Question 10.2.3: Performing Scalar Multiplication and Matrix Subtraction Let ...

a.    $3 A=3\left[\begin{array}{rrr}1 & 2 & 0 \\-1 & 3 & 1 \\2 & -1 & 4\end{array}\right]=\left[\begin{array}{ccc}3(1) & 3(2) & 3(0) \\3(-1) & 3(3) & 3(1) \\3(2) & 3(-1) & 3(4)\end{array}\right]=\left[\begin{array}{rrr}3 & 6 & 0 \\-3 & 9 & 3 \\6 & -3 & 12\end{array}\right]$

b.    $2 B=2\left[\begin{array}{rrr}2 & 1 & 3 \\1 & 0 & -2 \\-3 & 4 & 5\end{array}\right]=\left[\begin{array}{ccc}2(2) & 2(1) & 2(3) \\2(1) & 2(0) & 2(-2) \\2(-3) & 2(4) & 2(5)\end{array}\right]=\left[\begin{array}{rrr}4 & 2 & 6 \\2 & 0 & -4 \\-6 & 8 & 10\end{array}\right]$

c.  $3 A-2 B=3 A+(-1) 2 B$

$=\left[\begin{array}{rrr}3 & 6 & 0 \\-3 & 9 & 3 \\6 & -3 & 12\end{array}\right]+\left[\begin{array}{rrr}-4 & -2 & -6 \\-2 & 0 & 4 \\6 & -8 & -10\end{array}\right] \begin{aligned}&\text { Substitute from parts a and } b \\&\text { changing the sign of each entry } \\&\text { in } B .\end{aligned}$

$=\left[\begin{array}{ccc}3+(-4) & 6+(-2) & 0+(-6) \\-3+(-2) & 9+0 & 3+4 \\6+6 & -3+(-8) & 12+(-10)\end{array}\right]=\left[\begin{array}{rrr}-1 & 4 & -6 \\-5 & 9 & 7 \\12 & -11 & 2\end{array}\right]$