# Find each of the following

Question:

Let $A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{rr}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{rr}-2 & 5 \\ 3 & 4\end{array}\right]$

Find each of the following

(i) $A+B$

(ii) $A-B$

(iii) $3 A-C$

(iv) $A B$ (v) $B A$

Solution:

(i)

$A+B=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]+\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right]=\left[\begin{array}{cc}2+1 & 4+3 \\ 3-2 & 2+5\end{array}\right]=\left[\begin{array}{ll}3 & 7 \\ 1 & 7\end{array}\right]$

(ii)

$A-B=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]-\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right]=\left[\begin{array}{ll}2-1 & 4-3 \\ 3-(-2) & 2-5\end{array}\right]=\left[\begin{array}{cc}1 & 1 \\ 5 & -3\end{array}\right]$

(iii)

\begin{aligned} 3 A-C &=3\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]-\left[\begin{array}{rr}-2 & 5 \\ 3 & 4\end{array}\right] \\ &=\left[\begin{array}{ll}3 \times 2 & 3 \times 4 \\ 3 \times 3 & 3 \times 2\end{array}\right]-\left[\begin{array}{rr}-2 & 5 \\ 3 & 4\end{array}\right] \\ &=\left[\begin{array}{ll}6 & 12 \\ 9 & 6\end{array}\right]-\left[\begin{array}{rr}-2 & 5 \\ 3 & 4\end{array}\right] \\ &=\left[\begin{array}{ll}6+2 & 12-5 \\ 9-3 & 6-4\end{array}\right] \\ &=\left[\begin{array}{ll}8 & 7 \\ 6 & 2\end{array}\right] \end{aligned}

(iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as:

$A B=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right]=\left[\begin{array}{cc}2(1)+4(-2) & 2(3)+4(5) \\ 3(1)+2(-2) & 3(3)+2(5)\end{array}\right]$

$=\left[\begin{array}{cc}2-8 & 6+20 \\ 3-4 & 9+10\end{array}\right]=\left[\begin{array}{cc}-6 & 26 \\ -1 & 19\end{array}\right]$

(v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A. Therefore, BA is defined as:

$B A=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right]\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]=\left[\begin{array}{ll}1(2)+3(3) & 1(4)+3(2) \\ -2(2)+5(3) & -2(4)+5(2)\end{array}\right]$

$=\left[\begin{array}{lc}2+9 & 4+6 \\ -4+15 & -8+10\end{array}\right]=\left[\begin{array}{cc}11 & 10 \\ 11 & 2\end{array}\right]$