College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 24

Let

$\displaystyle C = \left[ \begin{array}{cc}
\displaystyle \frac{1}{2} & 3 \\
2 & \displaystyle \frac{3}{2} \\
-2 & 1
\end{array} \right] \qquad D = \left[ \begin{array}{cc}
1 & 4 \\
0 & -1 \\
2 & 0
\end{array} \right]$

Carry out the indicated operation $C - D$, or explain why it cannot be performed.

$\displaystyle C - D = \left[ \begin{array}{cc}
\displaystyle \frac{1}{2} & 3 \\
2 & \displaystyle \frac{3}{2} \\
-2 & 1
\end{array} \right] - \left[ \begin{array}{cc}
1 & 4 \\
0 & -1 \\
2 & 0
\end{array} \right] = \left[ \begin{array}{cc}
\displaystyle \frac{1}{2} - 1 & 3-4 \\
2-0 & \displaystyle \frac{3}{2}-(-1) \\
-2-2 & 1-0
\end{array} \right] = \left[ \begin{array}{cc}
\displaystyle \frac{-1}{2} & -1 \\
2 & \displaystyle \frac{5}{2} \\
-4 & 1
\end{array} \right]$

Comments

Popular posts from this blog

In what ways might RFID technology be used to serve customers better? What problems might arise? Do you think that the technology might be valuable when implanted in animals or people?

Calculus: Early Transcendentals, Chapter 9, 9.3, Section 9.3, Problem 18

Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 24