The values of \(m, n\), for which the system of equations \(x+y+z=4,\\ 2x+5y+5z=17,\\ x+2y+mz=n\) has infinitely many so…

Explanation

For infinitely many solutions, Cramer's rule requires

\(D=0,\quad D_1=0,\quad D_2=0,\quad D_3=0\).

The coefficient determinant is

\(D=\begin{vmatrix}
1&1&1\\
2&5&5\\
1&2&m
\end{vmatrix}\).

Expanding along the first row,

\(D=\begin{vmatrix}
5&5\\
2&m
\end{vmatrix}
-\begin{vmatrix}
2&5\\
1&m
\end{vmatrix}
+\begin{vmatrix}
2&5\\
1&2
\end{vmatrix}\)

\(=(5m-10)-(2m-5)+(4-5)\)

\(=3m-6\).

Since \(D=0\),

\(3m-6=0\),

\(m=2\).

Now

\(D_1=\begin{vmatrix}
4&1&1\\
17&5&5\\
n&2&2
\end{vmatrix}\).

Since the second and third columns are identical,

\(D_1=0\).

Next,

\(D_2=\begin{vmatrix}
1&4&1\\
2&17&5\\
1&n&2
\end{vmatrix}\).

Expanding along the first row,

\(D_2=
\begin{vmatrix}
17&5\\
n&2
\end{vmatrix}
-4\begin{vmatrix}
2&5\\
1&2
\end{vmatrix}
+\begin{vmatrix}
2&17\\
1&n
\end{vmatrix}\)

\(=(34-5n)-4(4-5)+(2n-17)\)

\(=21-3n\).

Since \(D_2=0\),

\(21-3n=0\),

\(n=7\).

Also,

\(D_3=\begin{vmatrix}
1&1&4\\
2&5&17\\
1&2&7
\end{vmatrix}\)

\(=\begin{vmatrix}
1&1&4\\
2&5&17\\
1&2&7
\end{vmatrix}\)

\(=0\)

after substituting \(n=7\), confirming consistency.

Thus

\(m=2,\quad n=7\).

Now,

\(m^2+n^2-mn
=2^2+7^2-(2)(7)\)

\(=4+49-14\)

\(=39\).