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
[tex]\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right][/tex] [tex]\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex] [tex]= \left[\begin{array}{ccc}26\\17\\23\end{array}\right][/tex].
Step 3:
The linear system B is represented as
[tex]\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right][/tex] [tex]\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex] [tex]= \left[\begin{array}{ccc}23\\17\\26\end{array}\right][/tex].
Step 4:
The linear system C is represented as
[tex]\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right][/tex] [tex]\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex] [tex]= \left[\begin{array}{ccc}4\\11\\12\end{array}\right][/tex].
Step 5:
The linear system D is represented as
[tex]\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right][/tex] [tex]\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex] [tex]= \left[\begin{array}{ccc}12\\11\\4\end{array}\right][/tex].
Step 6:
Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.
