How do you solve using gaussian elimination or gauss-jordan elimination, #y + 3z = 6#, #x + 2y + 4z = 9#, #2x + y + 6z = 11#?

Answer 1

Using Gaussian elimination
#color(white)("XXX")(x,y,z)= (1/7,9/7,11/7)#

Converting the given equations to augmented matrix (to avoid having to re-write the variable names) #{(0,1,3,6),(1,2,4,9),(2,1,6,11):}color(white)("XX"){:([1]),([2]),([3]):}#
Exchanging rows [1] and [2] #{(1,2,4,9),(0,1,3,6),(2,1,6,11):}color(white)("XX"){:([4]),([5]),([3]):}#

Note that at this point rows [4] and [5] are in Gaussian form and all that remains is to zero out the first two columns of row [3]

Subtracting #2# times row [4] from row [3] #{(1,2,4,9),(0,1,3,6),(0,-3,-2,-7):}color(white)("XX"){:([4]),([5]),([6]):}#
Adding #3# times row [5] to row [6] #{(1,2,4,9),(0,1,3,6),(0,0,7,11):}color(white)("XX"){:([4]),([5]),([7]):}#
Dividing row [7] by #7# #{(1,2,4,9),(0,1,3,6),(0,0,1,11/7):}color(white)("XX"){:([4]),([5]),([8]):}#
This gives us #color(white)("XXX")z=11/7#
Substituting back into row [5] #color(white)("XXX")y+3xx11/7=6 rarr y= 9/7#
Substituting back into row [4] #color(white)("XXX")x+2xx9/7+4xx11/7=9 rarr x=1/7#
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Answer 2

To solve the system of linear equations using Gaussian elimination or Gauss-Jordan elimination, we represent the equations in augmented matrix form and then perform row operations to transform the matrix into row-echelon form or reduced row-echelon form.

The given system of equations is:

[ \begin{align*} y + 3z &= 6 \ x + 2y + 4z &= 9 \ 2x + y + 6z &= 11 \ \end{align*} ]

Writing this system in augmented matrix form:

[ \left[ \begin{array}{ccc|c} 0 & 1 & 3 & 6 \ 1 & 2 & 4 & 9 \ 2 & 1 & 6 & 11 \ \end{array} \right] ]

Now, let's perform row operations to transform this matrix into row-echelon form:

  1. Multiply the first row by -1 and add it to the second row.
  2. Multiply the first row by -2 and add it to the third row.

This gives us:

[ \left[ \begin{array}{ccc|c} 0 & 1 & 3 & 6 \ 1 & 2 & 4 & 9 \ 0 & -1 & 0 & -1 \ \end{array} \right] ]

Now, continue with row operations to simplify the matrix:

  1. Multiply the second row by -1 and add it to the first row.

This gives:

[ \left[ \begin{array}{ccc|c} 1 & 0 & -1 & -3 \ 1 & 2 & 4 & 9 \ 0 & -1 & 0 & -1 \ \end{array} \right] ]

  1. Multiply the second row by 1 and add it to the third row.

This gives:

[ \left[ \begin{array}{ccc|c} 1 & 0 & -1 & -3 \ 0 & 1 & 4 & 6 \ 0 & -1 & 0 & -1 \ \end{array} \right] ]

Now, perform additional row operations to transform the matrix into reduced row-echelon form if desired. However, this matrix already provides enough information to find the solution.

From the matrix, we can see that:

[ \begin{align*} x &= -3 \ y &= 6 \ z &= 1 \ \end{align*} ]

So, the solution to the system of equations is ( x = -3 ), ( y = 6 ), and ( z = 1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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