Problem 35 - Algebra for College Students

Question

For Problems 27-40, use the method of matrix inverses to solve each system. $$ \left(\begin{array}{l} y=19-3 x \\ 9 x-5 y=1 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = -16 \) and \( y = -29 \).

Step 1: Write the System of Equations in Matrix Form

The given system of equations can be written in matrix form as follows: \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the matrix of coefficients, \( \mathbf{x} \) is the vector of variables, and \( \mathbf{b} \) is the constant vector.\[A = \begin{pmatrix} -3 & 1 \ 9 & -5 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 19 \ 1 \end{pmatrix}\]

Step 2: Find the Inverse of the Coefficient Matrix

To find \( \mathbf{x} \), we need to calculate the inverse of matrix \( A \). The inverse \( A^{-1} \) of a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\]Calculate the determinant first: \( \, \text{det}(A) = (-3)(-5) - (1)(9) = 15 - 9 = 6 \, \).Thus:\[A^{-1} = \frac{1}{6} \begin{pmatrix} -5 & -1 \ -9 & -3 \end{pmatrix}\]

Step 3: Multiply the Inverse with the Constant Matrix

Now multiply the inverse of matrix \( A \) by the constant vector \( \mathbf{b} \) to find \( \mathbf{x} \):\[\mathbf{x} = A^{-1} \mathbf{b} = \frac{1}{6} \begin{pmatrix} -5 & -1 \ -9 & -3 \end{pmatrix} \begin{pmatrix} 19 \ 1 \end{pmatrix}\]This gives:\[\begin{pmatrix} x \ y \end{pmatrix} = \frac{1}{6} \begin{pmatrix} (-5)(19) + (-1)(1) \ (-9)(19) + (-3)(1) \end{pmatrix} = \frac{1}{6} \begin{pmatrix} -95 - 1 \ -171 - 3 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} -96 \ -174 \end{pmatrix}\]

Step 4: Simplify to Find the Solution

Simplify the result from the matrix multiplication:\[\mathbf{x} = \begin{pmatrix} \frac{-96}{6} \ \frac{-174}{6} \end{pmatrix} = \begin{pmatrix} -16 \ -29 \end{pmatrix}\]Thus, the solution to the system is \( x = -16 \) and \( y = -29 \).

Key Concepts

Understanding Systems of Equations

A system of equations consists of two or more equations with shared variables. Solving a system means finding values for the variables that satisfy all the equations simultaneously.
This is fundamental in many fields like physics, engineering, and economics.
To solve such systems, we can use various methods, one of which employs matrix inverses.

• Consider two equations, each relating the variables in the system: in our example, these are equations involving \(x\) and \(y\).
• Each equation represents a line, and solving the system is equivalent to finding their intersection point.
• Using matrices can simplify calculations, especially for systems with more variables and equations.

In matrix terms, an equation like \(ax + by = c\) pairs with a second equation to be expressed as a matrix multiplication. Using matrix algebra, we express our system as \(A \mathbf{x} = \mathbf{b}\). Here, \(A\) is a matrix containing the coefficients of the variables, \(\mathbf{x}\) is the vector holding the variables, and \(\mathbf{b}\) is the output vector with constants from the right side of the equations. Understanding how to manipulate these setups forms the core of solving systems with matrices.

Basics of Matrix Algebra

Matrix algebra is the area of mathematics working with matrices, arrays of numbers arranged in rows and columns. This approach can solve multiple equations involving many variables.
Matrix multiplication is a critical operation here, enabling us to compactly represent and solve systems of equations.

• Mathematical operations such as addition, subtraction, and multiplication can be applied to matrices, similar to numbers.
• Each element of a resulting matrix from a multiplication reflects a sum of products of rows and columns from the two matrices involved.
• When solving equations using matrices, finding the inverse of a matrix, if it exists, is crucial.

In our system, matrix \(A\) contains the coefficients \(\begin{pmatrix} -3 & 1 \ 9 & -5 \end{pmatrix}\). Matrix algebra's laws guide how we compute the solution by isolating \(\mathbf{x}\) using the inverse of \(A\) and multiplying it with \(\mathbf{b}\): \(\mathbf{x} = A^{-1} \mathbf{b}\). This forms the groundwork to efficiently address more significant and complex systems you might encounter.

The Role of Determinants

A determinant is a unique number associated with a matrix, especially a square one. It plays a vital role in matrix algebra, particularly when determining if a matrix has an inverse.

• For a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is calculated as \(ad - bc\).
• If a matrix's determinant is zero, it doesn't have an inverse, and you cannot solve the associated system using matrix inverses.
• In our example, solving the equations required computing the determinant of \(A\) to find its inverse, ensuring it's non-zero.

Having calculated \(\text{det}(A) = 6\) in the example, we confirmed that \(A\) is invertible, allowing us to solve the equations using the inverse matrix method.
Determinants are not just mathematical tools but signposts indicating whether solving a system through matrix inverses is feasible. They emerge in multiple applications, from geometry to calculus, underlying many concepts involving area, volume, and transformations.