# 0.3 Matrices: homework  (Page 2/4)

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$x+y+2z=0$
$x+2y+z=0$
$2x+3y+3z=0$

Find three solutions to the following system of equations.

$x+2y+z=\text{12}$
$y=3$

(5, 3, 1), (4, 3, 2) (3, 3, 3)

For what values of $k$ the following system of equations have a). No solution? b). Infinitely many solutions?

$x+2y=5$
$2x+4y=k$
$x+3y-z=5$

( $5-3s+t$ , $s$ , $t$ )

Why is it not possible for a linear system to have exactly two solutions? Explain geometrically.

## Inverse matrices

In the next two problems, verify that the given matrices are inverses of each other.

$\left[\begin{array}{cc}7& 3\\ 2& 1\end{array}\right]\left[\begin{array}{cc}1& -3\\ -2& 7\end{array}\right]$
$\left[\begin{array}{ccc}1& -1& 0\\ 1& 0& -1\\ 2& 3& -4\end{array}\right]\left[\begin{array}{ccc}3& -4& 1\\ 2& -4& 1\\ 3& -5& 1\end{array}\right]$

In the following problems, find the inverse of each matrix by the row-reduction method.

$\left[\begin{array}{cc}3& -5\\ -1& 2\end{array}\right]$
$\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]$
$\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 4\\ 0& 0& 1\end{array}\right]$
$\left[\begin{array}{ccc}1& 1& -1\\ 1& 0& 1\\ 2& 1& 1\end{array}\right]$
$\left[\begin{array}{ccc}1& 2& –1\\ –1& –3& 2\\ –1& –1& 1\end{array}\right]$
$\left[\begin{array}{ccc}1& 1& 1\\ 3& 1& 0\\ 1& 1& 2\end{array}\right]$

In the following problems, first express the system as AX = B, and then solve using matrix inverses found in the preceding four problems.

$3x-5y=2$
$-x+2y=0$

(4, 2)

$\begin{array}{ccccccc}x& +& & & 2z& =& 8\\ & & y& +& 4z& =& 8\\ & & & & z& =& 3\end{array}$
$\begin{array}{ccccccc}x& +& y& -& z& =& 2\\ x& +& & & z& =& 7\\ 2x& +& y& +& z& =& \text{13}\end{array}$

(3, 3, 4)

$\begin{array}{ccccccc}x& +& y& +& z& =& 2\\ 3x& +& y& & & =& 7\\ x& +& y& +& 2z& =& 3\end{array}$

Why is it necessary that a matrix be a square matrix for its inverse to exist? Explain by relating the matrix to a system of equations.

If a matrix $M$ has an inverse, then the system of linear equations that has $M$ as its coefficient matrix has a unique solution. If a system of linear equations has a unique solution, then the number of equations must be the same as the number of variables. Therefore, the matrix that represents its coefficient matrix must be a square matrix.

Suppose we are solving a system $\text{AX}=B$ by the matrix inverse method, but discover $A$ has no inverse. How else can we solve this system? What can be said about the solutions of this system?

## Application of matrices in cryptography

In the following problems, the letters A to Z correspond to the numbers 1 to 26, as shown below, and a space is represented by the number 27.

 A B C D E F G H I J K L M 1 2 3 4 5 6 7 8 9 10 11 12 13 N O P Q R S T U V W X Y Z 14 15 16 17 18 19 20 21 22 23 24 25 26

In the next two problems, use the matrix A, given below, to encode the given messages.

$A=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$

In the two problems following, decode the messages that were encoded using matrix A.

Make sure to consider the spaces between words, but ignore all punctuation. Add a final space if necessary.

Encode the message: WATCH OUT!

$\left[\begin{array}{c}\text{71}\\ \text{24}\end{array}\right]\left[\begin{array}{c}\text{66}\\ \text{23}\end{array}\right]\left[\begin{array}{c}\text{78}\\ \text{35}\end{array}\right]\left[\begin{array}{c}\text{87}\\ \text{36}\end{array}\right]\left[\begin{array}{c}\text{114}\\ \text{47}\end{array}\right]$

Encode the message: HELP IS ON THE WAY.

Decode the following message:

64 23 102 41 82 32 97 35 71 28 69 32

RETURN HOME

Decode the following message:

105 40 117 48 39 19 69 32 72 27 37 15 114 47

In the next two problems, use the matrix B, given below, to encode the given messages.

$B=\left[\begin{array}{ccc}1& 0& 0\\ 2& 1& 2\\ 1& 0& -1\end{array}\right]$

In the two problems following, decode the messages that were encoded using matrix $B$ .

Make sure to consider the spaces between words, but ignore all punctuation. Add a final space(s) if necessary.

Encode the message using matrix $B$ :

LUCK IS ON YOUR SIDE.

$\left[\begin{array}{c}\text{12}\\ \text{51}\\ 9\end{array}\right]\left[\begin{array}{c}\text{11}\\ \text{67}\\ 2\end{array}\right]\left[\begin{array}{c}\text{19}\\ \text{95}\\ \text{14}\end{array}\right]\left[\begin{array}{c}\text{14}\\ \text{105}\\ -\text{11}\end{array}\right]\left[\begin{array}{c}\text{15}\\ \text{87}\\ -3\end{array}\right]\left[\begin{array}{c}\text{27}\\ \text{91}\\ \text{18}\end{array}\right]\left[\begin{array}{c}4\\ \text{67}\\ -\text{23}\end{array}\right]$

Encode the message using matrix $B$ :

MAY THE FORCE BE WITH YOU.

Decode the following message that was encoded using matrix B:

8 23 7 4 47 –2 15 102 –12 20 58 15 27 80 18 12 74 –7

HEAD FOR THE HILLS

Decode the following message that was encoded using matrix B:

12 69 –3 11 53 9 5 46 –10 18 95 –9 25 107 4 27 76 22 1 72 –26

#### Questions & Answers

how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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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.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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