# Math

K & R Builders build three models of houses, M1, M2, and M3, in three subdivisions, I, II, and III, located in three different areas of a city. The prices of the houses (in thousands of dollars) are given in matrix A.
M1 M2 M3 I 300 360 380 410 420 450 620 680 700 A = II III
K & R Builders has decided to raise the price of each house by 3% next year. Write a matrix B giving the new prices of the houses.

We are given a matrix which has the prices of models of buildings in three different areas. The prices of all houses will be increased by 3%; we are asked to write a matrix with the new prices.

The original price matrix will be:

`([300,360,380],[410,420,450],[620,680,700])`

where each column describes a...

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We are given a matrix which has the prices of models of buildings in three different areas. The prices of all houses will be increased by 3%; we are asked to write a matrix with the new prices.

The original price matrix will be:

`([300,360,380],[410,420,450],[620,680,700])`

where each column describes a model of the houses, and each row describes the area.

To increase the price by 3%, we would multiply each price by 1.03 (using the distributive property we take a price p, multiply by .03 to get the price increase, and add back the price or p+.03p=p(1+.03)=1.03p).

To multiply each element in a matrix by a scalar:

`1.03([300,360,380],[410,420,450],[620,680,700])`

we just multiply each entry by the scalar.

`B=([309,370.8,391.4],[422.3,432.6,463.5],[638.6,700.4,721])`

With technology (a graphing calculator, software, spreadsheet, etc.), this computation is easy. Once the original data is entered, we can then perform this multiplication without having to do the operation on each entry.

For example, in a TI-83/84 calculator you would enter the original matrix A. Then from the homescreen you enter matrix B=1.03 x matrix A, and the calculator will return the modified matrix in a single step.