If $x + y - z = 0,\,3x - \alpha y - 3z = 0,\,\,x - 3y + z = 0$ has non zero solution, then $\alpha = $
If $x + y - z = 0,\,3x - \alpha y - 3z = 0,\,\,x - 3y + z = 0$ has non zero solution, then $\alpha = $
- A
$-1$
- B
$0$
- C
$1$
- D
$-3$
Similar Questions
If the following system of linear equations
$2 x+y+z=5$
$x-y+z=3$
$x+y+a z=b$
has no solution, then :
If the following system of linear equations
$2 x+y+z=5$
$x-y+z=3$
$x+y+a z=b$
has no solution, then :
- [JEE MAIN 2021]
Consider system of equations in $x$ , $y$ and $z$
$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).
If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is
Consider system of equations in $x$ , $y$ and $z$
$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).
If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is
Evaluate the determinant $\Delta=\left|\begin{array}{rrr}1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0\end{array}\right|$
Evaluate the determinant $\Delta=\left|\begin{array}{rrr}1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0\end{array}\right|$
In a square matrix $A$ of order $3, a_{i i}'s$ are the sum of the roots of the equation $x^2 - (a + b)x + ab= 0$; $a_{i , i + 1}'s$ are the product of the roots, $a_{i , i - 1}'s$ are all unity and the rest of the elements are all zero. The value of the det. $(A)$ is equal to
The value of the determinant$\left| {\,\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}\,} \right|$is equal to
The value of the determinant$\left| {\,\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}\,} \right|$is equal to