Exercise 3.19 Solve each of the following systems of linear equations using Gaussian Elimination Methods. x+y=5 a. x-2y=-4 x+y+z=3 b. x-2y+3z=1 2x+y-z=2 x-y-z=0 C. 2x+y+2z=3

Let's solve each system of linear equations using Gaussian Elimination.<br /><br />a. <br />$x+y=5$<br />$x-2y=-4$<br /><br />Step 1: Write the augmented matrix for the system of equations.<br />\[<br />\begin{bmatrix}<br />1 & 1 & | & 5 \\<br />1 & -2 & | & -4<br />\end{bmatrix}<br />\]<br /><br />Step 2: Perform row operations to obtain a leading 1 in the first column of the first row.<br />\[<br />\begin{bmatrix}<br />1 & 1 & | & 5 \\<br />0 & -3 & | & -9<br />\end{bmatrix}<br />\]<br /><br />Step 3: Perform row operations to obtain a leading 1 in the second column of the second row.<br />\[<br />\begin{bmatrix}<br />1 & 1 & | & 5 \\<br />0 & 1 & | & 3<br />\end{bmatrix}<br />\]<br /><br />Step 4: Perform row operations to obtain a leading 1 in the first column of the second row.<br />\[<br />\begin{bmatrix}<br />1 & 0 & | & 2 \\<br />0 & 1 & | & 3<br />\end{bmatrix}<br />\]<br /><br />Step 5: Interpret the final augmented matrix to find the solution.<br />The solution is \(x = 2\) and \(y = 3\).<br /><br />b. <br />$x+y+z=3$<br />$x-2y+3z=1$<br />$2x+y-z=2$<br /><br />Step 1: Write the augmented matrix for the system of equations.<br />\[<br />\begin{bmatrix}<br />1 & 1 & 1 & | & 3 \\<br />1 & -2 & 3 & | & 1 \\<br />2 & 1 & -1 & | & 2<br />\end{bmatrix}<br />\]<br /><br />Step 2: Perform row operations to obtain a leading 1 in the first column of the first row.<br />\[<br />\begin{bmatrix}<br />1 & 1 & 1 & | & 3 \\<br />0 & -3 & 2 & | & -2 \\<br />2 & 1 & -1 & | & 2<br />\end{bmatrix}<br />\]<br /><br />Step 3: Perform row operations to obtain a leading 1 in the second column of the second row.<br />\[<br />\begin{bmatrix}<br />1 & 1 & 1 & | & 3 \\<br />0 & 1 & -\frac{2}{3} & | & \frac{2}{3} \\<br />2 & 1 & -1 & | & 2<br />\end{bmatrix}<br />\]<br /><br />Step 4: Perform row operations to obtain a leading 1 in the third column of the third row.<br />\[<br />\begin{bmatrix}<br />1 & 1 & 1 & | & 3 \\<br />0 & 1 & -\frac{2}{3} & | & \frac{2}{3} \\<br />0 & 0 & \frac{5}{3} & | & \frac{5}{3}<br />\end{bmatrix}<br />\]<br /><br />Step 5: Perform row operations to obtain a leading 1 in the third column of the third row.<br />\[<br />\begin{bmatrix}<br />1 & 1 & 1 & | & 3 \\<br />0 & 1 & -\frac{2}{3} & | & \frac{2}{3} \\<br />0 & 0 & 1 & | & 1<br />\end{bmatrix}<br />\]<br /><br />Step 6: Perform row operations to obtain a leading 1 in the first column of the first row.<br />\[<br />\begin{bmatrix}<br />1 & 1 & 0 & | & 2 \\<br />0 & 1 & -\frac{2}{3} & | & \frac{2}{3} \\<br />0 & 0 & 1 & | & 1<br />\end{bmatrix}<br />\]<br /><br />Step 7: Perform row operations to obtain a leading 1 in the second column of the second row.<br />\[<br />\begin{bmatrix}<br />1 & 1 & 0 & | & 2 \\<br />0 & 1 & 0 & | & 2 \\<br />0 & 0 & 1 & | & 1<br />\end{bmatrix}<br />\]<br /><br />Step 8: Interpret the final augmented matrix to find the solution.<br />The solution is \(x = 2\), \(y = 2\), and \(z = 1\).<br /><br />c. <br />$2x+y-z=3$<br />$x-y-z=0$<br />$2x+y+2z=3$<br /><br />Step 1: Write the augmented matrix for the system of equations.<br />\[<br />\begin{bmatrix}<br />2 & 1 & -1 & | & 3 \\<br />1 & -1 & -1 & | & 0 \\<br />2 & 1 & 2 & | & 3<br />\end{