Answer:
Option D
Step-by-step explanation:
We are given the following equations -
[tex]\begin{bmatrix}-5x-12y-43z=-136\\ -4x-14y-52z=-146\\ 21x+72y+267z=756\end{bmatrix}[/tex]
It would be best to solve this equation in matrix form. Write down the coefficients of each terms, and reduce to " row echelon form " -
[tex]\begin{bmatrix}-5&-12&-43&-136\\ -4&-14&-52&-146\\ 21&72&267&756\end{bmatrix}[/tex] First, I swapped the first and third rows.
[tex]\begin{bmatrix}21&72&267&756\\ -4&-14&-52&-146\\ -5&-12&-43&-136\end{bmatrix}[/tex] Leading coefficient of row 2 canceled.
[tex]\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ -5&-12&-43&-136\end{bmatrix}[/tex] The start value of row 3 was canceled.
[tex]\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ 0&\frac{36}{7}&\frac{144}{7}&44\end{bmatrix}[/tex] Matrix rows 2 and 3 were swapped.
[tex]\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\end{bmatrix}[/tex] Leading coefficient in row 3 was canceled.
[tex]\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&0&0&\frac{4}{9}\end{bmatrix}[/tex]
And at this point, I came to the conclusion that this system of equations had no solutions, considering it reduced to this -
[tex]\begin{bmatrix}1&0&-1&0\\ 0&1&4&0\\ 0&0&0&1\end{bmatrix}[/tex]
The positioning of the zeros indicated that there was no solution!
Hope that helps!
