Question

In C = 2A + 3B, the entry represented by c24 is . In D = 3A − 2B, the entry represented by d13 is .

Answer 1

Answer:

Step-by-step explanation:

When looking at the term c24 it is asking you to look at the row and Column of your new matrix c with row 2 and column 4. So after multiplying matrix a by 2 and b by 3 then adding you get matrix C. Going to the second row and the fourth column you find -12

C24= to -12

Then we go to find matrix D. We multiply matrix A by 3 and B by 2 then subtract. We have D13 so it is row 1 and column 3

So we stay on the top row and move to the third column to find 14

D13= 14

Answer 2

Answer:

See Explanation Below

Step-by-step explanation:

Question is incomplete; However, I'll assume values for matrices A and B. If you follow the steps highlighted below, you'll get your answer.

Assume that

$A = \left[\begin{array}{cccc}3&5&5&2\\-1&2&4&1\end{array}\right]$

$B = \left[\begin{array}{cccc}1&2&-7&4\\0&-5&1&0\end{array}\right]$

a. C = 2A + 3B

First, we calculate 2A.

2A is the multiplication of all cells in matrix A by 2

$2A = 2 * \left[\begin{array}{cccc}3&5&5&2\\-1&2&4&1\end{array}\right]$

$2A = \left[\begin{array}{cccc}6&10&10&4\\-2&4&8&2\end{array}\right]$

Using the same step as above, we'll solve for 3B

$3B = 3 * \left[\begin{array}{cccc}1&2&-7&4\\0&-5&1&0\end{array}\right]$

$3B = \left[\begin{array}{cccc}3&6&-21&12\\0&-15&3&0\end{array}\right]$

At this point, we can now add both matrices (2A + 3B)

$2A + 3B = \left[\begin{array}{cccc}6&10&10&4\\-2&4&8&2\end{array}\right]+ \left[\begin{array}{cccc}3&6&-21&12\\0&-15&3&0\end{array}\right]$

To get a resulting matrix, we have to add corresponding cells. This is shown below

$2A + 3B = \left[\begin{array}{cccc}6+3&10+6&10-21&4+12\\-2+0&4-15&8+3&2+0\end{array}\right]$

$2A + 3B = \left[\begin{array}{cccc}9&16&-11&16\\-2&-11&11&2\end{array}\right]$

Since, C = 2A + 3B

$C = \left[\begin{array}{cccc}9&16&-11&16\\-2&-11&11&2\end{array}\right]$

The entry represented c24 is the item in the 2nd row and the 4th column;

Going by the syntax

$C = \left[\begin{array}{cccc}1,1&1,2&1,3&1,4\\2,1&2,2&2,3&2,4\end{array}\right]$

By Comparison

C(2,4) = 2.

Hence c24 = 2

b. C = 2A + 3B

First, we calculate 2A.

3A is the multiplication of all cells in matrix A by 3

$3A = 3 * \left[\begin{array}{cccc}3&5&5&2\\-1&2&4&1\end{array}\right]$

$3A = \left[\begin{array}{cccc}9&15&15&6\\-3&6&12&3\end{array}\right]$

Using the same step as above, we'll solve for 2B

$2B = 2 * \left[\begin{array}{cccc}1&2&-7&4\\0&-5&1&0\end{array}\right]$

$2B = \left[\begin{array}{cccc}2&4&-14&8\\0&-10&2&0\end{array}\right]$

At this point, we can now subtract (3A - 2B)

$3A - 2B = \left[\begin{array}{cccc}9&15&15&6\\-3&6&12&3\end{array}\right] - \left[\begin{array}{cccc}2&4&-14&8\\0&-10&2&0\end{array}\right]$

To get a resulting matrix, we have to subtract corresponding cells. This is shown below

$3A - 2B = \left[\begin{array}{cccc}9-2&15-4&15+14&6-8\\-3-0&6+10&12-2&3-0\end{array}\right]$

$3A - 2B = \left[\begin{array}{cccc}7&11&29&2\\-3&16&10&3\end{array}\right]$

Since, D = 3A - 2B

$D = \left[\begin{array}{cccc}7&11&29&2\\-3&16&10&3\end{array}\right]$

The entry represented d13 is the item in the 1st row and the 3rd column;

Going by the syntax

$D = \left[\begin{array}{cccc}1,1&1,2&1,3&1,4\\2,1&2,2&2,3&2,4\end{array}\right]$

By Comparison

D(1,3) = 29

Hence d13 = 29