Subjects linear algebra

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1. The problem is to determine the solution set of the linear system represented by the augmented matrix: $$\begin{bmatrix} 1 & 1 & 1 & | & 5 \\ 1 & -1 & 5 & | & 3 \\ -2 & 2 & -10 & | & 0 \end{bmatrix}$$ 2. Write the system as equations: $$\begin{cases} x + y + z = 5 \\ x - y + 5z = 3 \\ -2x + 2y - 10z = 0 \end{cases}$$ 3. Add the first two equations: From eq1: $x + y + z = 5$ From eq2: $x - y + 5z = 3$ Subtract eq2 from eq1: $$ (x + y + z) - (x - y + 5z) = 5 - 3 $$ $$ x + y + z - x + y - 5z = 2 $$ $$ 2y - 4z = 2 $$ Simplify by dividing both sides by 2: $$ y - 2z = 1 \implies y = 1 + 2z $$ 4. Use the third equation: $$ -2x + 2y - 10z = 0 $$ Substitute $y = 1 + 2z$: $$ -2x + 2(1 + 2z) - 10z = 0 $$ $$ -2x + 2 + 4z - 10z = 0 $$ $$ -2x + 2 - 6z = 0 $$ Rearrange: $$ -2x = -2 + 6z $$ $$ x = 1 - 3z $$ 5. Use equation 1 to check consistency: $$ x + y + z = 5 $$ Substitute $x = 1 - 3z$ and $y = 1 + 2z$: $$ (1 - 3z) + (1 + 2z) + z = 5 $$ $$ 1 - 3z + 1 + 2z + z = 5 $$ $$ 2 + 0 = 5 $$ $$ 2 = 5 $$ This is a contradiction; hence the system is inconsistent. 6. Check if a mistake was made in step 3 during subtraction; re-examine step 3: $$ (x + y + z) = 5 $$ $$ (x - y + 5z) = 3 $$ Subtract eq2 from eq1: $$ (x + y + z) - (x - y + 5z) = 5 - 3 $$ $$ x + y + z - x + y - 5z = 2 $$ $$ 2y - 4z = 2 $$ Divide by 2: $$ y - 2z = 1 $$ consistent Double check substituting $y$ and $x$ back in 3rd equation: $$ -2x + 2y - 10z = 0 $$ Substitute $y = 1 + 2z$: $$ -2x + 2(1 + 2z) - 10z = 0 $$ $$ -2x + 2 + 4z - 10z = 0 $$ $$ -2x + 2 - 6z = 0 $$ $$ -2x = -2 + 6z $$ $$ x = 1 - 3z $$ Substitute $x$ and $y$ in first equation: $$ x + y + z = 5 $$ $$ (1 - 3z) + (1 + 2z) + z = 5 $$ $$ 1 - 3z + 1 + 2z + z = 5 $$ $$ 2 = 5 $$ Again a contradiction appears. 7. This contradiction means the system has no solution. \textbf{Answer:}\ The system has \textbf{no solutions}.