Problem 4 - Algebra 2

Question

Use a matrix equation to solve each system of equations. \(5 x-3 y=-30\) \(8 x+5 y=1\)

Step-by-Step Solution

Answer

The solution is \(x = -3\) and \(y = 5\).

Step 1: Write the System in Matrix Form

First, represent the given system of equations as a matrix equation. The system \(5x - 3y = -30\) and \(8x + 5y = 1\) can be rewritten in matrix form as: \[ \begin{bmatrix} 5 & -3 \ 8 & 5 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -30 \ 1 \end{bmatrix} \]Here, the coefficient matrix is \(A = \begin{bmatrix} 5 & -3 \ 8 & 5 \end{bmatrix}\), the variable vector is \(\mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}\), and the constant vector is \(\mathbf{b} = \begin{bmatrix} -30 \ 1 \end{bmatrix}\).

Step 2: Find the Inverse of the Coefficient Matrix

Calculate the inverse of the coefficient matrix \(A\), denoted as \(A^{-1}\), if it exists. The inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by: \[ \begin{bmatrix} a & b \ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]For our matrix \(A = \begin{bmatrix} 5 & -3 \ 8 & 5 \end{bmatrix}\), calculate the determinant \(ad-bc = 5 \times 5 - (-3) \times 8 = 25 + 24 = 49\). Thus, the inverse is: \[ A^{-1} = \frac{1}{49} \begin{bmatrix} 5 & 3 \ -8 & 5 \end{bmatrix} \]

Step 3: Multiply Inverse by Constant Vector

Multiply the inverse of the coefficient matrix \(A^{-1}\) by the constant vector \(\mathbf{b}\) to find the variable vector \(\mathbf{x}\). Thus, calculate: \[ \mathbf{x} = A^{-1} \mathbf{b} = \frac{1}{49} \begin{bmatrix} 5 & 3 \ -8 & 5 \end{bmatrix} \begin{bmatrix} -30 \ 1 \end{bmatrix} \]Perform the matrix multiplication: \[ \begin{bmatrix} 5 & 3 \ -8 & 5 \end{bmatrix} \begin{bmatrix} -30 \ 1 \end{bmatrix} = \begin{bmatrix} 5(-30) + 3(1) \ -8(-30) + 5(1) \end{bmatrix} = \begin{bmatrix} -150 + 3 \ 240 + 5 \end{bmatrix} = \begin{bmatrix} -147 \ 245 \end{bmatrix} \]Finally, calculate: \[ \mathbf{x} = \frac{1}{49} \begin{bmatrix} -147 \ 245 \end{bmatrix} = \begin{bmatrix} -3 \ 5 \end{bmatrix} \]This means \(x = -3\) and \(y = 5\).

Step 4: Verify the Solution

To ensure the solution is correct, substitute \(x = -3\) and \(y = 5\) back into the original equations.- For the first equation: \(5(-3) - 3(5) = -15 - 15 = -30\), which matches.- For the second equation: \(8(-3) + 5(5) = -24 + 25 = 1\), which also matches.Both equations are satisfied, confirming that the solution \(x = -3\) and \(y = 5\) is correct.

Key Concepts

System of Equations

A system of equations is simply a set of two or more equations that are solved together. Each equation informs the value of the unknowns being studied. In the given exercise, we are dealing with two equations in two variables, expressed as:

• \(5x - 3y = -30\)
• \(8x + 5y = 1\)

Our task is to find values for \(x\) and \(y\) that satisfy both equations at the same time. Systems of equations can be solved using methods such as substitution, elimination, or, as in this case, using matrix equations.

Inverse of a Matrix

The inverse of a matrix is similar to the inverse of a number. It effectively "undoes" or cancels the effect of the matrix when multiplied. Not all matrices have inverses, but when they do, the matrix inverse is a key tool in solving systems of equations.For a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the inverse \(A^{-1}\) is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]The term \(ad-bc\) is the determinant, and it must be non-zero for the inverse to exist. In this exercise, the coefficient matrix \( \begin{bmatrix} 5 & -3 \ 8 & 5 \end{bmatrix} \) has a determinant of 49, which allows us to find the inverse used in solving the system of equations.

Determinant

The determinant is a special number that can be calculated from a square matrix. It offers vital insights into the matrix's properties, including whether it is invertible. For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is calculated as \(ad-bc\).Let's discuss why it's important:

• A non-zero determinant indicates that the matrix is invertible.
• If the determinant is zero, the system may not have a unique solution.

In the exercise, our matrix \(\begin{bmatrix} 5 & -3 \ 8 & 5 \end{bmatrix}\) has a determinant of 49, confirming it is suitable for finding its inverse and thus helping to solve the given system of equations.

Matrix Multiplication

Matrix multiplication is a method of combining two matrices to produce a new matrix. It's a bit different from regular multiplication. In matrix multiplication, we multiply rows by columns.Suppose we have matrices \(A\) and \(B\), the product \(AB\) is calculated by taking each element in a row of \(A\) and multiplying it by each element in a column of \(B\), then summing these products.Here's how it works in the context of our exercise:

• We find the product \(A^{-1}b\), where \(A^{-1}\) is the inverse matrix and \(b\) is the constant vector.
• The multiplication results in a new matrix, or solution vector, which gives the values of the variables.
• In this case, multiplying \(A^{-1}\) and \(b\) ultimately provided us with the solution \(x = -3\) and \(y = 5\).