The values of m, n,for which the system of eq | vedclass

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(D) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the determinants

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$m^2+n^2-mn=39$

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(D) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the determinants $D_1, D_2, D_3$ must also be $0$.First,calculate $D$:$D = \begin{vmatrix} 1 & 1 & 1 \ 2 & 5 & 5 \ 1 & 2 & m \end{vmatrix} = 1(5m - 10) - 1(2m - 5) + 1(4 - 5) = 5m - 10 - 2m + 5 - 1 = 3m - 6$.Setting $D = 0$,we get $3m - 6 = 0 \Rightarrow m = 2$.Next,calculate $D_3$ with $m=2$:$D_3 = \begin{vmatrix} 1 & 1 & 4 \ 2 & 5 & 17 \ 1 & 2 & n \end{vmatrix} = 1(5n - 34) - 1(2n - 17) + 4(4 - 5) = 5n - 34 - 2n + 17 - 4 = 3n - 21$.Setting $D_3 = 0$,we get $3n - 21 = 0 \Rightarrow n = 7$.Now,check the options with $m=2$ and $n=7$:$m^2 + n^2 - mn = 2^2 + 7^2 - (2)(7) = 4 + 49 - 14 = 39$.Thus,the values satisfy $m^2 + n^2 - mn = 39$.

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 solutions,satisfy the equation :

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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 solutions,satisfy the equation :