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# how to solve thi matrics by using gauss elimination[(x+y-z=0),(2x-y+z=6),(3x+2y-4z=-4)]

my background is not mathematics

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### Linear Algebra Answers - Scribd

This is Cramer’s Rule for the system x + 2y = 6, 3x + y = 8. ... x + y − z = 10 2x − 2y + z = 0 −→ x ... (x, y, z) = (5, 5, 0). (f ) Here Gauss’ method ...

### Linear Algebra - Solutions - Answers to Exercises Linear ...

Linear Algebra - Solutions - Answers to Exercises Linear... SCHOOL Indian Institute of Technology, Chennai; COURSE TITLE MATHEMATIC MA2030; TYPE. Homework Help ...

### Jim Hefferon - Linear Algebra - Answers to Questions - Documents

Share Jim Hefferon - Linear Algebra - Answers to ... This is Cramer’s Rule for the system x + 2y = 6, 3x + y ... Gauss’ method ρ1 ↔ρ4 x + y − z = 10 2x ...

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### how to solve thi matrics by using gauss elimination[(x+y-z=0),(2x-y+z=6),(3x+2y-4z=-4)]

Matrix representation of problem: (  1 1 -1 | 0 ) (  2 -1 1 | 6 ) (  3 2 -4 | -4 ) In the following solution, only the rows that are changed are shown with each operation: R1+R2: ( 3 0 0 | 6 ); R1/3: ( 1 0 0 | 2 ); R2-2R1: ( 0 -1 1 | 2 ); R3-3R1: ( 0 2 -4 | -10 ); R3/2: ( 0 1 -2 | -5 ); R3+R2: ( 0 0 -1 | -3 ); -R3: ( 0 0 1 | 3 ): R2-R3: ( 0 -1 0 | -1 ): -R2: ( 0 1 0 | 1 ). Solution:  x=2; y=1; z=3. Check solution: ( 2 1 -3 | 0 ) ( 4 -1 3 | 6 ) ( 6 2 -12 | -4 ) Rows sum correctly.

### solve these equations using matrices without a calculator. x=y=z=6, 2x=y-4z=-15, 5x-3y+z=-10

x-y-z=6 2x-y-4z=-15 5x-3y+z=-10 Your equations, as written, had typos in them. I've assumed that some of the equal-signs should have been minus-signs and altered then appropriately. In matrix form the equations would be AX = b Where A is the matrix 1 -1 -1 2 -1 -4 5 -3 1 X is your unknown column vector [x y z] and b is a scalar column vector [ 6 -15 -10]. The solution is given by X = A^(-1)b, where A^(-1) is the inverse matrix of A. We now do Gauss-Jordan elimination on A to get its inverse. We write this out as, 1 -1 -1     |  1   0   0  --- [Row 1] 2 -1 -4     |  0   1   0  --- [Row 2] 5 -3 1      |  0   0   1  --- [Row 3] R2 - 2*R1, R3 - 5*R1,   1 -1 -1     |  1   0   0  --- [Row 1] 0  1 -2     |  -2  1   0  --- [Row 2] 0  2  6     |  -5  0   1  --- [Row 3] R1 + R2, R3 - 2*R2   1  0 -3     |  -1   1   0  --- [Row 1] 0  1 -2     |  -2   1   0  --- [Row 2] 0  0 10    |  -1  -2   1  --- [Row 3] R3/10,   1  0 -3     |  -1     1     0    --- [Row 1] 0  1 -2     |  -2     1     0    --- [Row 2] 0  0  1     | -0.1 -0.2  0.1  --- [Row 3] R1 + 3*R3, R2 + 2*R3,   1  0  0     |  -1.3   0.4   0.3  --- [Row 1] 0  1  0     |  -2.2   0.6   0.2  --- [Row 2] 0  0  1     |  -0.1  -0.2   0.1  --- [Row 3] Now that we have an identity marix on the lhs, then the rhs is the inverse matrix, so A^(-1) =    |  -1.3   0.4   0.3  |                  |  -2.2   0.6   0.2  |                  |  -0.1  -0.2   0.1  | And X = A^(-1)b        X  =    |  -1.3   0.4   0.3 | |   6  | = |-1.3*6 +0.4*(-15) + 0.3*(-10)  |                  |  -2.2   0.6   0.2  | | -15 |    |-2.2*6 +0.6*(-15) + 0.2*(-10) |                  |  -0.1  -0.2   0.1  | | -10 |    |-0.1*6 -0.2*(-15) + 0.1*(-10)  |        X  =   | -7.8 - 6 - 3   |                  | -13.2 - 9 - 2 |                  | -0.6 + 3 - 1  |        X  =   | -16.8  |                  | -24.2 |                  |    1.4 | The solution is: x = -16.8, y = -24.2, z = 1.4

### What is matricx

A matrix is a rectangular or tabular representation of quantities. Like a table, it has rows and columns and each "cell" is called an element. Matrices can be combined using addition, subtraction or multiplication, provided certain rules are followed about the sizes of the matrices involved. A matrix is a shorthand form of an expression that could consist of elements added together if written out in full. The advantage of using this shorthand form is that matrices can be used to solve problems, sometimes complex, that would be more tedious and difficult to solve using a "longhand" approach. There are rules that determine how matrices can be added, subtracted or multiplied, and matrices occupy a complete branch of mathematics of their own. The given matrix is a 1 by 4 matrix (one row and 4 columns) and it could represent the longhand: w+5x+4y+z (for example), where w, x, y, z are variables and the numbers are coefficients. But it could represent a 4-dimensional vector (a difficult concept to visualise) which is involved with other similar matrices representing vectors to solve a problem using vector analysis. Matrices are used in many branches of science to solve quite difficult physical problems.

