Problem 36 - A Graphical Approach to Precalculus with Limits

Question

Solve each system by using the matrix inverse method. $$\begin{aligned} x+3 y &=-12 \\ 2 x-y &=11 \end{aligned}$$

Step-by-Step Solution

Answer

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

Step 1: Write the system in matrix form

The given system of equations can be written in matrix form as \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the column matrix of variables, and \( \mathbf{b} \) is the column matrix of constants. Thus, we have:\[ \begin{bmatrix} 1 & 3 \ 2 & -1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -12 \ 11 \end{bmatrix} \]

Step 2: Find the inverse of the coefficient matrix

To solve the system using the inverse matrix method, we need to find the inverse of matrix \( A \).To find the inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), we use the formula:\[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]For our matrix \( A = \begin{bmatrix} 1 & 3 \ 2 & -1 \end{bmatrix} \):\[ ad - bc = (1)(-1) - (3)(2) = -1 - 6 = -7 \]Thus, the inverse is:\[ A^{-1} = \frac{1}{-7} \begin{bmatrix} -1 & -3 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{7} & \frac{3}{7} \ \frac{2}{7} & -\frac{1}{7} \end{bmatrix} \]

Step 3: Multiply the inverse of the coefficient matrix by the constants matrix

Using the inverse matrix, we solve for \( \mathbf{x} \) as follows:\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{bmatrix} \frac{1}{7} & \frac{3}{7} \ \frac{2}{7} & -\frac{1}{7} \end{bmatrix} \begin{bmatrix} -12 \ 11 \end{bmatrix} \]Perform matrix multiplication:\[ \begin{bmatrix} \frac{1}{7} & \frac{3}{7} \ \frac{2}{7} & -\frac{1}{7} \end{bmatrix} \begin{bmatrix} -12 \ 11 \end{bmatrix} = \begin{bmatrix} \frac{1}{7}(-12) + \frac{3}{7}(11) \ \frac{2}{7}(-12) + -\frac{1}{7}(11) \end{bmatrix} \]Calculate each element:\[ \begin{bmatrix} \frac{-12}{7} + \frac{33}{7} \ \frac{-24}{7} - \frac{11}{7} \end{bmatrix} = \begin{bmatrix} \frac{21}{7} \ \frac{-35}{7} \end{bmatrix} = \begin{bmatrix} 3 \ -5 \end{bmatrix} \]

Step 4: Interpret the result

The resulting matrix \( \begin{bmatrix} 3 \ -5 \end{bmatrix} \) gives the values for \( x \) and \( y \). Thus, the solution to the system of equations is:\( x = 3 \) and \( y = -5 \).

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more linear equations set together, sharing the same variables. Each equation represents a line when graphed in two dimensions. The point where these lines intersect represents the solution to the system. Finding the solution involves determining the values of the variables that satisfy all equations simultaneously.
In the given example, the equations are expressed with two variables, \(x\) and \(y\):

• \(x + 3y = -12\)
• \(2x - y = 11\)

To solve this system using the matrix inverse method, we first need to express it in matrix form, a step which re-arranges and combines the coefficients into a matrix equation. This approach allows us to use matrix operations to find the solution.
By rewriting the system as \(A \mathbf{x} = \mathbf{b}\), we consolidate the system into a single matrix equation format. This sets the stage for the rest of the inverse method procedure.

Matrix Operations

Matrix operations are essential mathematical procedures used for manipulating matrices. These operations include addition, subtraction, multiplication, and finding determinants, among others. In the context of solving systems of linear equations, multiplication and finding the inverse are particularly important.
Matrix multiplication involves taking the rows of the first matrix and the columns of the second to produce a new matrix. This operation is essential when applying the inverse matrix method, as it allows us to rearrange and solve for the unknown variables.
In our specific exercise, we multiply the inverse of the coefficient matrix \(A^{-1}\) with the constants matrix \(\mathbf{b}\) to solve for the variable matrix \(\mathbf{x}\). This step involves sequential calculations:

• Calculate each position of the resulting matrix by summing the products of corresponding entries from the inverse matrix and the constants matrix.
• The matrix multiplication produces the vector, or column matrix, that represents the solutions for \(x\) and \(y\).

Understanding these operations smoothens the transition from setting up matrices to solving them effectively.

Inverse of a Matrix

The inverse of a matrix is akin to finding the reciprocal of a number. It is a computed matrix that, when multiplied by the original matrix, yields the identity matrix. The identity matrix is like the number 1 in regular multiplication, and acts as a neutral element in matrix operations.
The formula for finding the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is:
\[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]
For matrices larger than 2x2, finding the inverse can be more complex and involves advanced techniques.
In solving our system of equations, after transitioning to the matrix form, we calculate the inverse of the coefficient matrix \(A\). This is crucial because:

• It allows us to isolate the variable matrix \(\mathbf{x}\) when rearranging the matrix equation \(A \mathbf{x} = \mathbf{b}\).
• Without the inverse, solving the matrix equation directly for \(\mathbf{x}\) would not be feasible.

Therefore, mastering the calculation and application of matrix inverses is essential for applying the matrix inverse method to solve systems of linear equations efficiently.