(12x + y + z = 26

Mathematics

1 answer:

mel-nik [20]2 years ago

4 0

Option D. D has the matrix of constants [[12], [11], [4]].

Step-by-step explanation:

Step 1:

With the given equations, we can form matrices to represent them.

The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.

Step 2:

The linear system A is represented as

$\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]$ .

Step 3:

The linear system B is represented as

$\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]$ .

Step 4:

The linear system C is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]$ .

Step 5:

The linear system D is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]$ .