Microsoft Word Free Math Add-In - Gonzaga...

Microsoft Word Free Math Add-In

Linear Algebra

Vectors and Matrices

To write a vector of the form A=(123) , type \doubleA followed with the space bar to get the optional

character font for A and use the Matrix feature.

Similarly, the vector B=( 4−53 ) requires \doubleB for the font for B. After entry of a vector or matrix,

select Calculate to place the parentheses.

Example 1: Add two vectors.

(123)+( 4−53 )The command Calculate gives the resultant vector,

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( 5−36 )Unfortunately, the input A+B does not give the answer. The software does not assign the vectors to what was previously defined.

Example 2: Find the magnitude of a vector.

magnitude ( {3 ,−2,1 })

Enter a vector as a string with { }. Separate components of the vector with a comma. Use the command Calculate to find the answer:

√14

Example 3: Find the inner product.

inner ({3 ,−2,1} , {5,0,2 })

Separate vectors with a comma. Use the command Calculate to yield:

17

Example 4: Find the cross product where A=←2 ,4 ,5>¿ and B=¿3 ,2 ,1> .

Enter the vectors as strings. Separate vectors with a comma. Use the command cross. The strings are {−2,4,5 }∧{3,2,1 }. The input is:

cross ({−2,4,5 } , {3,2,1 })

Select Calculate to give the vector that is the cross product.

{−6 ,17 ,−16 }

Example 5: Calculate the determinant of a 3 x 3 matrix.

( 1 1 22 −2 4

−3 3 6)Highlight, right-click, and select Calculate Determinant to find the answer of −48for our example.

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Using the other options, find the trace of the matrix is 5, the inverse matrix is (12

0 −16

12

−14

0

0 18

112

) and the transpose of the matrix is (1 2 −3

1 −2 32 4 6 ).

Example 6: Given a 3 x 3 matrix and its inverse, show the product gives the identity matrix.

(12

0 −16

12

−14

0

0 18

112

)( 1 1 22 −2 4

−3 3 6)No symbol is needed between the matrices. The input window needs to be one blue window for both matrices as shown below:

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The command Calculate gives the answer of the identity matrix below:

(1 0 00 1 00 0 1)

Example 7: Find the identity 8 x 8 matrix.

identityMatrix (8)

The command identityMatrix(n) gives the identity n x n amatrix where n is from one to fifteen. The Calculate option gives our answer:

(1 0 0 0 0 0 0 00 1 0 0 0 0 0 00 0 1 0 0 0 0 00 0 0 1 0 0 0 00 0 0 0 1 0 0 00 0 0 0 0 1 0 00 0 0 0 0 0 1 00 0 0 0 0 0 0 1

)Example 8: Find the Boolean product of(0 0 1

1 0 01 1 0)

3

.

There are a few ways to do this. For one approach, enter the 3 x 3 matrix 0 0 11 0 01 1 0

and do not press

Calculate. The insertion of a pasted object with spaces or parentheses often gives an error message.

Without pressing Calculate, the matrix, 0 0 11 0 01 1 0

will contain no parentheses. Cut and paste this

matrix and place as the ‘base’. Insert the three as a superscript using this feature:

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The command Calculate from the drop-down menu gives the desired result of:

(1 0 11 1 01 1 1)

The superscript may be toggled to be 7, and the new Boolean product can be found to be:

0 0 11 0 01 1 0

7

=(3 2 22 1 24 2 3)

Another approach is to enter the matrix and superscript directly without cutting and pasting to give:

0 0 11 0 01 1 0

3

If there is a desire for the parentheses to be displayed, enter the matrix and then press Calcluate to give:

(0 0 11 0 01 1 0)

With the cursor inside the blue box, press ^3 followed by a space bar.

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This will move the three to the desired location and give:

(1 0 11 0 01 1 0)

3

Press Calculate to simplify.

Example 9: Row reduce the 2 x 2 matrix.

(3 16 2)

Use the command reduce followed by the matrix.

reduce3 16 2

Calculate will give the answer:

(1 13

0 0 )Example 10: Reduce a 3 x 3 matrix to row reduced echelon form (Lipschutz, 1968,p. 53).

( 6 3 −4−4 1 −61 2 −5)

Enter the 3 x 3 matrix using Matrix:

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The input will look like this:

6 3 −4−4 1 −61 2 −5

Press Calculate to obtain the parentheses.

( 6 3 −4−4 1 −61 2 −5)

Right-click to bring open a blue box. Place the cursor within the blue box and in front of the parentheses.

Type reduce.

reduce( 6 3 −4−4 1 −61 2 −5 )

The command Calculate will yield:

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(1 0 79

0 1 −269

0 0 0)

Example 11: Reduce the matrix to row reduced echelon form (Lipschutz, 1968, p.52).(1 −2 3 −12 −1 2 23 1 2 3 )

There is a problem entering this size of matrix from the menu since it is larger than a 3 x 3 matrix.

Enter each row as a string. Separate each string with a comma. Recall, a string requires the notation, { and }. With Insert New Equation, type:

{1 ,−2,3 ,−1 } , {2 ,−1,2,2 }, {3,1,2,3 }Calculate will place { } around the input.

