For the system of linear equations $a x+y+z=1$,$x+a y+z=1$,$x+y+a z=\beta$,which one of the following statements is $NOT$ correct?

  • A
    It has infinitely many solutions if $a=2$ and $\beta=-1$
  • B
    It has no solution if $a=-2$ and $\beta=1$
  • C
    $x+y+z=\frac{3}{4}$ if $a=2$ and $\beta=1$
  • D
    It has infinitely many solutions if $a=1$ and $\beta=1$

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The bookshop of a particular school has $10$ dozen chemistry books,$8$ dozen physics books,and $10$ dozen economics books. Their selling prices are Rs. $80$,Rs. $60$,and Rs. $40$ each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Consider the following system of equations: $\alpha x + 2y + z = 1$; $2\alpha x + 3y + z = 1$; $3x + \alpha y + 2z = \beta$. For some $\alpha, \beta \in \mathbb{R}$. Which of the following is $NOT$ correct?

Let $S$ be the set of all column matrices $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$ such that $b_1, b_2, b_3 \in \mathbb{R}$ and the system of equations (in real variables)
$-x+2y+5z=b_1$
$2x-4y+3z=b_2$
$x-2y+2z=b_3$
has at least one solution. Then,which of the following system$(s)$ (in real variables) has (have) at least one solution for each $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right] \in S$?
$(A)$ $x+2y+3z=b_1, 4y+5z=b_2$ and $x+2y+6z=b_3$
$(B)$ $x+y+3z=b_1, 5x+2y+6z=b_2$ and $-2x-y-3z=b_3$
$(C)$ $-x+2y-5z=b_1, 2x-4y+10z=b_2$ and $x-2y+5z=b_3$
$(D)$ $x+2y+5z=b_1, 2x+3z=b_2$ and $x+4y-5z=b_3$

If the solution of the system of simultaneous linear equations $x+y-z=6$,$3x+2y-z=5$ and $2x-y-2z+3=0$ is $x=\alpha, y=\beta, z=\gamma$,then $\alpha+\beta=$

The system of linear equations $\lambda x + y + z = 3$, $x - y - 2z = 6$, and $-x + y + z = \mu$ has:

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