Problem 4 - Intermediate Algebra

Question

For Problems \(1-22\), solve each of the systems and use matrices as we did in the examples of this section. $$ \left(\begin{array}{r} 7 x-9 y=53 \\ x-3 y=11 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = 5 \) and \( y = -2 \).

Step 1: Write the System in Matrix Form

The given system of equations is: \[ 7x - 9y = 53 \] \[ x - 3y = 11 \]. We can write this system in matrix form as \( A\mathbf{x} = \mathbf{b} \) where \( A = \begin{pmatrix} 7 & -9 \ 1 & -3 \end{pmatrix} \), \( \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix} \), and \( \mathbf{b} = \begin{pmatrix} 53 \ 11 \end{pmatrix} \).

Step 2: Find the Inverse of Matrix A

To solve the system using matrices, we use the inverse of matrix \( A \). The formula for the inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is \( A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \) given \( ad-bc eq 0 \). For our matrix \( A \), we calculate the determinant: \( 7(-3) - (-9)(1) = -21 + 9 = -12 \), which is not zero. Thus, \( A \) is invertible.

Step 3: Calculate the Inverse of A

Using the inverse formula, we find \( A^{-1} = \frac{1}{-12} \begin{pmatrix} -3 & 9 \ -1 & 7 \end{pmatrix} \). Simplifying, \( A^{-1} = \begin{pmatrix} \frac{1}{4} & -\frac{3}{4} \ \frac{1}{12} & -\frac{7}{12} \end{pmatrix} \).

Step 4: Solve for x using A^{-1}

Now, we solve for \( \mathbf{x} \) by calculating \( \mathbf{x} = A^{-1} \mathbf{b} \). First, substitute the matrix values: \[ A^{-1} \mathbf{b} = \begin{pmatrix} \frac{1}{4} & -\frac{3}{4} \ \frac{1}{12} & -\frac{7}{12} \end{pmatrix} \begin{pmatrix} 53 \ 11 \end{pmatrix} \].

Step 5: Calculate the Matrix Product

Perform the matrix multiplication: \[ \begin{pmatrix} \frac{1}{4}(53) - \frac{3}{4}(11) \ \frac{1}{12}(53) - \frac{7}{12}(11) \end{pmatrix} = \begin{pmatrix} \frac{53}{4} - \frac{33}{4} \ \frac{53}{12} - \frac{77}{12} \end{pmatrix} = \begin{pmatrix} 5 \ -2 \end{pmatrix} \].

Step 6: Write the Solution

The matrix product gives us the values of \( x \) and \( y \). Therefore, the solution to the system of equations is \( x = 5 \) and \( y = -2 \).

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations with a set of variables. Our goal is to find the values of the variables that satisfy all the equations simultaneously.
For example, consider the system of equations given by:

• \(7x - 9y = 53\)
• \(x - 3y = 11\)

In these equations, \(x\) and \(y\) are the variables we need to solve for.
Systems of equations can be solved using different methods, including substitution, elimination, and using matrices. Matrices provide a way to organize the coefficients and constants in a compact form, enabling solutions through matrix operations such as finding the inverse.
By converting our system into matrix form, we set ourselves up to apply powerful matrix algebra techniques to find the solution quickly and efficiently.

Matrix Multiplication

Matrix multiplication is a fundamental operation in matrix algebra that allows us to connect matrices and compute results such as solving systems of equations. To better understand, consider two matrices, \(A\) and \(B\). The product of the matrices, \(AB\), is calculated by taking the dot product of rows from \(A\) with columns from \(B\).
In the case of solving our system of equations, once we have the inverse matrix \(A^{-1}\), we apply matrix multiplication to find \(\mathbf{x}\).
Let’s look at the multiplication in our solution:

• The inverse matrix \(A^{-1}\) was found to be \( \begin{pmatrix} \frac{1}{4} & -\frac{3}{4} \ \frac{1}{12} & -\frac{7}{12} \end{pmatrix} \).
• The matrix \(\mathbf{b}\) is \( \begin{pmatrix} 53 \ 11 \end{pmatrix} \).

Thus, we compute \(A^{-1}\mathbf{b}\) using matrix multiplication. Each element of the resulting vector is a result of multiplying and summing products from corresponding elements in \(A^{-1}\) and \(\mathbf{b}\). For the first element, we calculate:\[\frac{1}{4} \times 53 - \frac{3}{4} \times 11 = 5\]Similarly, for the second element:\[\frac{1}{12} \times 53 - \frac{7}{12} \times 11 = -2\].This multiplication reveals the values \(x = 5\) and \(y = -2\) that solve the system of equations.

Matrix Inverse

The inverse of a matrix \(A\), denoted \(A^{-1}\), is crucial in solving systems of linear equations. When multiplying the inverse matrix by the original matrix, it results in the identity matrix, similar to how the product of a number and its reciprocal equals one.
To find the inverse of a 2x2 matrix, ensure the determinant is non-zero. The formula used is:\[A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\]Here, \(ad-bc\) is the determinant of matrix \(A\), calculated in our example system as \(-12\).
Thus, for our specific matrix:

• \(A = \begin{pmatrix} 7 & -9 \ 1 & -3 \end{pmatrix}\)
• The inverse \(A^{-1}\) is \( \frac{1}{-12} \begin{pmatrix} -3 & 9 \ -1 & 7 \end{pmatrix} = \begin{pmatrix} \frac{1}{4} & -\frac{3}{4} \ \frac{1}{12} & -\frac{7}{12} \end{pmatrix} \)

By calculating \(A^{-1}\), we convert the matrix equation \(A\mathbf{x} = \mathbf{b}\) into \(\mathbf{x} = A^{-1}\mathbf{b}\), allowing us to solve for \(\mathbf{x}\). The concept of a matrix inverse is key, as it translates the problem into one we can easily compute using multiplication.