### what is the answer to 2x-y-2z=1, 4x+y-z=5, x+y+4z=-5 using matrices

what is the answer to 2x-y-2z=1, 4x+y-z=5, x+y+4z=-5 using matrices how to solve this math problem using matrices?   2  -1  -2  │  1   4   1  -1  │  5   1   1   4  │ -5 Multiply line 2 by 2.   2  -1  -2  │  1   8   2  -2  │ 10   1   1   4  │ -5 Subtract line 2 from line 1, replace line 1.  -6  -3   0  │ -9   8   2  -2  │ 10   1   1   4  │ -5 Multiply line 2 by 2.  -6  -3   0   │ -9  16   4  -4  │ 20    1   1   4  │ -5 Add line 3 to line 2, replace line 2.  -6   -3   0  │ -9  17   5   0  │ 15    1   1   4  │ -5 Multiply line 3 by 6.  -6  -3   0  │  -9  17   5   0  │  15   6   6  24  │ -30 Add line 1 to line 3, replace line 3.  -6  -3   0  │  -9  17   5   0  │  15   0   3  24  │ -39 Multiply line 1 by 5; multiply line 2 by 3. -30 -15   0  │ -45  51  15   0  │  45    0   3  24  │ -39 Add line 2 to line 1, replace line 1.  21    0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Divide line 1 by 21.   1     0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Multiply line 1 by 51.  51    0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Subtract line 1 from line 2, replace line 2.  51   0   0  │   0   0  15   0  │  45   0   3  24  │ -39 Divide line 2 by 5.  51   0   0  │   0    0   3   0  │   9   0   3  24  │ -39 Subtract line 2 from line 3, replace line 3.  51   0   0  │   0    0   3   0  │   9   0   0  24  │ -48 Three-in-one: divide line 1 by 51, divide line 2 by 3, divide line 3 by 24.   1   0   0  │   0   0   1   0  │   3   0   0   1  │  -2 This shows that x = 0, y = 3, z = -2

### x+2y-37=7, x+y+7=2, 2x-y+27=1

I assume that there is a missing z in each equation. If that is the case, the solution using matrices and determinants is x=255/119, y=67/119, z=-12/119. CHECK Plug these values into the original equations: (255+134+444)/119=833/119=7; (255+67-84)/119=238/119=2; (510-67-324)/119=119/119=1. So all equations are satisfied. The solution is correct.

### solve 6x+5y=13 and 3x-2y=-16 using matrices

solve 6x+5y=13 and 3x-2y=-16 using matrices. We can write down the equations as, 6x + 5y = 13 3x - 2y = -16 In matrix notation, the set of equations is written as, | 6   5 | | x |= | 13 | | 3  -2 | | y |   |-16 | or, AX = b The solution is given by: X = A^(-1)b, where A^(-1) is the inverse matrix of the matrix A. For a 2 x 2 matrrix, calculating the inverse is quite simple. If A is the matrix | a  b | | c  d | Then the inverse, A^(-1), is given by (1/det(A))*| d -b |, where det(A) = ad - bc  (note that the a and d have been interchanged)                  |-c  a |                                        ( and c and b have their signs changed) Since we have A = | 6   5 |, with a = 6, b = 5, c = 3, d = -2, then                                | 3  -2 | det(A) = ad - bc = 6*(-2) - 5*3 = -12 - 15 = -27 det(A)=-27 A^(-1) = (-1/27)*|-2  -5 |, and X = A^(-1)b                           |-3   6 | So, X = (-1/27)*|-2  -5 || 13 | = (-1/27)|-26 + 80 | = (-1/27)|  54 | = | -2 |                          |-3   6 ||-16 |               |-39 - 96 |               |-135 |   |  5 | So, X = | x | = | -2 |              | y |    |  5 | The solution is: x = -2, y = 5

### using gauss elimination what are the steps taken to solve 5x + 3y equal to 6 and 6x - 2y equal to 10

line 1...5x+3y=6 line 2... 6x-2y=10, or 3x-y=5, or y=3x-5 me dont no nuthun bout NE "Gaus Liminate" insted, plug line 2 intu line1...5x +3*(3x-5)=6 5x +9x -15=6 14x=6+15=21 x=21/14 x=1.5 y=3x-5...3*1.5 -5=4.5-5... x=-0.5 chek...5x+3y=6...5*1.5 +3*(-0.5)=7.5-1.5=6

3 7 8 2 1 1

### rewrite the equations as matrices, -19x=-4y+12, and 4y=-3x+100

rearrange the equations to become: -19x + 4y = 12 3x + 4y = 100 augmented matrix is: -19 4 : 12 3 4 : 100 Determinant A = -19*4 - 3*4 = -88 Determinant Ax (replace the x column with the constants) resulting matrix: 12 4 100 4 so Determinant Ax = 12*4 - 100*4 = -352 x = Determinant Ax / Determinant A x = -352 / -88 x = 4 Determinant Ay (replace y column with the constants) resulting matrix: -19 12 3 100 Determinant Ay = -19*100 - 3*12 = -1936 y = Determinant Ay/Determinant A = -1936/-88 y = 22 checking: -19 ( 4 ) = -4 (22) + 12 - 76 = -76 correct 4 (22) = -3 (4) + 100 88 = 88 correct