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Lab A Lab B
Computers 15 27
Computer Tables 16 34
Chairs 16 34

Converting the data to a matrix, we have

C 2013 = [ 15 16 16 27 34 34 ]

To calculate how much computer equipment will be needed, we multiply all entries in matrix C by 0.15.

( 0.15 ) C 2013 = [ ( 0.15 ) 15 ( 0.15 ) 16 ( 0.15 ) 16 ( 0.15 ) 27 ( 0.15 ) 34 ( 0.15 ) 34 ] = [ 2.25 2.4 2.4 4.05 5.1 5.1 ]

We must round up to the next integer, so the amount of new equipment needed is

[ 3 3 3 5 6 6 ]

Adding the two matrices as shown below, we see the new inventory amounts.

[ 15 16 16 27 34 34 ] + [ 3 3 3 5 6 6 ] = [ 18 19 19 32 40 40 ]

This means

C 2014 = [ 18 19 19 32 40 40 ]

Thus, Lab A will have 18 computers, 19 computer tables, and 19 chairs; Lab B will have 32 computers, 40 computer tables, and 40 chairs.

Scalar multiplication

Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given

A = [ a 11 a 12 a 21 a 22 ]

the scalar multiple c A is

c A = c [ a 11 a 12 a 21 a 22 ]      = [ c a 11 c a 12 c a 21 c a 22 ]

Scalar multiplication is distributive. For the matrices A , B , and C with scalars a and b ,

a ( A + B ) = a A + a B ( a + b ) A = a A + b A

Multiplying the matrix by a scalar

Multiply matrix A by the scalar 3.

A = [ 8 1 5 4 ]

Multiply each entry in A by the scalar 3.

3 A = 3 [ 8 1 5 4 ] =   [ 3 8 3 1 3 5 3 4 ] =   [ 24 3 15 12 ]
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Given matrix B , find −2 B where

B = [ 4 1 3 2 ]

−2 B = [ −8 −2 −6 −4 ]

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Finding the sum of scalar multiples

Find the sum 3 A + 2 B .

A = [ 1 −2 0 0 −1 2 4 3 −6 ]  and  B = [ −1 2 1 0 −3 2 0 1 −4 ]

First, find 3 A , then 2 B .

3 A = [ 3 1 3 ( −2 ) 3 0 3 0 3 ( −1 ) 3 2 3 4 3 3 3 ( −6 ) ] = [ 3 −6 0 0 −3 6 12 9 −18 ]
2 B = [ 2 ( −1 ) 2 2 2 1 2 0 2 ( −3 ) 2 2 2 0 2 1 2 ( −4 ) ] = [ −2 4 2 0 −6 4 0 2 −8 ]

Now, add 3 A + 2 B .

3 A + 2 B = [ 3 −6 0 0 −3 6 12 9 −18 ] + [ −2 4 2 0 −6 4 0 2 −8 ]               = [ 3 2 −6 + 4 0 + 2 0 + 0 −3 6 6 + 4 12 + 0 9 + 2 −18 −8 ]               = [ 1 −2 2 0 −9 10 12 11 26 ]
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Finding the product of two matrices

In addition to multiplying a matrix by a scalar, we can multiply two matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If A is an   m   ×   r   matrix and B is an   r   ×   n   matrix, then the product matrix A B is an   m   ×   n   matrix. For example, the product A B is possible because the number of columns in A is the same as the number of rows in B . If the inner dimensions do not match, the product is not defined.

We multiply entries of A with entries of B according to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers.

To obtain the entries in row i of A B , we multiply the entries in row i of A by column j in B and add. For example, given matrices A and B , where the dimensions of A are 2   ×   3 and the dimensions of B are 3   ×   3 , the product of A B will be a 2   ×   3 matrix.

A = [ a 11 a 12 a 13 a 21 a 22 a 23 ]  and  B = [ b 11 b 12 b 13 b 21 b 22 b 23 b 31 b 32 b 33 ]

Multiply and add as follows to obtain the first entry of the product matrix A B .

  1. To obtain the entry in row 1, column 1 of A B , multiply the first row in A by the first column in B , and add.
    [ a 11 a 12 a 13 ] [ b 11 b 21 b 31 ] = a 11 b 11 + a 12 b 21 + a 13 b 31
  2. To obtain the entry in row 1, column 2 of A B , multiply the first row of A by the second column in B , and add.
    [ a 11 a 12 a 13 ] [ b 12 b 22 b 32 ] = a 11 b 12 + a 12 b 22 + a 13 b 32
  3. To obtain the entry in row 1, column 3 of A B , multiply the first row of A by the third column in B , and add.
    [ a 11 a 12 a 13 ] [ b 13 b 23 b 33 ] = a 11 b 13 + a 12 b 23 + a 13 b 33

Questions & Answers

preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
ayesha Reply
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Sandra Reply
Find the domain of the function in interval or inequality notation f(x)=4-9x+3x^2
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Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of ?105°F??105°F? occurs at 5PM and the average temperature for the day is ?85°F.??85°F.? Find the temperature, to the nearest degree, at 9AM.
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rotation by 80 of (x^2/9)-(y^2/16)=1
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thanks the domain is good but a i would like to get some other examples of how to find the range of a function
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what is the standard form if the focus is at (0,2) ?
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a²=4
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Roy Reply
A bridge is to be built in the shape of a semi-elliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center
Abdulfatah Reply
Practice Key Terms 5

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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