Problem 57 - Precalculus with Limits

Question

In Exercises \(55-58\) , write the system of linear equations represented by the augmented matrix. Then use back. substitution to solve. (Use variables \(x, y,\) and \(z,\) if applicable.) $$\left[\begin{array}{rrrrr}{1} & {-1} & {2} & {\vdots} & {4} \\ {0} & {1} & {-1} & {\vdots} & {2} \\ {0} & {0} & {1} & {\vdots} & {-2}\end{array}\right]$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 8\), \(y = 0\), and \(z = -2\).

Step 1: Write out the system of equations

The given augmented matrix is a 3x4 matrix. The columns represent the coefficients of the variables \(x\), \(y\), and \(z\) respectively and the last column is the constants. With these, we can write out the system of linear equations as follows: \\(1x - 1y + 2z = 4\\) \\(0x + 1y - 1z = 2\\) \\(0x + 0y + 1z = -2\\) We can simplify this to become: \\(x - y + 2z = 4\\) \\(y - z = 2\\) \\(z = -2\\)

Step 2: Back-substitution

With the simplified system of equations, we can now use the back-substitution method. This involves substituting \(z\) into the second equation to get \(y\), and then substituting \(z\) and \(y\) into the first equation to get \(x\). Starting with the final equation \(z = -2\), we substitute \(z\) into the second equation to get \\(y = 2 + z = 2 + (-2) = 0\\) Now we substitute \(y = 0\) and \(z = -2\) into the first equation to get \\(x = 4 + y - 2z = 4 + 0 - 2*(-2) = 4 + 4 = 8\\)

Step 3: Solution

So, we have determined that the solution of the system of equations is \(x = 8\), \(y = 0\), and \(z = -2\).