Use Gauss-Jordan elimination to find the solution to the following system of equations. —x+ y+ z= —2 —x +4y— 14z= — 17 5x-3y-15z= 0 Select the correct choice below and fill in any answer boxes within...

Solution 1:
X+5Y=23
6X+2Y=-2
The matrix is represented as:
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To solve expression we use format:
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Hence the correct choice is:
Part A: solution is x=-2 and y=5
Solution 2:
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Hence inverse of matrix is :
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Solution 3:
Equations are written in matrix form:
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Augmented matrix is:
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R3R3+5*R1
R2R2-R1
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[



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R2R2/3
R3R3/2
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R3R3-R2
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[



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R1R1-R2
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Since the elements of last row are zero hence infinitely many solution in terms of Z
Solution 4:
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Multiplying the matrices on left hand of equality:
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Hence the equations are:
Solution 5:
Matrix A is 1X4
That means 1 row and 4 columns
Matrix A is 4X3
That means 4 rows and 3 columns
Since number of columns of matrix A is equal to the number of rows of matrix B hence they can be
multiplied.
When these are multiplied:
Then the new matrix formed is having:
1 row and 3 columns
Hence dimension of new matrix is: 1X3
Solution 6:
The given matrix is:
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Hence the number of rows is 1
Number of columns is 2
Hence the size of the matrix is:
1X2
The matrix is a row matrix as it has only one row
Solution 7:
The given matrix is
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Pivoting around 2nd row 3rd column:
Diving row 2 by 2
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Replacing R3 by R3-2*R2
Replacing R1 by R1-4*R2
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R2R2+R3/2
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R1R1-R3
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