Precalculus (6th Edition) Blitzer

$\left[ \begin{matrix} 1 \\ -1 \\ 2 \\ \end{matrix} \right],\left[ \begin{matrix} 1 \\ -1 \\ 1 \\ \end{matrix} \right],\left[ \begin{matrix} 3 \\ -1 \\ -1 \\ \end{matrix} \right],\left[ \begin{matrix} -3 \\ 0 \\ 1 \\ \end{matrix} \right],\left[ \begin{matrix} 2 \\ -1 \\ 1 \\ \end{matrix} \right]$
Consider the system of equations of exercise 21: The equations are: \begin{align} & x+y-z=-2 \\ & 2x-y+z=5 \\ & -x+2y+2z=1 \end{align} Follow the steps given below to solve the provided equations: Step 1: Open the Ti-83 calculator and press the $\left[ 2\text{nd} \right]$ key then press $\left[ \text{MATRIX} \right]$ key. Step 2: Press $\left[ 2\text{nd} \right]$ key, then press $\left[ \text{MATRIX} \right]$ key. Use right arrow and select $\left[ \text{EDIT} \right]$. Enter $\left[ 1 \right]$ for matrix $\left[ \text{A} \right]$ then enter the order of the matrix then press $\left[ \text{ENTER} \right]$ key and enter the elements into the matrix. Step 3: Press $\left[ 2\text{nd} \right]$ and $\left[ \text{MATRIX} \right]$ key then go to “MATH” and select “rref (“. Step 4: Press $\left[ \text{ENTER} \right]$, then press $\left[ 2\text{nd} \right]$ keyand 1, then press $\left[ \text{ENTER} \right]$ key. The solution is: $\left[ \begin{matrix} 1 \\ -1 \\ 2 \\ \end{matrix} \right]$ Consider the matrix of exercise 22: The equations are: \begin{align} & x-2y-z=2 \\ & 2x-y+z=4 \\ & -x+y-2z=-4 \end{align} Follow the steps given below to solve the provided equations: Step 1: Open the Ti-83 calculator and press the $\left[ 2\text{nd} \right]$ key then press $\left[ \text{MATRIX} \right]$ key. Step 2: Press $\left[ 2\text{nd} \right]$ key, then press $\left[ \text{MATRIX} \right]$ key. Use right arrow and select $\left[ \text{EDIT} \right]$. Enter $\left[ 1 \right]$ for matrix $\left[ \text{A} \right]$ then enter the order of the matrix then press $\left[ \text{ENTER} \right]$ key and enter the elements into the matrix. Step 3: Press $\left[ 2\text{nd} \right]$ and $\left[ \text{MATRIX} \right]$ key then go to “MATH” and select “rref (“. Step 4: Press $\left[ \text{ENTER} \right]$, then press $\left[ 2\text{nd} \right]$ key and 1, then press $\left[ \text{ENTER} \right]$ key. The solution is: $\left[ \begin{matrix} 1 \\ -1 \\ 1 \\ \end{matrix} \right]$ Consider the matrix of exercise 23: The equations are: \begin{align} & x+3y=0 \\ & x+y+z=1 \\ & 3x-y-z=11 \end{align} Follow the steps given below to solve the provided equations: Step 1: Open the Ti-83 calculator and press the $\left[ 2\text{nd} \right]$ key then press $\left[ \text{MATRIX} \right]$ key. Step 2: Press $\left[ 2\text{nd} \right]$ key, then press $\left[ \text{MATRIX} \right]$ key. Use right arrow and select $\left[ \text{EDIT} \right]$. Enter $\left[ 1 \right]$ for matrix $\left[ \text{A} \right]$ then enter the order of the matrix then press $\left[ \text{ENTER} \right]$ key and enter the elements into the matrix. Step 3: Press $\left[ 2\text{nd} \right]$ and $\left[ \text{MATRIX} \right]$ key then go to “MATH” and select “rref (“. Step 4: Press $\left[ \text{ENTER} \right]$, then press $\left[ 2\text{nd} \right]$ key and 1, then press $\left[ \text{ENTER} \right]$ key. The solution is: $\left[ \begin{matrix} 3 \\ -1 \\ -1 \\ \end{matrix} \right]$ Consider the matrix of exercise 24: The equations are: \begin{align} & 3y-z=-1 \\ & x-5y-z=-4 \\ & -3x+6y+2z=11 \end{align} Follow the steps given below to solve the provided equations: Step 1: Open the Ti-83 calculator and press the $\left[ 2\text{nd} \right]$ key then press $\left[ \text{MATRIX} \right]$ key. Step 2: Press $\left[ 2\text{nd} \right]$ key, then press $\left[ \text{MATRIX} \right]$ key. Use right arrow and select $\left[ \text{EDIT} \right]$. Enter $\left[ 1 \right]$ for matrix $\left[ \text{A} \right]$ then enter the order of the matrix then press $\left[ \text{ENTER} \right]$ key and enter the elements into the matrix. Step 3: Press $\left[ 2\text{nd} \right]$ and $\left[ \text{MATRIX} \right]$ key then go to “MATH” and select “rref (“. Step 4: Press $\left[ \text{ENTER} \right]$, then press $\left[ 2\text{nd} \right]$ key and 1, then press $\left[ \text{ENTER} \right]$ key. The solution is: $\left[ \begin{matrix} -3 \\ 0 \\ 1 \\ \end{matrix} \right]$ Consider the matrix of exercise 25: The equations are: \begin{align} & 2x-y-z=4 \\ & x+y-5z=-4 \\ & x-2y=4 \end{align} Follow the steps given below to solve the provided equations: Step 1: Open the Ti-83 calculator and press the $\left[ 2\text{nd} \right]$ key then press $\left[ \text{MATRIX} \right]$ key. Step 2: Press $\left[ 2\text{nd} \right]$ key, then press $\left[ \text{MATRIX} \right]$ key. Use right arrow and select $\left[ \text{EDIT} \right]$. Enter $\left[ 1 \right]$ for matrix $\left[ \text{A} \right]$ then enter the order of the matrix then press $\left[ \text{ENTER} \right]$ key and enter the elements into the matrix. Step 3: Press $\left[ 2\text{nd} \right]$ and $\left[ \text{MATRIX} \right]$ key then go to “MATH” and select “rref (“. Step 4: Press $\left[ \text{ENTER} \right]$, then press $\left[ 2\text{nd} \right]$ key and 1, then press $\left[ \text{ENTER} \right]$ key. The solution is: $\left[ \begin{matrix} 2 \\ -1 \\ 1 \\ \end{matrix} \right]$ The solution of the provided systems are: $\left[ \begin{matrix} 1 \\ -1 \\ 2 \\ \end{matrix} \right],\left[ \begin{matrix} 1 \\ -1 \\ 1 \\ \end{matrix} \right],\left[ \begin{matrix} 3 \\ -1 \\ -1 \\ \end{matrix} \right],\left[ \begin{matrix} -3 \\ 0 \\ 1 \\ \end{matrix} \right],\left[ \begin{matrix} 2 \\ -1 \\ 1 \\ \end{matrix} \right]$