Solve the linear system represented by the Augmentad matrix. $$ \left[\begin{array}{ccccc}1 & -7 & -5 & 8 & 10 \ 0 & 1 & -3 & 3 & -5 \ 0 & 0 & 0 & 1 & 4\end{array} ight] $$

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Solve the linear system represented by the Augmentad matrix. $$ \left[\begin{array}{ccccc}1 & -7 & -5 & 8 & 10 \ 0 & 1 & -3 & 3 & -5 \ 0 & 0 & 0 & 1 & 4\end{array}ight] $$

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We are given the augmented matrix

[ 1  -7  -5   8 | 10 ]
  [ 0   1  -3   3 | -5 ]
  [ 0   0   0   1 | 4  ]

This corresponds to the following system of linear equations, where x₁, x₂, x₃, and x₄ are unknowns:

1) x₁ − 7x₂ − 5x₃ + 8x₄ = 10  
 2)   x₂ − 3x₃ + 3x₄ = -5  
 3)           x₄ = 4

Step 1. From Equation 3, we immediately have:
  x₄ = 4

Step 2. Substitute x₄ = 4 into Equation 2:
  x₂ − 3x₃ + 3(4) = -5  
  x₂ − 3x₃ + 12 = -5  
  x₂ − 3x₃ = -17  
Thus, we can express x₂ in terms of a free parameter (let x₃ = t where t is any real number):
  x₂ = -17 + 3t      (Equation A)

Step 3. Substitute x₄ = 4 and x₂ = -17 + 3t into Equation 1:
  x₁ − 7(-17 + 3t) − 5t + 8(4) = 10  
  x₁ + 119 − 21t − 5t + 32 = 10  
  x₁ + 151 − 26t = 10  
Solve for x₁:
  x₁ = 10 − 151 + 26t  
  x₁ = 26t − 141

Step 4. Write the complete solution using the free parameter t ∈ ℝ:
  x₁ = 26t − 141  
  x₂ = 3t − 17  
  x₃ = t      (t is a free parameter)  
  x₄ = 4

Thus, the solution set of the given linear system is:

{ (26t − 141, 3t − 17, t, 4) : t ∈ ℝ }
The solution to the system is: - $$ x₁ = 26t - 141 $$ - $$ x₂ = 3t - 17 $$ - $$ x₃ = t $$ - $$ x₄ = 4 $$ where $$ t $$ is any real number.

Solve the linear system represented by the Augmentad matrix. \( \left[\begin{array}{ccccc}1 & -7 & -5 & 8 & 10 \\ 0 & 1 & -3 & 3 & -5 \\ 0 & 0 & 0 & 1 & 4\end{array}\right] \)

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