Questions on Algebra: Matrices, determinant, Cramer rule answered by real tutors!

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We want to try to end up with a matrix 
that looks like this:



with 0's in the three lower left hand positions:


We can get a 0 where the 2 is by multiplying Row 1 by -2
and adding it to row 2:

That instruction is written as

-2·R1+1·R2->R2





Notice that Row 2 will be simpler if we divide it through by 5,

That instruction is written R2->R2





We can get a 0 where the 1 is in the lower left corner by 
multiplying Row 1 by -1 and adding it to row 3:

That instruction is written as

-1·R1+1·R3->R3





We can get a 0 where the -1 is in the bottom row by 
multiplying Row 2 by 1 and adding it to row 3:

That instruction is written as 1R2+1R3->R3





Now that we have 0's in the lower lefthand corner,
we convert the matrix back to a system of equations
in x, y and z:



or just



Now we use back substitution.

From the third equation, z=1, we substitute
that into the middle equation, getting:

y-z = -3
y-1 = -3
  y = -2

Then substitute y=-2 and z=1 in the 1st equation:

x-2(-2)+(1) = 6
      x+4+1 = 6
        x+5 = 6
          x = 1

Solution (x,y,z) = (1,-2,1)

Edwin

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For this square matrix, the determinant is equal to

2 det A - 5 det B + 0 det C - 4 det D

where A,B,C,D are the 3x3 matrices whose entries are neither in a row or column of the top entry (e.g. matrix A would contain -2 0 2/-1 1 6/0 3 -2). Additionally, you would have to find the determinant of 3x3 matrices, which reduces to a several 2x2 matrices.

Since this problem requires a lot of "bashing" (not the correct word usage, but math people use it all the time), you could use a calculator. Most graphing calculators can evaluate determinants of square matrices.

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here's a reference on cramer's rule.
i'll do one.
you do the rest.
the reference for applying cramer's rule can be found here:
http://www.purplemath.com/modules/cramers.htm
you will need to be able to find the value of the determinant / matrix.
the reference for that can be found here:
http://www.purplemath.com/modules/determs.htm

your first problem is:
        x   +    2y   =    2
       3x   +     y   =    3 
you set up a matrix as shown below:
        1        2     |       2
        3        1     |       3
cramer's rule says:
x = Dx / D
y = Dy / D
first you want to find D.
That's the left part or our matrix which is equal to:
       1       2
       3       1
the value for D is equal to 1*1 - 2*3 = 1 - 6 = -5
next we want to find Dx
Dx is found by replacing the first column in D with the result column.
the matrix for Dx becomes: 
      2        2
      3        1
the value for Dx is equal to 2*1 - 2*3 = 2 - 6 = -4
next we want to find Dy
Dy is found by replacing the second column in D with the result column.
the matrix for Dy becomes:
      1        2
      3        3
the value for Dy is equal to 1*3 - 2*3 = 3 - 6 = -3
we have:
D = -5
Dx = -4
Dy = -3
we now solve for x and y as follows:
x = Dx / D = -4 / -5 = 4/5
y = Dy / D = -3 / -5 = 3/5
our final answer is:
x = 4/5
y = 3/5
i will solve this system of equations the standard way to confirm that these answers are what we should expect.
the original equations are:
        x   +    2y   =    2
       3x   +     y   =    3
multiply the second equation by -1 to get:
         x   +     2y   =     2
       -6x   -     2y   =    -6
add these equations together to get:
       -5x              =    -4
divide both sides of this equation by -5 to get:
         x              =    4/5
substitute 4/5 for x in the first equation to get:
        x        2y   =    2
this becomes:
        4/5   +  2y   =    2
subtract 4/5 from both sides of this equation to get:
                 2y   =    2 - 4/5
simplify to get:
                 2y   =    6/5
divide both sides of this equation by 2 to get:
                  y   =    6/5 * 1/2
this becomes:
                  y   =    3/5
our answer doing it the conventional way are:
x = 4/5
y = 3/5

this agrees with our answer that we derives using cramer's rule, so the rule is confirmed as being accurately processed.
you should be able to do the rest on your own.
let me know if you have problems.

