How do you solve #x + 2y + z = 6#, #2x - y - z = 0#, and #3x + 2y +z = 10# using matrices?

Answer 1

#x=2#, #y=0# and #z=4#

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,2,1,|,6),(2,-1,-1,|,0),(3,2,1,|,10))#
I have written the equations not in the sequence as in the question in order to get #1# as pivot.

Perform the folowing operations on the rows of the matrix

#R2larrR2-2R1# ; #R3larrR3-3R1#
#A=((1,2,1,|,6),(0,-5,-3,|,-12),(0,-4,-2,|,-8))#
#R1larrR1-(R3)/2# ; #R2larrR2-R3#
#A=((1,0,0,|,2),(0,-1,-1,|,-4),(0,-4,-2,|,-8))#
#R3larrR3-4R2#
#A=((1,0,0,|,2),(0,-1,-1,|,-4),(0,0,2,|,8))#
#R2larr(R2)/(-1)# ; #R3larr(R3)/2#
#A=((1,0,0,|,2),(0,1,1,|,4),(0,0,1,|,4))#
#R2larrR2-R3#
#A=((1,0,0,|,2),(0,1,0,|,0),(0,0,1,|,4))#
Thus #x=2#, #y=0# and #z=4#
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Answer 2

#x=2#
#y=0#
#z=4#

Given -

#x+2y+z=6#
#2x-y-z=0#
#3x+2y+z=10#
Answer is developed from the template I created in Excel

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Answer 3

You can solve this system of equations using matrices by representing the coefficients and constants in matrix form and then applying matrix operations to find the values of (x), (y), and (z). First, arrange the coefficients and constants into matrices:

[A = \begin{bmatrix} 1 & 2 & 1 \ 2 & -1 & -1 \ 3 & 2 & 1 \end{bmatrix}]

[X = \begin{bmatrix} x \ y \ z \end{bmatrix}]

[B = \begin{bmatrix} 6 \ 0 \ 10 \end{bmatrix}]

Then, use the matrix equation (AX = B), where (A) is the coefficient matrix, (X) is the column matrix containing the variables, and (B) is the constant matrix. To solve for (X), you can use the formula (X = A^{-1}B), where (A^{-1}) is the inverse of matrix (A).

After finding the inverse of matrix (A), denoted as (A^{-1}), multiply it by matrix (B) to find (X):

[X = A^{-1}B]

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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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