Problem 64 - Precalculus

Question

apply matrix algebra to solve the system of linear equations. $$2 x-3 y=12$$ $$x+y=1$$

Step-by-Step Solution

Answer

The solution is \( x = 3 \) and \( y = -2 \).

Step 1: Write the System in Matrix Form

Convert the system of equations into matrix form. The given system of equations is:\[ \begin{align*} 2x - 3y &= 12 \ x + y &= 1 \end{align*} \] Rewriting in matrix form, we get: \[ \begin{bmatrix} 2 & -3 \ 1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 12 \ 1 \end{bmatrix} \] where \( A = \begin{bmatrix} 2 & -3 \ 1 & 1 \end{bmatrix} \), \( \vec{x} = \begin{bmatrix} x \ y \end{bmatrix} \), and \( \vec{b} = \begin{bmatrix} 12 \ 1 \end{bmatrix} \).

Step 2: Calculate the Inverse of Matrix A

Find the inverse of the coefficient matrix \( A \). For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is calculated as: \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \] For \( A = \begin{bmatrix} 2 & -3 \ 1 & 1 \end{bmatrix} \), the determinant \( \det(A) = (2)(1) - (-3)(1) = 5 \). Thus, \( A^{-1} = \frac{1}{5} \begin{bmatrix} 1 & 3 \ -1 & 2 \end{bmatrix} \).

Step 3: Compute the Solution Using the Inverse

Multiply the inverse of matrix \( A \) by the vector \( \vec{b} \) to find \( \vec{x} \). We have: \[ \vec{x} = A^{-1} \vec{b} \] Substitute \( A^{-1} \) and \( \vec{b} \): \[ \vec{x} = \frac{1}{5} \begin{bmatrix} 1 & 3 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 12 \ 1 \end{bmatrix} \] Perform the matrix multiplication: \[ \vec{x} = \frac{1}{5} \begin{bmatrix} (1)(12) + (3)(1) \ (-1)(12) + (2)(1) \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 15 \ -10 \end{bmatrix} = \begin{bmatrix} 3 \ -2 \end{bmatrix} \]

Step 4: Interpret the Result

The vector \( \vec{x} = \begin{bmatrix} 3 \ -2 \end{bmatrix} \) gives the solution to the system of equations. This means \( x = 3 \) and \( y = -2 \).

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations that have the same set of variables. These equations must be linear, meaning they graph as straight lines. The solution to these systems is a set of values for the variables that satisfy all the equations simultaneously. For example, in the system given in the exercise:

• Equation 1: \(2x - 3y = 12\)
• Equation 2: \(x + y = 1\)

We need to find values for \(x\) and \(y\) that make both equations true. By solving these equations together, you find a point where both lines intersect on a graph, which represents the solution. This process allows for practical applications such as determining how different constraints can be balanced in real-world problems.

Inverse Matrix

In matrix algebra, an inverse matrix can be thought of as a matrix that, when multiplied with the original matrix, results in an identity matrix. This is somewhat similar to how multiplying a number by its reciprocal results in 1. An inverse matrix only exists for square matrices with a non-zero determinant. For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the inverse can be found if the determinant is non-zero.To calculate the inverse:

• Find the determinant (\(ad - bc\)). If it is zero, the matrix does not have an inverse.
• If the determinant is non-zero, use it to calculate the inverse matrix.

In the original problem, the matrix \(A = \begin{bmatrix} 2 & -3 \ 1 & 1 \end{bmatrix}\) has a determinant of 5 and thus, has an inverse given by:\[A^{-1} = \frac{1}{5} \begin{bmatrix} 1 & 3 \ -1 & 2 \end{bmatrix}\].

Matrix Form

Matrix form is a way of expressing a system of linear equations using matrices to simplify calculations. In matrix form, the system \(A\vec{x} = \vec{b}\) represents:

• \(A\) as the coefficient matrix, which is built from the coefficients of the variables in the equations.
• \(\vec{x}\) as the variable vector, representing the variables involved in the equations.
• \(\vec{b}\) as the constant matrix or vector, comprising the constants from the equations.

For our exercise, the matrix form is:\[ \begin{bmatrix} 2 & -3 \ 1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 12 \ 1 \end{bmatrix} \]This expression in matrix form allows us to use matrix operations to solve the system.

Determinant

The determinant is a scalar value that can be computed from the elements of a square matrix. It provides useful properties, such as determining if a matrix has an inverse. For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is calculated as \(ad - bc\). If the determinant is zero, the matrix does not have an inverse, which implies that the system of equations it represents may not have a unique solution.In the exercise's example, the coefficient matrix \(\begin{bmatrix} 2 & -3 \ 1 & 1 \end{bmatrix}\) has a determinant of 5:\[(2)(1) - (-3)(1) = 5\]Since the determinant is not zero, the matrix is invertible, which confirms that the system of equations has a unique solution. The determinant is a crucial concept in understanding the solvability of systems represented in matrix form.