19.5 - Solved Examples

In the previous section, we saw negative matrix and difference of two matrices. We also saw multiplication of matrix by a scalar and a solved example. In this section, we will see a few more solved examples.

Solved example 19.10
Find X and Y if:
X + Y = $\left[\begin{array}{r}       
5    &{2}    \\
0    &{9}    \\
\end{array}\right]       
$ and X − Y = $\left[\begin{array}{r}       
3    &{6}    \\
0    &{-1}    \\
\end{array}\right]$.
Solution:

Fig.19.14

Solved example 19.10
Find the values of x and y from the following equation:
$2\left[\begin{array}{r}       
x    &{5}    \\
7    &{y-3}    \\
\end{array}\right] +
\left[\begin{array}{r}       
3    &{-4}    \\
1    &{2}    \\
\end{array}\right] =
\left[\begin{array}{r}       
7    &{6}    \\
15    &{14}    \\
\end{array}\right]       
$
Solution:

Fig.19.15

◼ Remarks:
5,6: Equations are obtained by equating corresponding elements.
7,8: x and y values are obtained by solving the equations.

Solved example 19.11
Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely, Basmati, Permal and Naura. The sale (in rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B.

Fig.19.16

(i) Find the combined sales in September and October for each farmer in each variety.
(ii) Find the decrease in sales from September to October.
(iii) If both farmers receive 2% profit on gross sales, compute the profit for each farmer and for each variety sold in October.
Solution:
Part (i):
The combined sales for the two months can be obtained by adding A and B. We get:

Fig.19.17

• Let us write a sample information that can be obtained from (A+B):
When the two months of September and October are taken together, Gurcharan Singh achieved a sale of Rs.20000/- in the case of Naura rice.

Part (ii):
The decrease in sales can be obtained by subtracting B from A. We get:

Fig.19.18

• Let us write two sample information that can be obtained from (A−B):

1. Compared to September, Ramkishan experienced a decrease of Rs.10000/- in the sale of Permal rice in october. 

2. Compared to September, Gurcharan Singh did not experience any increase or decrease in the sale of Naura rice in october.

Part (iii):

First we will see an example. It can be written in steps:
1. Consider the matrix for October.
2. We see that, Ramkishan achieved a sale of Rs.5000/- in the case of Basmati rice. This amount includes the profit obtained.
3. Given that, profit is 2%.
• So if x is the original cost, then (x  × 1.02) = 5000
4. If p is the amount obtained as profit, then (5000 - x) = p
5. Solving the equations in (3) and (4), we get:
p = 5000 × 0.02
6. It is clear that:
To find the profit amount (or simply profit), we need to multiply the sales amount by 0.02.
7. To find the profit for each product, for each farmer, in just one step, we multiply the October matrix by 0.02. We get:

Fig.19.19

• Let us write a sample information that can be obtained from 0.02B:
In the month of October, Gurcharan Singh obtained a profit of Rs.400/- in the case of Basmati rice.

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In the next section, we will see multiplication of matrices. 

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