If the system of linear equations
$2x + y - z = 3$
$x - y - z = \alpha$
$3x + 3y + \beta z = 3$
has infinitely many solutions,then $\alpha + \beta - \alpha \beta$ is equal to .... .

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $5$

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Similar Questions

Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations
$x+2y+3z=\alpha$
$4x+5y+6z=\beta$
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is consistent. Let $|M|$ represent the determinant of the matrix
$M=\begin{bmatrix} \alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$
Let $P$ be the plane containing all those $(\alpha, \beta, \gamma)$ for which the above system of linear equations is consistent,and $D$ be the square of the distance of the point $(0,1,0)$ from the plane $P$.
$(1)$ The value of $|M|$ is
$(2)$ The value of $D$ is

The number of solutions of the equations $x + 4y - z = 0,$ $3x - 4y - z = 0,$ and $x - 3y + z = 0$ is

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Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ be a $2 \times 2$ real matrix with $\det(A) = 1$. If the equation $\det(A - \lambda I_2) = 0$ has imaginary roots (where $I_2$ is the identity matrix of order $2$), then:

If the system of equations $x + y + z = 6$,$x + 2y + 3z = 10$,and $x + 2y + \lambda z = 0$ has a unique solution,then $\lambda$ is not equal to:

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