## Precalculus (6th Edition) Blitzer

Take $A=\left[ \begin{matrix} 4 & 0 \\ -3 & 5 \\ 0 & 1 \\ \end{matrix} \right], \\ B=\left[ \begin{matrix} 5 & 1 \\ -2 & -2 \\ \end{matrix} \right]\text{, and }\\ C=\left[ \begin{matrix} 1 & -1 \\ -1 & 1 \\ \end{matrix} \right].$ The steps to enter and calculate the matrices by using a graphical utility are given below: Step 1: Go to the matrix above the $\left[ {{x}^{-1}} \right]$ key. Step 2: Press the arrow to the right of EDIT to allow for entering the matrix. Step 3: Type in the dimensions of the matrix and enter the values and press ENTER. Step 4: Repeat the process for the second matrix. Step 5: Press the arrow to the right to EDIT and choose a new name. Step 6: Type in the dimensions of the matrix and enter the values and press ENTER. Step 7: Return to the home screen. Go to MATRIX to get the names of the matrices for adding, subtracting, or multiplying. Step 8: Choose the mathematical operation (addition, subtraction, or multiplication) and press ENTER to get the answer on the screen. The difference of A−C cannot be obtained because the orders of the matrices are different. The difference of B−A cannot be obtained because the orders of the matrices are different. The solutions calculated in Exercises 37–44 are as follows: (37) $\left[ \begin{matrix} 17 & 7 \\ -5 & -11 \\ \end{matrix} \right]$ (38) $\left[ \begin{matrix} -5 & -7 \\ -1 & 9 \\ \end{matrix} \right]$ (39) $\left[ \begin{matrix} 11 & -1 \\ -7 & -3 \\ \end{matrix} \right]$ (40) $\left[ \begin{matrix} 24 & 0 \\ -33 & -5 \\ -3 & -1 \\ \end{matrix} \right]$ (41) Since A and C are of different order, their difference cannot be calculated. (42) Since B and A are of different order, their difference cannot be calculated. (43) $\left[ \begin{matrix} 16 & -16 \\ -12 & 12 \\ 0 & 0 \\ \end{matrix} \right]$ (44) $\left[ \begin{matrix} 28 & 12 \\ -56 & -24 \\ -7 & -3 \\ \end{matrix} \right]$ Since, the solutions calculated manually and the ones calculated using a graphing utility are the same, therefore they are verified to be the same.