Problem 46 - Precalculus: Mathematics for Calculus

Question

Solving a Linear System as a Matrix Equation Solve the system of equations by converting to a matrix equation and using the inverse of the coefficient matrix, as in Example \(6 .\) $$\left\\{\begin{aligned}x+2 y+z &+3 w=0 \\\y+z+w &=1 \\\y+\quad w &=2 \\\x+2 y &+2 w=3\end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = 0\), \(y = 0\), \(z = -1\), and \(w = 1\).

Step 1: Write the System as Coefficient Matrix and Vectors

Identify the coefficients of the variables in each equation to form the coefficient matrix \(A\). Define the variables vector \(\mathbf{x}\) and the constants vector \(\mathbf{b}\). The system can be represented as \(A\mathbf{x} = \mathbf{b}\).\[A = \begin{bmatrix} 1 & 2 & 1 & 3 \ 0 & 1 & 1 & 1 \ 0 & 1 & 0 & 1 \ 1 & 2 & 0 & 2 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \ z \ w \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 0 \ 1 \ 2 \ 3 \end{bmatrix} \]

Step 2: Matrix Equation

Write the system of equations in the matrix equation form: \(A\mathbf{x} = \mathbf{b}\).

Step 3: Find the Inverse of the Coefficient Matrix

Calculate the inverse of the matrix \(A\), denoted as \(A^{-1}\). Multiply both sides of the matrix equation by \(A^{-1}\) to find \(\mathbf{x} = A^{-1}\mathbf{b}\). If the inverse does not exist, the matrix equation has no unique solution. In this case, upon calculation, the inverse does exist. The inverse is:\[ A^{-1} = \begin{bmatrix} 2 & 0 & -2 & 1 \ 0 & 2 & -1 & -1 \ 0 & 1 & 0 & -1 \ -1 & -2 & 2 & 0 \end{bmatrix} \]

Step 4: Multiply Inverse by Constants Vector

Calculate the product of \(A^{-1}\) and \(\mathbf{b}\) to find the vector \(\mathbf{x}\).\[\mathbf{x} = A^{-1}\mathbf{b} = \begin{bmatrix} 2 & 0 & -2 & 1 \ 0 & 2 & -1 & -1 \ 0 & 1 & 0 & -1 \ -1 & -2 & 2 & 0 \end{bmatrix} \begin{bmatrix} 0 \ 1 \ 2 \ 3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ -1 \ 1 \end{bmatrix} \]

Step 5: Interpret the Solution

The solution vector \(\mathbf{x}\) gives the values of the variables: \(x = 0\), \(y = 0\), \(z = -1\), and \(w = 1\).

Key Concepts

Linear Systems

Linear systems are collections of equations that align in a straight line system. Each equation represents a line, plane, or hyperplane in a multidimensional space. The challenge is to find a common point shared among these lines, which is the solution to the system. For example, in the given problem, we have a set of equations that define relationships among variables like \(x, y, z,\) and \(w\). Our goal is to find specific values for these variables that satisfy all equations simultaneously.
To solve a system of linear equations efficiently, matrix methods are often used. This involves expressing the system as a single matrix equation and finding solutions through various matrix operations. We aim to find where the lines intersect by using these operations, and in this case, these intersections are represented by matrices.

Inverse Matrices

Inverse matrices are a crucial tool in solving linear systems. When dealing with matrix equations of the form \(A\mathbf{x} = \mathbf{b}\), finding \(\mathbf{x}\) involves the concept of an inverse matrix. An inverse matrix \(A^{-1}\) is such that when you multiply \(A\) by \(A^{-1}\), you get the identity matrix, akin to multiplying a number by its inverse to get one. It's like division in matrix mathematics but not all matrices have an inverse. For a matrix to have an inverse, it must be square (same number of rows and columns) and its determinant must not be zero.
In the provided solution, the inverse of the coefficient matrix \(A\) is \(A^{-1}\), which was calculated to find the solution. Once the inverse is determined, it simplifies finding the solution vector \(\mathbf{x}\) by transforming the equation to \(\mathbf{x} = A^{-1}\mathbf{b}\). This direct method often offers a neat and computationally efficient way to handle equations as compared to other methods.

Coefficient Matrix

The coefficient matrix is a central element in forming and solving matrix equations from systems of linear equations. It is called a coefficient matrix because it consists entirely of all the coefficients from the variables in each equation. In our exercise, this matrix is denoted by \(A\) and is formulated by extracting these coefficients into a structured format.

• Each row of the coefficient matrix represents an equation, and each column represents a variable.
• Matrix size will be equivalent to the number of equations by the number of variables involved.

The matrix \(A\) in our problem is given as \(\begin{bmatrix} 1 & 2 & 1 & 3 \ 0 & 1 & 1 & 1 \ 0 & 1 & 0 & 1 \ 1 & 2 & 0 & 2 \end{bmatrix}\). Solving the linear system requires using this matrix to find and multiply its inverse, which ultimately helps in deriving the variable values.

Vector Spaces

Vector spaces are fundamental in linear algebra where sets of vectors are structured to allow vector addition and scalar multiplication. In our context, vector spaces relate to the concept of representing linear systems as matrices and vectors. These spaces provide a framework where tuples of solutions like \(\begin{bmatrix}x \ y \ z \ w \end{bmatrix}\) exist.
In matrix equations, a solution vector can be seen as a point in a multi-dimensional vector space defined by the equations in the system.

• The column space or range is formed by all possible linear combinations of the coefficient matrix's columns.
• The null space of a matrix is related to solutions where output equals zero, indicating no unique solution or infinite possible solutions if non-trivial.

Our solution vector \(\mathbf{x}\), \(\begin{bmatrix}0 \ 0 \ -1 \ 1 \end{bmatrix}\), showcases specific coordinates in a vector space that simultaneously satisfies the equations described by the matrix.