
Here’s how a system of equations can be set up:
- Identify the variables common to both equations - usually represented by letters like \(x\) and \(y\).
- Construct equations that relate these variables - often forming a straight line when graphed.
2x2 Matrix
A 2x2 matrix is a simple grid that consists of four numbers arranged into two rows and two columns. Each position in the matrix is significant and corresponds to a specific operation in mathematical expressions. For example, in a system of equations, a 2x2 matrix can represent the coefficients of the variables involved.
Consider the following matrix from the solution: \[A = \begin{pmatrix} 6 & -1 \ 3 & 2 \end{pmatrix}\]This matrix corresponds to the coefficients in the given system of equations.
The key properties of a 2x2 matrix include:
Consider the following matrix from the solution: \[A = \begin{pmatrix} 6 & -1 \ 3 & 2 \end{pmatrix}\]This matrix corresponds to the coefficients in the given system of equations.
The key properties of a 2x2 matrix include:
- Its determinant, which is calculated as \((ad-bc)\) for a matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \).
Matrix Multiplication
Matrix multiplication involves combining two matrices to produce a new matrix. This operation is essential when transitioning from the matrix equation \(AX = B\) to solve for \(X\), especially using matrix inverses. The process follows strict rules:
- To multiply a matrix \(A\) of size \(m \times n\) with another matrix \(B\) of size \(n \times p\), the number of columns in \(A\) must match the number of rows in \(B\).
- The resulting matrix will have dimensions \(m \times p\).
- Each element in the resulting matrix is calculated by taking the dot product of rows of the first matrix with columns of the second.