Problem 68 - Intermediate Algebra

Question

Use matrices to solve the system. $$ \left\\{\begin{array}{l} w+x+y+z=0 \\ w-2 x+y-3 z=-3 \\ 2 w+3 x+y-2 z=-1 \\ 2 w-2 x-2 y+z=-12 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(w = 2\), \(x = 3\), \(y = -6\), \(z = -5\).

Step 1: Write the System in Matrix Form

First, write the given system of equations in matrix form: \[\begin{bmatrix}1 & 1 & 1 & 1 \1 & -2 & 1 & -3 \2 & 3 & 1 & -2 \2 & -2 & -2 & 1 \end{bmatrix}\begin{bmatrix} w \ x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ -3 \ -1 \ -12 \end{bmatrix}\] This can be expressed as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the column matrix of variables, and \(B\) is the column matrix of constants.

Step 2: Use Inverse of Matrix A

Find the inverse of matrix \(A\), if it exists. The inverse \(A^{-1}\) is given by the formula:\[{A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A)}\]Calculate the determinant of \(A\):\[\text{det}(A) = 1(-2(-2) + (-1)(1)) - 1(1(-2) + (-2)(1)) + 1(3(1) - (-2)(-2)) - 1( (3)(-2) - (1)(2) ) = 10\]Since \(\text{det}(A) eq 0\), the inverse exists.

Step 3: Calculate the Inverse of Matrix A

Using the formula for the inverse, compute \(A^{-1}\):\[A^{-1} = \frac{1}{10} \begin{bmatrix}-4 & -7 & 1 & -5 \2 & 3 & 0 & 1 \0 & 1 & -5 & -4 \-1 & 1 & 0 & 2 \end{bmatrix}\]

Step 4: Solve for X Using Matrix Multiplication

Multiply the inverse of the matrix \(A\) by matrix \(B\) to find the solution \(X\):\[X = A^{-1}B = \frac{1}{10} \begin{bmatrix}-4 & -7 & 1 & -5 \2 & 3 & 0 & 1 \0 & 1 & -5 & -4 \-1 & 1 & 0 & 2 \end{bmatrix} \begin{bmatrix}0 \-3 \-1 \-12 \end{bmatrix} = \begin{bmatrix}2 \3 \-6 \-5 \end{bmatrix}\]Thus, the solution is \(w = 2\), \(x = 3\), \(y = -6\), and \(z = -5\).

Key Concepts

System of Equations

In mathematics, a system of equations is a collection of two or more equations with the same set of variables. When solving these systems, we aim to find values for each variable that satisfy all equations simultaneously.
For example, in the given system:

• \(w+x+y+z=0\)
• \(w-2x+y-3z=-3\)
• \(2w+3x+y-2z=-1\)
• \(2w-2x-2y+z=-12\)

We have four equations and four unknowns (\(w, x, y, z\)). By converting this system into a matrix format, we can employ matrix algebra techniques such as finding the inverse of a matrix to find solutions efficiently.

Determinant

The determinant is a special number that can be calculated from a square matrix. It provides critical information about the matrix, such as whether the matrix is invertible (i.e., has an inverse).
When the determinant of a matrix is zero, the matrix does not have an inverse, and the system of equations may have no unique solution.
For the matrix \(A\) from our system of equations:\[A = \begin{bmatrix} 1 & 1 & 1 & 1 \1 & -2 & 1 & -3 \2 & 3 & 1 & -2 \2 & -2 & -2 & 1 \end{bmatrix}\]Calculating the determinant, we found \(\text{det}(A) = 10\). Since it is not zero, the matrix has an inverse, and the system has a unique solution.

Inverse of a Matrix

Finding the inverse of a matrix is a crucial step in solving systems of linear equations. The inverse helps us isolate the variable matrix when solving the equation \(AX = B\).
To find the inverse of matrix \(A\), we use the formula:\[A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A)\]In this formula, \(\text{adj}(A)\) is the adjugate of the matrix \(A\), and \(\text{det}(A)\) is the determinant we discussed previously.
Upon calculating, the inverse of our specific matrix was found as:\[A^{-1} = \frac{1}{10} \begin{bmatrix}-4 & -7 & 1 & -5 \2 & 3 & 0 & 1 \0 & 1 & -5 & -4 \-1 & 1 & 0 & 2 \end{bmatrix}\] This matrix can now be used to solve the system of equations.

Matrix Multiplication

Matrix multiplication is the process of multiplying two matrices to produce another matrix. This operation follows specific rules for row and column alignment.
When solving a system of equations using matrices, we multiply the inverse of the coefficient matrix by the constant matrix to find the solution matrix.
For our problem, after finding \(A^{-1}\), we calculated \[X = A^{-1}B\]Here, \(B\) is the constants' matrix from the system of equations:\[B = \begin{bmatrix}0 \ -3 \ -1 \ -12 \end{bmatrix}\]Using matrix multiplication, we get:\[X = \begin{bmatrix}2 \ 3 \ -6 \ -5 \end{bmatrix}\]This result gives the values for \(w, x, y,\) and \(z\) that satisfy the original system of equations.