{ {1 ,−2 ,3 ,−1 }, {2 ,−1 ,2,2 } , {3 ,1 ,2 ,3 }}

With the cursor in the blue box, type matrix

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Type in here matrix.


The input line is:

¿{{1,−2 ,3 ,−1 } , {2 ,−1 ,2 ,2 }, {3 ,1,2 ,3 }¿}Right-click and select Calculate gives our matrix:

(1 −2 3 −12 −1 2 23 1 2 3 )

With the cursor within the blue box and in front of the first parenthesis, type reduce.

Right click and press Calculate to execute:

reduce(1 −2 3 −12 −1 2 23 1 2 3 )

The answer is:

(1 0 0 15

7

0 1 0 −47

0 0 1 −107

)Example 12: Find the inverse of a matrix.If the matrix is a 2 x 2 or 3 x 3 matrix, the pull-down menu will contain the option, Invert Matrix.

However, for larger matrices the option will appear with the command, matrix, in the Insert New Equation line. Enter each row of the matrix as a string using { and }. Separate the elements with a comma. Consider the matrix,

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(2 5 −3 −2

−2 −3 2 −51 3 −2 2

−1 −6 4 3)

To enter the matrix, type:{ {2,5 ,−3 ,−2 } , {−2 ,−3,2 ,−5 }, {1,3 ,−2,2 }, {−1 ,−6,4,3 }}

Place the cursor in the blue box and in front of the desired matrix.

Type matrix. ¿{{2,5 ,−3 ,−2 } , {−2,−3 ,2,−5 } , {1 ,3 ,−2 ,2 } , {−1,−6 ,4 ,3 }¿}

With the insertion of matrix, the pop-up box changes and the desired option of Invert Matrix appears.

The answer is:

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(0 −7

4−134

−34

2 174

314

134

3 234

414

194

0 14

34

14

)Example 13: Solve a system of linear equations.

2 x+6 y−z=7

x+2 y−z=−1

5 x+7 y−4 z=9

Type each equation in a separate Insert New Equation line. If the last equation had no term of 7y, then a place-holder of 0y would have been needed. Now, highlight all three and right-click. The menu is:

Select Solve for x, z, y. The answer is:

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{ x∧¿12z∧¿11y∧¿−1

An alternate way is to set-up an augmented matrix. Insert:

¿ {{2,6 ,−1 ,7 }, {1 ,2 ,−1,−1 } , {5 ,7 ,−4 ,9 }¿}

Right-click on Calculate to give:

(2 6 −1 71 2 −1 −15 7 −4 9 )

Type reduce.

reduce(2 6 −1 71 2 −1 −15 7 −4 9 )

Right-click to bring up the menu and the option, Calculate. The output is:

(1 0 0 120 1 0 −10 0 1 11 )

Interpret the answer to be:

{ x∧¿12y∧¿−1z=11

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Place cursor in blue box at the front.


References

Lipschutz, Seymour. , Schaum’s Outline Series Theory and Problems of Linear Algebra, McGraw-Hill, New York, 1968.

Zill, D. and Cullen, M. Advanced Engineering Mathematics, third edition, Jones and Bartlett, Sudbury, Massachusetts, 2006).

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Linear Algebra

Eigenvalues

Example 1: Find the eigenvalues of the matrix of coefficients.

(1−λ 43 2−λ)

There are some common symbols in Basic Math.

An alternate way to reach them is to type a back-slash (\) followed by what you would like. Pressing the space bar will insert the object. For example, \pi followed by the space bar will insert π. Lambda is not on our list. We can insert lambda with the following input:

¿

To execute, use the space bar to toggle this to λ.

Insert the desired matrix using the ribbon icon Matrix.

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Selecting Calculate from the menu will place the parentheses.

(1−λ 43 2−λ)

Right-click to bring up the following menu:

Select Calculate Determinant to yield:

λ2−3 λ−10

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Cut and paste this expression and set equal to zero.

λ2−3 λ−10=0

The menu will allow you to solve for lambda.

The answer is:

λ=5λ=−2

Example 2: Find the characteristic equation of the matrix A and the eigenvalues of the matrix of coefficients.

(1 5 −93 2 44 7 −2)−λ(

1 0 00 1 00 0 1)

The input must be within one blue frame as shown below:

No operation is needed between lambda and the identity matrix. A short-cut to insert the identity matrix is the command IdentityMatrix (n) where n is the number of rows of the square matrix. The

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option, Calculate, will insert the parentheses and the 3 x 3 identity matrix. Simplify from the pull-down menu will give:

(− λ+1 5 −93 − λ+2 44 7 − λ−2)

Right-click and press Calculate Determinant. Cut and paste the result into a new equation line.

−λ3+ λ2+11 λ−39

Set equal to zero to form an equation, and select Solve for λ.

−λ3+ λ2+11 λ−39=0

The answer is:

λ≈−4.0654239158404λ≈2.5327119579202−1.7828251932075 iλ≈2.5327119579202+1.7828251932075 i

References

Lipschutz, Seymour. Schaum’s Outline Series Theory and Problems of Linear Algebra, McGraw-Hill, New York, 1968.

Zill, D. and Cullen, M. Advanced Engineering Mathematics, third edition, Jones and Bartlett, Sudbury, Massachusetts, 2006).

©2009 Nord

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