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This is a 2-by-2 determinant, meaning that it has two rows and two columns.
.
The value of a 2-by-2 determinant is found by multiplying along the diagonals.
.
The first diagonal to multiply along is the one that goes from the upper left corner to the lower right corner. This diagonal multiplication gives you the product of 2 times 2, which, of course, equals +4.
.
The second diagonal to multiply along is the one that goes from the upper right corner to the lower left corner. This diagonal multiplication gives you the product of 3 times 3 which is +9.
.
The final step is to note the product of the first diagonal multiplication (down and to the right - in this problem +4) and subtract from it the product of the second diagonal (down and to the left - in this problem +9).
.
So the value of this determinant is given by:
.
+4 - (+9) = 4 - 9 = -5
.
The answer to this problem is that the value of the given determinant is -5.
.
In summary, to get the value of a 2-by-2 determinant, first find the product obtained by multiplying down and to the right and second, subtract from that the product obtained by multiplying down and to the left.
.
I hope this helps you to understand the rules for evaluating 2-by-2 determinants.
.

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sum of two consecutive odd integers is 36.find them
1st: 2x-1
2nd: 2x+1
---
Equation:
sum = 36
4x = 36
x = 9
---
1st: 2x-1 = 17
2nd: 2x+1 = 19
==============
Cheers,
Stan H.

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Write as the 2x2 coefficient matrix multiplied by the 2x1 variable 
matrix and set it equal to the 2x1 constant matrix:

 = 

To find the inverse of the 2x2 coefficient matrix: 

1. Swap the upper left and lower right emements: 

2. Change the signs of the upper right and lower left elements: 

3. Calculate the determinant, either of the original matrix or the one that is the result of step 3 for their determinants are the same: 
   
original matrix's determinant:  (1)(-3)-(2)(2) = -3-4 = -7
step 3 matrix;s determinant:  (-3)(1) - (-2)(-2) = -3-4 = -7

4. Divide each term in the matrix by the value of the determinant:



Left-multiply both sides of the matrix equation by this inverse:

 = 

 = 

 = 

 = 

 = 

 = 

 = 

x=2, y=1

Edwin

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Let
Adults = x
Children = 510-x
Total receipts = 810

Admission price for Adults = 2
Admission price for Children = 1


(Admission price for Adults)(Adults)+ (Admission price for Children)(Children) = Total receipts
(2)(x)+(1)(510-x)=810
2x+510-x=810
x=810-510
x=300

Adults = x = 300
Children = 510-x = 510-300 = 210

Check
======
300*2+210*1=810
600+210=810
810=810

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x + y = 3k
x - y = k


Add the equations term by term just as they are and the terms in y will cancel:

 x + y = 3k
 x - y =  k
-----------
2x     = 4k

Divide both sides by 2

     x = 2k


Multiply the second equation x - y = k through by -1 making it
-x + y = -k

Then add it to the first equation term by term and the terms in x will cancel:

 x + y = 3k
-x + y = -k
-----------
    2y = 2k

Divide both sides by 2

     y = k

Answer (x,y) = (2k,k)

Edwin

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3x-2y=2
2x-y=2
----------
Multiply the 2nd eqn by 2, then subtract.

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2 x + 6 y = 17 .............1
2 x -10 y = 9 .............2
Eliminate y
multiply (1)by 5
Multiply (2) by 3
10 x 30 y = 85
6 x -30 y = 27
Add the two equations
16 x = 112
/ 16
x = 7
plug value of x in (1)
2 x + 6 y = 17
14 + 6 y = 17
6 y = 17-14
6 y = 3
y = 0.50

m.ananth@hotmail.ca

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2 x + 6 y = 17 .............1
2 x -10 y = 9 .............2
Eliminate y
multiply (1)by 5
Multiply (2) by 3
10 x 30 y = 85
6 x -30 y = 27
Add the two equations
16 x = 112
/ 16
x = 7
plug value of x in (1)
2 x + 6 y = 17
14 + 6 y = 17
6 y = 17-14
6 y = 3
y = 0.50

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This matrix is in "row echelon" form because each
row's left-most non-zero element is 1 and they
move to the right as we go down the matrix.  So the
matrix is an abbreviation for this system:


Removing all the 1 coefficients and the 0 terms:


We use "back substitution".

