Problem 25 - Algebra and Trigonometry

Question

Solve the system of equations by converting to a matrix equation and using the inverse of the coefficient matrix, as in Example 6. Use the inverses from Exercises 7–10, 15, 16, 19, and 21. \(\left\\{\begin{aligned} 2 x+5 y &=2 \\\\-5 x-13 y &=20 \end{aligned}\right.\)

Step-by-Step Solution

Answer

The solution is \(x = -126\) and \(y = 50\).

Step 1: Write the System as a Matrix Equation

The given system of equations is: \(2x + 5y = 2\) and \(-5x - 13y = 20\). We can express this system as a matrix equation of the form \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. Here, \(A = \begin{bmatrix} 2 & 5 \ -5 & -13 \end{bmatrix}\), \(X = \begin{bmatrix} x \ y \end{bmatrix}\), and \(B = \begin{bmatrix} 2 \ 20 \end{bmatrix}\). So, the matrix equation is \(\begin{bmatrix} 2 & 5 \ -5 & -13 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 2 \ 20 \end{bmatrix}\).

Step 2: Find the Inverse of Matrix A

From Exercise 7-10, we know the inverse of matrix \(A\). The inverse \(A^{-1}\) is used to solve the system by converting the matrix equation into \(X = A^{-1}B\). When the inverse is already given, we can proceed to apply it. However, we need to calculate here: To find \(A^{-1}\), we use the formula for 2x2 matrices, which is:\[ A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]For \(A = \begin{bmatrix} 2 & 5 \ -5 & -13 \end{bmatrix}\), \(a = 2\), \(b = 5\), \(c = -5\), \(d = -13\). The determinant \(ad - bc = (2)(-13) - (5)(-5) = -26 + 25 = -1\).So, \(A^{-1} = \begin{bmatrix} -13 & -5 \ 5 & 2 \end{bmatrix}\).

Step 3: Multiply A Inverse by B

Now, we compute \(X = A^{-1}B\): \[ \begin{bmatrix} -13 & -5 \ 5 & 2 \end{bmatrix} \begin{bmatrix} 2 \ 20 \end{bmatrix} = \begin{bmatrix} (-13)(2) + (-5)(20) \ (5)(2) + (2)(20) \end{bmatrix} \] This calculation yields:\[ \begin{bmatrix} -26 - 100 \ 10 + 40 \end{bmatrix} = \begin{bmatrix} -126 \ 50 \end{bmatrix}\].

Step 4: Interpret the Result as Values for x and y

The resulting matrix \(\begin{bmatrix} -126 \ 50 \end{bmatrix}\) corresponds to the values of \(x\) and \(y\). Thus, \(x = -126\) and \(y = 50\).

Key Concepts

Understanding a System of Equations

A system of equations involves multiple equations that share common variables. Solving such systems entails finding values for the variables that satisfy all the equations simultaneously. For instance, in the system of equations:

• \( 2x + 5y = 2 \)
• \( -5x - 13y = 20 \)

we aim to discover the specific values of \(x\) and \(y\) that make both equations true.

There are various ways to solve them, including substitution, elimination, and matrix methods. Using matrices can be particularly efficient, especially as the number of equations increases.

In this approach, we transform the equations into a matrix, allowing us to use advanced techniques like calculating the inverse, which simplifies everything into compact operations.

Inverse of a Matrix: The Key to Solving Matrix Equations

The inverse of a matrix plays a vital role in solving matrix equations. For a matrix \(A\) to be invertible, it must be square (same number of rows and columns) and have a non-zero determinant. Here, the determinant is the magic number that tells us whether a matrix has an inverse or not.

To find the inverse of a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), we use the formula:

\[ A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]

This formula involves swapping the diagonal elements, changing the signs of off-diagonal elements, and dividing by the determinant \(ad - bc\). Using the inverse, we can solve the equation \(AX = B\) by transforming it into \(X = A^{-1}B\). This manipulation efficiently provides the variable values we're looking for.

A critical note is that if the determinant is zero, the matrix is non-invertible, meaning no unique solution exists for the system.

Matrix Multiplication: Combining Matrices to Solve Equations

Matrix multiplication enables us to combine and solve matrix equations. This technique requires strict adherence to size compatibility, meaning the number of columns in the first matrix must match the number of rows in the second. For two matrices \(A\) (of size \(m \times n\)) and \(B\) (of size \(n \times p\)), their product \(AB\) will result in a new matrix of size \(m \times p\).

Let's consider the matrix product of inverse \(A^{-1}\) and constant matrix \(B\):

• Multiply the first row of \(A^{-1}\) with the first column of \(B\).
• Repeat for all rows and columns as needed.

This systematic multiplication results in a single matrix containing solutions, providing us with values for the variables. In our example, multiplying the inverse matrix and matrix \(B\) calculates:\[ \begin{bmatrix} -13 & -5 \ 5 & 2 \end{bmatrix} \begin{bmatrix} 2 \ 20 \end{bmatrix} = \begin{bmatrix} (-13)(2) + (-5)(20) \ (5)(2) + (2)(20) \end{bmatrix} = \begin{bmatrix} -126 \ 50 \end{bmatrix} \]

This illustrates how, through multiplication, each component adds up to ultimately solve the original system of equations.