Problem 26 - Intermediate Algebra

Question

Use matrices to solve each system of equations. $$ \left\\{\begin{array}{l} x+y=3 \\ x-y=-1 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \) and \( y = 2 \).

Step 1: Write the System in Matrix Form

We start by expressing the system of equations in matrix form. We do this by identifying the coefficients of the variables and the constants on the right-hand side. The system \( x + y = 3 \) and \( x - y = -1 \) can be written as a matrix equation: \[\begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 3 \ -1 \end{bmatrix}\] where the matrix \( A = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \) contains the coefficients, the vector \( \begin{bmatrix} x \ y \end{bmatrix} \) represents the variables, and \( \begin{bmatrix} 3 \ -1 \end{bmatrix} \) is the constants vector.

Step 2: Find the Inverse of the Coefficient Matrix

Next, we need to find the inverse of the coefficient matrix \( A \). The formula for the inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For our matrix \( A = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \), we have \( a=1, b=1, c=1, d=-1 \). Calculate the determinant \( ad-bc = 1(-1) - 1(1) = -1 - 1 = -2 \). Thus, the inverse is:\[A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & -1 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}\]

Step 3: Solve for the Variable Vector

Now that we have the inverse of the coefficient matrix, we can solve for the vector of variables \( x, y \). Multiply the inverse of \( A \) with the constants vector:\[\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} 3 \ -1 \end{bmatrix}\]Carrying out this multiplication, we get:\[\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \cdot 3 + \frac{1}{2} \cdot (-1) \ \frac{1}{2} \cdot 3 - \frac{1}{2} \cdot (-1) \end{bmatrix} = \begin{bmatrix} 1 \ 2 \end{bmatrix}\]Thus, \( x = 1 \) and \( y = 2 \).

Step 4: Verify the Solution

To confirm our solution, substitute \( x = 1 \) and \( y = 2 \) back into the original equations:- For the first equation, \( x + y = 1 + 2 = 3 \), which matches.- For the second equation, \( x - y = 1 - 2 = -1 \), which also matches.Since both equations are satisfied, our solution \( x = 1 \) and \( y = 2 \) is correct.

Key Concepts

Understanding Matrix Algebra

Matrix algebra is a powerful tool that allows us to work with systems of equations in a concise and efficient manner. In our given problem, we use matrices to represent the system of equations as a matrix equation. This involves organizing and aligning our coefficients, variables, and constants into matrix form. Such a transformation helps in visualizing multidimensional data as linear transformations, aiding in calculations.

Once expressed in matrix form, the system

• The coefficient matrix \( A \) holds the coefficients from the system's equations.
• The variable matrix is a column vector \( \begin{bmatrix} x \ y \end{bmatrix} \) containing the variables to solve for.
• The constants matrix \( \begin{bmatrix} 3 \ -1 \end{bmatrix} \) contains the solutions to each equation in the system.

Understanding how to construct matrices and operate on them is key in using matrix algebra to solve systems of equations, which can be applied to complex problems in engineering, computer science, physics, and more.

Finding the Inverse Matrix

The inverse of a matrix is somewhat like the reciprocal of a number. It reverses the transformation done by the original matrix. Finding the inverse is crucial for solving matrix equations because it allows us to isolate our variables.

For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse matrix \( A^{-1} \) is calculated using the formula:\[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]The term \((ad-bc)\) is known as the determinant, and it must not be zero for the inverse to exist. In our solution, the inverse of \( A \) was used to solve for \( x \) and \( y \). The inverse matrix effectively 'undoes' the coefficients' matrix impact, allowing direct computation of the solution vector.

Calculating and applying the inverse is a fundamental aspect of linear algebra, making it an essential skill when dealing with systems of linear equations.

Determinant Calculation

A determinant is a unique number that can be computed from a square matrix. It's a scalar value providing essential properties about the matrix, such as whether the matrix has an inverse. For a 2x2 matrix, the determinant is calculated as \( ad-bc \).

In determining whether a matrix can be inverted, the determinant is crucial:

• If the determinant is zero, the matrix does not have an inverse and is classified as 'singular'.
• If non-zero, the matrix is 'non-singular', meaning it has an inverse.

In our example, the determinant \( ad-bc \) was \(-2\), signaling the matrix is invertible. This computation not only allows inversion but also contributes fundamentally to assessing matrices' behaviors.

Determinants play a core role across linear algebra including geometry transformations, where they scale the area or volume transformed by the matrix, giving insights into the matrix's effect on data.