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zhenek [66]
3 years ago
11

PLEASE HELP MEEEE HURRRY!!! :)

Mathematics
1 answer:
sammy [17]3 years ago
6 0

Answer:

Option D

Step-by-step explanation:

We are given the following equations -

\begin{bmatrix}-5x-12y-43z=-136\\ -4x-14y-52z=-146\\ 21x+72y+267z=756\end{bmatrix}

It would be best to solve this equation in matrix form. Write down the coefficients of each terms, and reduce to " row echelon form " -

\begin{bmatrix}-5&-12&-43&-136\\ -4&-14&-52&-146\\ 21&72&267&756\end{bmatrix}  First, I swapped the first and third rows.

\begin{bmatrix}21&72&267&756\\ -4&-14&-52&-146\\ -5&-12&-43&-136\end{bmatrix}  Leading coefficient of row 2 canceled.  

\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ -5&-12&-43&-136\end{bmatrix}  The start value of row 3 was canceled.

\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ 0&\frac{36}{7}&\frac{144}{7}&44\end{bmatrix}       Matrix rows 2 and 3 were swapped.

\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\end{bmatrix}      Leading coefficient in row 3 was canceled.

\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&0&0&\frac{4}{9}\end{bmatrix}

And at this point, I came to the conclusion that this system of equations had no solutions, considering it reduced to this -

\begin{bmatrix}1&0&-1&0\\ 0&1&4&0\\ 0&0&0&1\end{bmatrix}

The positioning of the zeros indicated that there was no solution!

<u><em>Hope that helps!</em></u>

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