The bottom, fourth, equation is already solved for z,

so we substitute -2 for z in the third equation:













Now we substitute 3 for y and -2 for z in the
second equation:








Finally we substitute 2 for x, 3 for y 
and -2 for z in the first equation:






Solution (w,x,y,z) = (1,2,3,-2)

Edwin

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3x-5y+4z=43
3x-2y-2z=10
x-y+3z=25
-----
Use any method you know to get:
x = 8
y = 1
z = 6
===========
Cheers,
Stan H.
=================

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The dimensions are the number of rows by the number of columns. There are 3 rows two columns here so its a 3x2
To remember it I either think of a Remote Control Car (an RC Car) or RC Cola to remember its rows first then columns. Also, if you think about x,y, we do going across first and then going up and down.

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Could you give me about six matrices with solutions of 5,9,10,12,6, and 18?

Matrices?  Matrices do not equal to numbers.  Do you mean determinants?:

 = 11·3 - 4·7 = 33 - 28 = 5

 = 5·5 - 4·4 = 25 - 16 = 9

 = 7·4 - 6·3 = 28 - 18 = 10

 = 4·2 - (-2)·2 = 8 + 4 = 12

 = 6·1 - 0·0 = 6 - 0 = 6

 = -3·9 - (-9)·5 = -27 + 45 = 18.

Edwin

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Augment the matrix with the identity matrix with the 
same dimensions (4×4):



We will use row operations until we get the identity on the left.

Multiply row 1 by -1 and add it to row 2:



Multiply row 1 by -1 and add it to row 3:



Multiply row 1 by -1 and add it to row 4:



Multiply row 2 by -1 and add it to row 1:



Multiply row 2 by 2 and add it to row 3:




Multiply row 2 by -2 and add it to row 4:



Add row 3 to row 1:



Multiply row 3 by 2 and add to row 4:



To avoid fractions multiply row 3 by -2 and row 2 by 3:



Add row 3 to row 2:



Again to avoid fractions multiply row 4 by 2 and row 1 by -3:


 
Add row 4 to row 1:



To avoid fractions multiply row 2 by 6:



Add row 4 to row 2:




To avoid fractions multiply row 4 by 2 and row 3 by 3:



Add row 4 to row 3:



Now get 1's on the diagonal:

Divide row 1 by -3
Divide row 2 by 18
Divide row 3 by 18
Divide row 4 by 12



Now that we have the identity on the left, the 
augmented part is the inverse:



Edwin

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That is at least an hour of work rendering the complete solution on this site. How much are you willing to pay?

John

My calculator said it, I believe it, that settles it

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You're absolutely right! (x, y, z) = (1+k, 2-k, k)

Edwin

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perform matrix row operation and write the new matrix

1  2  2  2
0  1 -1  2   -5R2+R3
0  5  4  1

-5R2+R3
 
-5R2 means to do this work to the side 

1. Multiply row #2 by -5 :  -5(0  1 -1  2) = (0 -5  5 -10)

and then +R3 means
 
2. Add that to row #3  :     (0 -5  5 -10) + (0  5  4  1) = (0  0  9  -9)
         
3. Replace row #3 by (0  0  9  -9)

1  2  2  2
0  1 -1  2   
0  0  9 -9

That's the new matrix.

