#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question (71) math

Answer: Bill of A=Rs 157.80, bill of B=Rs 167.40 and bill of C=Rs 281.40

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given: Three shopkeepers A, B and C.

A purchases 12 dozen notebooks, 5 dozen pens and 6 dozen pencils.

B purchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils.

C purchase 11 dozen notebooks, 13 dozen pens and 8 dozen pencils.

Cost of notebook=40 paise = Rs 0.40

Cost of pen=Rs 1.25

Cost of pencil=35 paise = Rs 0.35

The number of items purchased by A, B and C are represented in matrix form as,

[As we know 1 dozen=12 quantity]

$\text { Notebooks } \begin{array}{l} \text { Pens } \end{array} \text { Pencils }\\$

\begin {aligned} & \begin{array}{l} X=\begin{array}{c} A \\ B \\ C \end{array}\left[\begin{array}{ccc} 12 \times 12 & 5 \times 12 & 6 \times 12 \\ 10 \times 12 & 6 \times 12 & 7 \times 12 \\ 11 \times 12 & 13 \times 12 & 8 \times 12 \end{array}\right] \end{array} \end {aligned}

$X=\begin{array}{c} A \\ B \\ C \end{array}\left[\begin{array}{ccc} 144 & 60 & 72 \\ 120 & 72 & 84 \\ 132 & 156 & 96 \end{array}\right]$

Now, matrix formed by the cost of each item is given by,

$Y = \left[\begin{array}{l} 0.40 \\ 1.25 \\ 0.35 \end{array}\right]$

Individual bill can be calculated by

$\\X Y=\left[\begin{array}{lll} 144 & 60 & 72 \\ 120 & 72 & 84 \\ 132 & 156 & 96 \end{array}\right]\left[\begin{array}{l} 0.40 \\ 1.25 \\ 0.35 \end{array}\right] \\\\ X Y=\left[\begin{array}{l} 157.80 \\ 167.40 \\ 281.40 \end{array}\right]$

So,

Bill of A=Rs 157.80

Bill of B=Rs 167.40

Bill of C=Rs 281.40

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