Consider the system of linear equations in x,y,z : gathered x+2y+tz=0 6x+y+5tz=0 3x+t^2y+f…

Question

Consider the system of linear equations in $x,y,z$ :

$\begin{gathered} x+2y+tz=0 \ 6x+y+5tz=0 \ 3x+{t}^{2}y+f(t)z=0 \end{gathered}$

where $f:R	o R$ is a differentiable function. If this system has infinitely many solutions for all $t\in R$, then $f$

Accepted Answer

For a homogeneous system of three linear equations in $x,y,z$ to have infinitely many solutions, the coefficient matrix must be singular for every $t\in R$.

So, we form the coefficient matrix:

$A=((\begin{vmatrix} 1 \quad  2 \quad  t \ 6 \quad  1 \quad  5t \ 3 \quad  {t}^{2} \quad  f(t) \end{gathered}))|$

Since the system has infinitely many solutions for all $t$, we must have

$det(A)=0\, t\in R.$

Now compute the determinant.

$det(A)=|\begin{gathered} 1 \quad  2 \quad  t \ 6 \quad  1 \quad  5t \ 3 \quad  {t}^{2} \quad  f(t) \end{vmatrix}$

Expanding along the first row,

$det(A)=1|\begin{array}{cc} 1 & 5t \ {t}^{2} & f(t) \end{array}|-2|\begin{array}{cc} 6 & 5t \ 3 & f(t) \end{array}|+t|\begin{array}{cc} 6 & 1 \ 3 & {t}^{2} \end{array}|$

Now evaluate each minor:

$|\begin{array}{cc} 1 & 5t \ {t}^{2} & f(t) \end{array}|=f(t)-5{t}^{3}$

$|\begin{array}{cc} 6 & 5t \ 3 & f(t) \end{array}|=6f(t)-15t$

$|\begin{array}{cc} 6 & 1 \ 3 & {t}^{2} \end{array}|=6{t}^{2}-3$

So,

$det(A)=(f(t)-5{t}^{3})-2(6f(t)-15t)+t(6{t}^{2}-3)$

$=f(t)-5{t}^{3}-12f(t)+30t+6{t}^{3}-3t$

$=-11f(t)+{t}^{3}+27t$

Since this is zero for all $t$,

$-11f(t)+{t}^{3}+27t=0$

Hence,

$f(t)=\frac{{t}^{3}+27t}{11}$

Now differentiate:

${f}^{'}(t)=\frac{3{t}^{2}+27}{11}=\frac{3({t}^{2}+9)}{11}$

Since ${t}^{2}+9>0$ for all $t\in R$, we get

${f}^{'}(t)>0\, t\in R.$

Therefore, $f$ is strictly increasing on $R$.

Hence, the correct option is:

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