Find the following matrices where $A=\left[\begin{array}{ll}1 & 1 \\ 4 & 6\end{array}\right]$ and $B=\left[\begin{array}{ll}8 & 9 \\ 5 & 6\end{array}\right]$. a. $A+B$ b. A-B c. $-5 \mathrm{~A}$ d. $3 A+2 B$

Step 1 ：Given matrices A and B as \(A=\left[\begin{array}{ll}1 & 1 \\ 4 & 6\end{array}\right]\) and \(B=\left[\begin{array}{ll}8 & 9 \\ 5 & 6\end{array}\right]\)

Step 2 ：For part a, we need to find the sum of A and B. This is done by adding the corresponding elements of the two matrices. So, \(A+B = \left[\begin{array}{ll}1+8 & 1+9 \\ 4+5 & 6+6\end{array}\right] = \left[\begin{array}{ll}9 & 10 \\ 9 & 12\end{array}\right]\)

Step 3 ：For part b, we need to find the difference of A and B. This is done by subtracting the corresponding elements of B from A. So, \(A-B = \left[\begin{array}{ll}1-8 & 1-9 \\ 4-5 & 6-6\end{array}\right] = \left[\begin{array}{ll}-7 & -8 \\ -1 & 0\end{array}\right]\)

Step 4 ：For part c, we need to find the product of -5 and A. This is done by multiplying each element of A by -5. So, \(-5A = \left[\begin{array}{ll}-5*1 & -5*1 \\ -5*4 & -5*6\end{array}\right] = \left[\begin{array}{ll}-5 & -5 \\ -20 & -30\end{array}\right]\)

Step 5 ：For part d, we need to find the sum of 3 times A and 2 times B. This is done by first multiplying each element of A by 3 and each element of B by 2, and then adding the resulting matrices. So, \(3A+2B = \left[\begin{array}{ll}3*1+2*8 & 3*1+2*9 \\ 3*4+2*5 & 3*6+2*6\end{array}\right] = \left[\begin{array}{ll}19 & 21 \\ 22 & 30\end{array}\right]\)

Step 6 ：So, the final answers are \(\boxed{A+B = \left[\begin{array}{ll}9 & 10 \\ 9 & 12\end{array}\right]}\), \(\boxed{A-B = \left[\begin{array}{ll}-7 & -8 \\ -1 & 0\end{array}\right]}\), \(\boxed{-5A = \left[\begin{array}{ll}-5 & -5 \\ -20 & -30\end{array}\right]}\), and \(\boxed{3A+2B = \left[\begin{array}{ll}19 & 21 \\ 22 & 30\end{array}\right]}\)