Answer

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**Hint:**Here, we will use the matrix multiplication method states below: It \[{\text{A}}\] is a \[m \times n\] matrix and \[{\text{B}}\] is a \[n \times p\] matrix, as shown below: \[{\text{A = }}\left( {\begin{array}{*{20}{c}}

{{a_{11}}}& \ldots &{{a_{1n}}} \\

\vdots & \ddots & \vdots \\

{{a_{m1}}}& \cdots &{{a_{mn}}}

\end{array}} \right)\] and

\[{\text{B = }}\left( {\begin{array}{*{20}{c}}

{{b_{11}}}& \ldots &{{b_{1p}}} \\

\vdots & \ddots & \vdots \\

{{b_{n1}}}& \cdots &{{b_{np}}}

\end{array}} \right)\]

Then the product \[{\text{C = A}} \times {\text{B}}\] of these two matrices will be as shown below:

\[{\text{C = }}\left( {\begin{array}{*{20}{c}}

{{c_{11}}}& \ldots &{{c_{1p}}} \\

\vdots & \ddots & \vdots \\

{{c_{m1}}}& \cdots &{{a_{mp}}}

\end{array}} \right)\]

Such that

\[{c_{ij}} = {a_{i1}}{b_{1j}} + {a_{i2}}{b_{2j}} + ..... + {a_{in}}{b_{nj}} = \sum\limits_{k = 1}^n {{a_{ik}}{b_{kj}}} \] for \[i = 1,........m\] and \[j = 1,........p\].

**Complete step by step answer:**

Step (1): Now, we have the unit cost of paper bags, scrapbooks, and pastel sheets, respectively \[{\text{Rs}}.20\], \[{\text{Rs}}{\text{.15}}\] and \[{\text{Rs}}{\text{.10}}\] , So the matrix formation will be of one row and three columns as below:

\[ \Rightarrow \left[ {{\text{20 }}15{\text{ }}10} \right]\] ……… (1)

Step (2): Now, we have school A who sold

\[25\] paper bags,

\[10\] scrapbooks, and

\[30\] pastel sheets, school B sold

\[20\] paper bags, \[15\] scrapbooks, and \[30\] pastel sheets while School C sold \[25\] paper bags, \[18\] scrapbooks, and \[35\] pastel sheets.

We will arrange the sale of each school in the column of a matrix \[\left( {3 \times 3} \right)\]as shown below:

\[ \Rightarrow \left( {\begin{array}{*{20}{c}}

{25}&{20}&{25} \\

{10}&{15}&{18} \\

{30}&{30}&{35}

\end{array}} \right)\] …………….. (2)

Step 3: Now, the amount raised by the school A, B, and C is equal to the product of the unit cost matrix which \[\left( {1 \times 3} \right)\] is and the sale matrix \[\left( {3 \times 3} \right)\].

Since, the number of columns in both the matrices (1) and (2) are the same which is \[3\], by doing multiplication of both matrices we get:

Amount raised by each school\[ = \left[ {20{\text{ }}15{\text{ }}10} \right]\left( {\begin{array}{*{20}{c}}

{25}&{20}&{25} \\

{10}&{15}&{18} \\

{30}&{30}&{35}

\end{array}} \right)\]

By doing multiplication of each element from the matrix (1) with each element of matrix (2) column-wise, we get:

Amount raised by each school\[ = \left[ {20 \times 25 + 15 \times 10 + 10 \times 30{\text{ }}20 \times 20 + 15 \times 15{\text{ }}10 \times 30 + 20 \times 25 + 15 \times 18 + 10 \times 35} \right]\]

Amount raised by each school\[ = \left[ {950{\text{ }}975{\text{ }}1120} \right]\]

Thus, the total cost \[ = {\text{Rs}}.950 + {\text{Rs}}.975 + {\text{Rs}}.1120\]

Total cost= \[{\text{Rs}}.3045\]

**\[\because \] The total cost will be equal to \[{\text{Rs}}.3045\].**

**Note:**Students need to take care while solving the product of any two matrices. While doing the multiplication you should remember the below points:

The number of columns of the 1st matrix must equal the number of rows of the 2nd matrix. And the result will have the same number of rows as the 1st matrix, and the same number of columns as the 2nd matrix.

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