Edwin

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4 5 6 7
3 4 5 6
2 3 4 5
1 2 3 4

Multiply row 2 by -1 and add it to row 1

1 1 1 1
3 4 5 6
2 3 4 5
1 2 3 4

Multiply row 3 by -1 and add it to row 2

1 1 1 1
1 1 1 1
2 3 4 5
1 2 3 4

Multiply row 4 by -1 and add it to row 3

1 1 1 1
1 1 1 1
1 1 1 1
1 2 3 4

Multiply row 1 by -1 and add it to row 2

1 1 1 1
0 0 0 0
1 1 1 1
1 2 3 4

Multiply row 1 by -1 and add it to row 3

1 1 1 1
0 0 0 0
0 0 0 0
1 2 3 4

Swap rows 2 and 4

1 1 1 1
1 2 3 4
0 0 0 0
0 0 0 0

Multiply row 1 by -1 and add it to row 2

1 1 1 1
0 1 2 3
0 0 0 0
0 0 0 0

Multiply row 2 by -1 and add it to row 1

1 0 -1 -2
0 1  2  3
0 0  0  0
0 0  0  0

That's in row reduced echelon form.  The number of
not-all-zero rows is 2, so its rank is 2.

Edwin

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Let be the number of tradesmen and be the number of laborers.
and
---> (multiplying both sides times 4 to get rid of decimals)
=
Now you have your matrices:
=
= , and
=
The equation with matrices is

Now you have to find the inverse matrix and calculate
=
Whichever way you were taught to calculate it
= and
= =
= =
So the number of tradesmen is , and
the number of laborers is .

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integral [(9x+4)/(x^2-7x+15)] dx
---
Use partial fraction decompostion to get::
------------
= [(A/(x-5))+(B/(x-3)] = (9x/4)/(x^2-7x+15)
-----
A(x-3)+B(x-5) = 9x+4
----
A+B = 9
-3A-5B = 4
-----
3A+3B = 27
-3A-5B = 4
--------
-2B = 31
B = -31/2
---
A = 49/2
---------------
integra[((49/2)/(x-5) - (31/2)/(x-3)]
----
= (49/2)ln(x-5) - (31/2)ln(x-3)
----
Check the following site to see how this all works.
===============
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/
partialfracsoldirectory/PartialFracSol.html#SOLUTION 1
=====================
Cheers,
Stan H.
==============

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2 14
y= - x + --
3 3

If x=0 ,y= 14/3
-------------------
That is true.
And if x = 3, y = 20/3
--------------------------
Plot those 2 points and draw a line thru them.

----
Cheers,
Stan H.
====

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Here is the formula for multiplying a "3 down by 2 across" matrix 
by a "2 down by 3 across" matrix.  The matrix you get when you 
multiply twn matrices gets its "down dimension" from the first
matrix and its "across dimension" from the second matrix: 

· = 

Use that formula and you'll get this answer, which is "3 down by 3 across":

· = 

Edwin

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find the products of the following equation in clock arithmetic: 12 x 3
-----
That's not an equation, there's no equal sign.
There's only one product.
12*3 = 36
Don't know what you mean by clock arithmetic.

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your equations are:
2x+3y+z=13 (equation 1)
3x+2y+4z=17 (equation 2)
4x+5y+2z=24 (equation 3)
multiply equation 1 by -4 and add it to equation 2 to get:
-8x - 12y -4z = -52 plus:
3x + 2y + 4z = 17 equals:
-5x -10y = -35 (equation 4)
multiply equation 1 by -2 and add it to equation 3 to get:
-4x - 6y - 2z = -26 plus:
4x + 5y + 2z = 24 equals:
-y = -2
multiply both sides of this equation by -1 to get:
y = -2
that's one of the values you are looking for.
substitute for y in equation 4 to get:
-5x - 10(2) = -35
simplify to get:
-5x - 20 = -35
add 20 to both sides of the equation to get:
-5x = -15
divide both sides of the equation by -5 to get:
x = 3
that's the second of the values you are looking for.
so far you have:
x = 3
y = 2
substitute for x and y in equation 1 to get:
2(3) + 3(2) + z = 13
simplify to get:
6 + 6 + z = 13
combine like terms to get:
12 + z = 13
substract 12 from both sides of the equation to get:
z = 1
that's the last of the values you are looking for.
you have:
x = 3
y = 2
z = 1
substitute in all 3 original equations to confirm these solutions apply to all of the equations in the system.
your original equations are:
2x+3y+z=13
3x+2y+4z=17
4x+5y+2z=24
after substitution, these equations become:
2(3) + 3(2) + 1 = 13
3(3) + 2(2) + 4(1) = 17
4(3) + 5(2) + 2(1) = 24
after simplification, these equations become:
6 + 6 + 1 = 13
9 + 4 + 4 = 17
12 + 10 + 2 = 24
after combining like terms, these equations becomes:
13 = 13
17 = 17
24 = 24
all equations are true confirming the values for x and y and z are good.
answer is:
x = 3
y = 2
z = 1

