7. Systems of Equations & Matrices

Introduction to Matrices

7. Systems of Equations & Matrices

Introduction to Matrices

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Multiple Choice

Solve the system of equations by using row operations to write a matrix in REDUCED row-echelon form.
$4x+2y+3z=6$
$x+y+z=3$ $5x+y+2z=5$

$x=0,y=-3,z=4$

$x=-\frac14,y=-\frac{19}{4},z=\frac{11}{2}$

$x=1,y=4,z=-2$

$x=\frac12,y=\frac12,z=1$

Verified step by step guidance

Write the system of equations as an augmented matrix. The system is: $ \begin{align*} 4x + 2y + 3z &= 6 \\ x + y + z &= 3 \\ 5x + y + 2z &= 5 \end{align*} $. The augmented matrix is: $ \begin{bmatrix} 4 & 2 & 3 & | & 6 \\ 1 & 1 & 1 & | & 3 \\ 5 & 1 & 2 & | & 5 \end{bmatrix} $.

Use row operations to get a leading 1 in the first row, first column. This can be done by swapping Row 1 and Row 2: $ \begin{bmatrix} 1 & 1 & 1 & | & 3 \\ 4 & 2 & 3 & | & 6 \\ 5 & 1 & 2 & | & 5 \end{bmatrix} $.

Eliminate the first column below the leading 1 by replacing Row 2 with Row 2 minus 4 times Row 1, and Row 3 with Row 3 minus 5 times Row 1: $ \begin{bmatrix} 1 & 1 & 1 & | & 3 \\ 0 & -2 & -1 & | & -6 \\ 0 & -4 & -3 & | & -10 \end{bmatrix} $.

Get a leading 1 in the second row, second column by dividing Row 2 by -2: $ \begin{bmatrix} 1 & 1 & 1 & | & 3 \\ 0 & 1 & \frac{1}{2} & | & 3 \\ 0 & -4 & -3 & | & -10 \end{bmatrix} $.

Eliminate the second column below the leading 1 by replacing Row 3 with Row 3 plus 4 times Row 2: $ \begin{bmatrix} 1 & 1 & 1 & | & 3 \\ 0 & 1 & \frac{1}{2} & | & 3 \\ 0 & 0 & -1 & | & 2 \end{bmatrix} $.