If the system of equations
$2x + 7y + \lambda z = 3$
$3x + 2y + 5z = 4$
$x + \mu y + 32z = -1$
has infinitely many solutions,then $(\lambda - \mu)$ is equal to

  • A
    $38$
  • B
    $39$
  • C
    $34$
  • D
    $15$

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The sum of three numbers is $6$. Thrice the third number when added to the first number gives $7$. On adding three times the first number to the sum of the second and third numbers,we get $12$. The product of these numbers is:

If $x=\alpha, y=\beta, z=\gamma$ is the solution of the system of equations $2x+3y+z=-1$, $3x+y+z=4$, and $x-3y-2z=1$, then the value of $\beta$ is:

Let $p, q, r$ be nonzero real numbers that are,respectively,the $10^{\text{th}}$,$100^{\text{th}}$,and $1000^{\text{th}}$ terms of a harmonic progression. Consider the system of linear equations:
$x+y+z=1$
$10x+100y+1000z=0$
$qrx + pry + pqz = 0$
$List-I$ $List-II$
$(I)$ If $\frac{q}{r}=10$,then the system of linear equations has $(P)$ $x=0, y=\frac{10}{9}, z=-\frac{1}{9}$ as a solution
$(II)$ If $\frac{p}{r} \neq 100$,then the system of linear equations has $(Q)$ $x=\frac{10}{9}, y=-\frac{1}{9}, z=0$ as a solution
$(III)$ If $\frac{p}{q} \neq 10$,then the system of linear equations has $(R)$ infinitely many solutions
$(IV)$ If $\frac{p}{q}=10$,then the system of linear equations has $(S)$ no solution
$(T)$ at least one solution

The correct option is:

Let $[\lambda]$ be the greatest integer less than or equal to $\lambda$. The set of all values of $\lambda$ for which the system of linear equations $x+y+z=4$,$3x+2y+5z=3$,$9x+4y+(28+[\lambda])z=[\lambda]$ has a solution is:

For a system of simultaneous linear equations, if $A X=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$, $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1\end{array}\right]$ and $\operatorname{det} A>0$, then $X=$

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