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If the determinant is zero, the matrix does not have and inverse. (And if the matrix does not have and inverse, then determinant is zero).
The determinant of is

-->

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Solve the following systems of equations:
x+2y+3z+4s+5t=87
17z-5y+203.5=-(t+5s)
4t+23s-14z+21y+13x=17.5
8y+4z+2s+10t=30x
7.5x+8y+10t=18z+9s+738
I need helping showing work on how to get these answers.
----------------
There are several methods.
There's elimination, eliminating one variable at a time until you solve for 1 of them.
---------
There's inverting the matrix.
-------
I would use determinants if I were to solve it.
A 5 by 5 is very tedious and error prone.
I would use an online solver, just type it in and get the answers. If you have to solve it manually, use a solver to check your work.

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I have not heard of an echelon method, but I have heard translations of the word echelon from the French into other languages to refer to a strategy to solve systems of linear equations.
You want to keep an equation involving both variables and replace the other equation with a combination of the two where one of the variables has been eliminated.
I would keep because it's nice and simple and replace the other equation, using the sum of the two
-->
The system is now


Even better, we can further simplify, dividing both sides by 7.
-->
So we could say that we now have the system


Then substituting we get -->
Maybe you were trying to eliminate instead of y to give your modified system the same shape as the one in the book.
We can do that too, if you get points for eliminating variables in alphabetical order.
Just multiply times -6 to get and add it to
to get
-->
Then you could simplify it by dividing both sides by -7 to get

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Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 2 variables

First let . This is the matrix formed by the coefficients of the given system of equations. Take note that the right hand values of the system are and which are highlighted here: These values are important as they will be used to replace the columns of the matrix A. Now let's calculate the the determinant of the matrix A to get . Remember that the determinant of the 2x2 matrix is . If you need help with calculating the determinant of any two by two matrices, then check out this solver. Notation note: denotes the determinant of the matrix A. --------------------------------------------------------- Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix (since we're replacing the 'x' column so to speak). Now compute the determinant of to get . Once again, remember that the determinant of the 2x2 matrix is To find the first solution, simply divide the determinant of by the determinant of to get: So the first solution is --------------------------------------------------------- We'll follow the same basic idea to find the other solution. Let's reset by letting again (this is the coefficient matrix). Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix (since we're replacing the 'y' column in a way). Now compute the determinant of to get . To find the second solution, divide the determinant of by the determinant of to get: So the second solution is ==================================================================================== Final Answer: So the solutions are and giving the ordered pair (-18/7, 20/7) Once again, Cramer's Rule is dependent on determinants. Take a look at this 2x2 Determinant Solver if you need more practice with determinants.

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Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables

First let . This is the matrix formed by the coefficients of the given system of equations. Take note that the right hand values of the system are , , and and they are highlighted here: These values are important as they will be used to replace the columns of the matrix A. Now let's calculate the the determinant of the matrix A to get . To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver. Notation note: denotes the determinant of the matrix A. --------------------------------------------------------- Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix (since we're replacing the 'x' column so to speak). Now compute the determinant of to get . Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant. To find the first solution, simply divide the determinant of by the determinant of to get: So the first solution is --------------------------------------------------------- We'll follow the same basic idea to find the other two solutions. Let's reset by letting again (this is the coefficient matrix). Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix (since we're replacing the 'y' column in a way). Now compute the determinant of to get . To find the second solution, divide the determinant of by the determinant of to get: So the second solution is --------------------------------------------------------- Let's reset again by letting which is the coefficient matrix. Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix Now compute the determinant of to get . To find the third solution, divide the determinant of by the determinant of to get: So the third solution is ==================================================================================== Final Answer: So the three solutions are , , and giving the ordered triple (9, 3, -3) Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.

