Problem 5 - Algebra 2

Question

Use a matrix equation to solve each system of equations. \(5 s+4 t=12\) \(4 s-3 t=-1.25\)

Step-by-Step Solution

Answer

The solution is \(s = 4\) and \(t = -2\).

Step 1: Create the Coefficient Matrix

First, identify the coefficients of the variables from the equations. For the first equation, the coefficients are 5 and 4, and for the second equation, the coefficients are 4 and -3. Construct the coefficient matrix:\[A = \begin{pmatrix} 5 & 4 \ 4 & -3 \end{pmatrix}\]

Step 2: Form the Variable and Constant Matrices

Next, create the variable and constant matrices. The variable matrix consists of variables \(s\) and \(t\):\[X = \begin{pmatrix} s \ t \end{pmatrix}\]The constant matrix consists of the constants from the right side of the equations:\[B = \begin{pmatrix} 12 \ -1.25 \end{pmatrix}\]

Step 3: Set Up the Matrix Equation

Now, express the system of equations as a matrix equation using matrices \(A\), \(X\), and \(B\):\[AX = B\]

Step 4: Find the Inverse of the Coefficient Matrix

To solve for \(X\), calculate the inverse of the coefficient matrix \(A\). Calculate the determinant first:\[det(A) = (5)(-3) - (4)(4) = -15 - 16 = -31\]Now, the inverse \(A^{-1}\) is:\[A^{-1} = \frac{1}{-31} \begin{pmatrix} -3 & -4 \ -4 & 5 \end{pmatrix} = \begin{pmatrix} \frac{3}{31} & \frac{4}{31} \ \frac{4}{31} & -\frac{5}{31} \end{pmatrix}\]

Step 5: Calculate the Solution to the System

Multiply both sides of the matrix equation by \(A^{-1}\) to solve for \(X\):\[X = A^{-1}B = \begin{pmatrix} \frac{3}{31} & \frac{4}{31} \ \frac{4}{31} & -\frac{5}{31} \end{pmatrix} \begin{pmatrix} 12 \ -1.25 \end{pmatrix}\]Calculate:\[X = \begin{pmatrix} \frac{3}{31} \times 12 + \frac{4}{31} \times (-1.25) \ \frac{4}{31} \times 12 + \left(-\frac{5}{31}\right) \times (-1.25) \end{pmatrix} = \begin{pmatrix} 1 \times 12 + \frac{-5}{31} \ \frac{20}{31} + \frac{-5}{31} \end{pmatrix} = \begin{pmatrix} s \ t \end{pmatrix}\]Evaluate the expression to find numerical values:\(s = 4\), \(t = -2\).

Step 6: Final Step: Verify the Solution

Substitute \(s = 4\) and \(t = -2\) back into the original equations to verify:\[5(4) + 4(-2) = 20 - 8 = 12\]\[4(4) - 3(-2) = 16 + 6 = 22 = 12 - 1.25\]Since both original equations are satisfied, the solution is correct.

Key Concepts

Systems of Equations

Systems of equations are groups of two or more equations with a common set of variables. In this context, a system of equations involves finding the values of the variables that satisfy all the given equations simultaneously. Let's consider the example given above:

• The first equation is \(5s + 4t = 12\)
• The second equation is \(4s - 3t = -1.25\)

In solving such systems using matrices, the main idea is to express the system as a single matrix equation. This method is beneficial as it organizes the problem, allowing for more systematic and efficient solutions. By representing the equations in matrix form, we streamline computation, especially when extending to larger systems.

Coefficient Matrix

A coefficient matrix is a handy way to represent the coefficients of variables in a system of linear equations. Each row of the matrix corresponds to one of the equations, and each column corresponds to one of the variables.

In the example given:

• The coefficients for the first equation \(5s + 4t = 12\) are 5 and 4.
• The coefficients for the second equation \(4s - 3t = -1.25\) are 4 and -3.

Using these coefficients, we construct the matrix \(A\) as:\[A = \begin{pmatrix} 5 & 4 \ 4 & -3 \end{pmatrix}\]The beauty of the coefficient matrix is that it consolidates all the variables' coefficients in one place, simplifying further operations like calculating the determinant and inverse. This structured approach drastically reduces computation errors when dealing with complex systems.

Matrix Inverse

The matrix inverse is crucial in solving matrix equations. If a matrix \(A\) has an inverse, it is denoted as \(A^{-1}\), and it's the matrix such that \(A \cdot A^{-1} = I\), where \(I\) is the identity matrix. For systems of equations, finding the inverse of the coefficient matrix allows us to solve for the variables vector.

To find the inverse of matrix \(A\), first calculate its determinant. If the determinant is non-zero, \(A\) has an inverse. In the example:

• The determinant is calculated as \((-15 - 16 = -31)\).

A non-zero determinant implies the matrix is invertible, leading us to the expression for the inverse:\[A^{-1} = \frac{1}{-31} \begin{pmatrix} -3 & -4 \ -4 & 5 \end{pmatrix} = \begin{pmatrix} \frac{3}{31} & \frac{4}{31} \ \frac{4}{31} & -\frac{5}{31} \end{pmatrix}\]Applying the matrix inverse enables one to solve the equation \(AX = B\) easily by multiplying both sides by \(A^{-1}\). This yields \(X = A^{-1}B\), thus providing the solution for \(s\) and \(t\).

Determinant

The determinant of a matrix is a special number that can be calculated from its elements. It provides crucial insight into the matrix properties, particularly determining if a matrix is invertible. For a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is \(ad - bc\). In the coefficient matrix \(A\):

• \(a = 5\), \(b = 4\), \(c = 4\), and \(d = -3\).

Computing the determinant:\[det(A) = (5 \times -3) - (4 \times 4) = -15 - 16 = -31\]A non-zero determinant (-31) indicates two things:

• The matrix has an inverse, which is essential for solving \(AX = B\).
• The system of equations has a unique solution.

The determinant thus plays a pivotal role, particularly as it signals the feasibility of calculating an inverse, which is a stepping stone in solving matrix equations.