Problem 64 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use matrices to solve the system of equations, if possible. Use Gaussian elimination with back-substitution. $$\left\\{\begin{array}{rr} x-4 y+3 z-2 w= & 9 \\ 3 x-2 y+z-4 w= & -13 \\ -4 x+3 y-2 z+w= & -4 \\ -2 x+y-4 z+3 w= & -10 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution of the system of equations is \(x = 43/7, y = -89/21, z = -37/21, w = 9/7\).

Step 1: Writing the equations in matrix form

We write the system of equations as a matrix: \[ \left[ \begin{array}{cccc|c} 1 & -4 & 3 & -2 & 9 \\ 3 & -2 & 1 & -4 & -13 \\ -4 & 3 & -2 & 1 & -4 \\ -2 & 1 & -4 & 3 & -10 \end{array} \right] \]

Step 2: Applying Gaussian elimination

We now perform row operations to reach an upper triangular matrix, first normalize row 1 and then eliminate x in rows 2, 3, 4: \[ \left[ \begin{array}{cccc|c} 1 & -4 & 3 & -2 & 9 \\ 0 & 10 & -8 & -10 & -40 \\ 0 & -19 & -10 & -9 & -32 \\ 0 & -9 & -10 & -1 & -8 \end{array} \right] \] then eliminate y in rows 3 and 4: \[ \left[ \begin{array}{cccc|c} 1 & -4 & 3 & -2 & 9 \\ 0 & 1 & -0.8 & -1 & -4 \\ 0 & 0 & 1.2 & 3.6 & -0.4 \\ 0 & 0 & -0.2 & -8 & -8 \end{array} \right] \] finally, normalize row 3 and eliminate z in row 4: \[ \left[ \begin{array}{cccc|c} 1 & -4 & 3 & -2 & 9 \\ 0 & 1 & -0.8 & -1 & -4 \\ 0 & 0 & 1 & 3 & -1/3 \\ 0 & 0 & 0 & -7 & -9 \end{array} \right] \]

Step 3: Apply back-substitution

Now that we have the upper triangular matrix, we can get the values for each variable by back-substitution: From the 4th row, \(w = -9/-7 = 9/7\). Substituting \(w\) into 3rd row, we get \(z = -1/3 - 3*(9/7) = -37/21\). Substituting \(w\) and \(z\) into 2nd row, we get \(y = -4 - (-0.8*(-37/21) - (-1*(9/7)) = -89/21\). Finally, substituting \(w, z, y\) into 1st row, we get \(x = 9 - (-4*-89/21 + 3*-37/21 - 2*9/7) = 43/7\).

Key Concepts

Matrices

Matrices are mathematical objects that are organized into rows and columns. They are often used to represent systems of equations. In the context of Gaussian elimination, a matrix can be used to simplify solving these systems. For example, given the set of equations we had, we converted them into a matrix form as follows:
\[\begin{bmatrix} 1 & -4 & 3 & -2 & | & 9 \3 & -2 & 1 & -4 & | & -13 \-4 & 3 & -2 & 1 & | & -4 \-2 & 1 & -4 & 3 & | & -10 \end{bmatrix}\]Each row of this matrix corresponds to one of the equations, and each column represents the coefficients of a variable from those equations. By converting systems of equations to matrices, they become easier to manipulate and solve systematically.

Systems of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. Solving a system of equations involves finding values for those variables that satisfy all the given equations simultaneously.
In the exercise, we dealt with a system of four equations involving four variables: \(x, y, z,\) and \(w\). Systems like these can often be complex, but using methods such as matrices and Gaussian elimination simplifies the process.

• A consistent system has at least one solution.
• An inconsistent system has no solution.
• A dependent system has infinitely many solutions.

When employing matrices for such calculations, it is crucial to ensure that the system is consistent to find a unique solution.

Back-Substitution

Back-substitution is a method used to solve a system of equations once it has been transformed into an upper triangular matrix.
This process involves starting from the last equation in the matrix and solving for one variable at a time, working backward through the remaining equations. In our exercise:

• First, solve for \(w\) from the last equation.
• Next, substitute \(w\) into the third equation, and solve for \(z\).
• Then, use the values of \(w\) and \(z\) to solve for \(y\) in the second equation.
• Finally, substitute \(w, z, y\) to find \(x\).

Back-substitution is effective because it simplifies the process of finding exact variable values from a stepwise simplified matrix form. It's akin to peeling an onion, layer by layer.

Upper Triangular Matrix

An upper triangular matrix is a type of matrix in which all elements below the main diagonal are zero. This form is key in simplifying the process of solving systems of equations, as it allows for easy back-substitution.
In Gaussian elimination, transforming a given matrix into an upper triangular matrix involves performing a series of row operations. Here's what we did:

• Normalized and swapped rows to position pivot elements strategically.
• Used row operations to eliminate variables, creating zeros below the pivot.

Ultimately, the original equations were simplified, revealing clear solutions through back-substitution. Achieving an upper triangular matrix is critical because it simplifies the arithmetic required to solve for each variable in turn.