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[Always skip a space before and after <  and > so that the computer won't
think it's an HTML tag and delete it.]

If -3 < f < 4 and -2 < g < 1,then what is the range of possible values of fg.

f is defined for these values:

---------o===========================o--------
-5  -4  -3  -2  -1   0   1   2   3   4   5   6


g is defined for these values:

-------------o===========o--------------------
-5  -4  -3  -2  -1   0   1   2   3   4   5   6

When f is very near -3 and g is very near -2, their product
fg is very near +6, so +6 is the least upper bound.


When f is very near 4 and g is very near -2, their product
fg is very near -8, so -8 is the greatest lower bound.

Therefore:
                -8 < fg < 6  


Edwin

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The way to multiply two 2×2 matrices is like this:

 = 

Use that rule to multiply the two 2×2 matrices on the left:

 = 

 = 

 = 

Equate corresponding elements of the matrices on the left
and right:



From those you get x=5, y = 2, z = -3

Edwin

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X-2y+3t-2w=0
3x-7y-2t+4w=0
4x+3y+5t+3w=0
---------------
There are 4 variables and 3 equations.
--> no unique solution

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1 2 3 4
6 7 8 9
-1 -2 -3 -4
-6 -7 -8 -9
---------------
calculate the cofactor of the element 9 of this determinant:
---------
If you mean where the 9 is, element 2,4:
---
= Det of
1 2 3
-1 -2 -3
-6 -7 -8
-----------
= 1(16-21) - 2(8-18) + 3(7-12) = -5 + 20 - 15
= zero

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Okay so when doing cofactors, think of it as a checkerboard I find helps...
The top left is positive, and then every other changes sign. So that means when finding the cofactor of 9, it will be positive.

From there we take out the row and column that the 9 is in and find the determinant . This means we find the determinent of A(that's what I will call this next matrix)
1 2 3
-1 -2 -3
-6 -7 -8
So we just need to find this determinant.
Well going along the top row, we have
detA=1*detB-2*detC+2*detD
where B=
-2 -3
-7 -8
so detB=(-2*-8)-(-3*-7)=16-21=-5
And C=
-1 -3
-6 -8
So detC=(-1*-8)-(-3*-6)=8-18=-10
and D=
-1 -2
-6 -7
So detD=(-1*-7)-(-2*-6)=7-12=-5

Since detA=cofactor of 9=1*detB-2*detC+2*detD
We can plug in the values we have figured out and solve for this. I will leave this up to you :)

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ax + by = c
dx + ey = f

By Cramer's rule:

Delta =  = ae - bd

Dx =  = ce - bf 

Dy =  = af - cd

x =  =  = 

y =  =  = 

Denominators cannot be 0, so Delta =  = ae - bd ≠ 0.
and both  and  have unique
values when that denominator is not zero.

------------------------------------------------------

In case you need to explain why Cramer's rule works:
 
To eliminate y, multiply the first equation by e and the
second equation by -b, and add the equations vertically
term by term:

     aex + bey =  ce
    -bdx - bey = -bf
-----------------------
 aex-bdx       =  ce-bf  
(ae-bd)x       =  ce-bf
       x = 

To eliminate x, multiply the first equation by -d and the
second equation by a, and add the equations vertically
term by term:

  -adx   - bdy = -cd
   adx   + aey =  af
-----------------------
       aey-bdy =  af-cd  
      (ae-bd)y =  af-cd
             y = 

a and y are the same using the elimination method as they are 
using Cramer's rule.
  
Edwin

You can put this solution on YOUR website!


You have worded the question incorrectly because if or then the given system is either consistent/dependent (same line) or inconsistent (parallel lines).

Calculate the determinant:



If the determinant is zero, then the system is either consistent/dependent (same line) or inconsistent (parallel lines).






or



So or yields no unique solution.

In fact, the set of values of for which your system yields a unique solution is all real numbers EXCEPT 0 and 2. Note, however, that the unique solution is, in every case, the origin

John

My calculator said it, I believe it, that settles it