Evaluate the determinants
$\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$
Evaluate the determinants
$\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$
Let $A=\left[\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right]$
By expanding along the first row, we have:
$|A|=0\left|\begin{array}{ll}0 & -3 \\ 3 & 0\end{array}\right|-1\left|\begin{array}{ll}-1 & -3 \\ -2 & 0\end{array}\right|+2\left|\begin{array}{ll}-1 & 0 \\ -2 & 3\end{array}\right|$
$=0-1(0-6)+2(-3-0)$
$=-1(-6)+2(-3)$
$=6-6=0$
Similar Questions
For what value of $k$ to the following system of equations possess a non-trivial solution ?
$x + ky + 3z = 0$ ; $3x + ky + 2z = 0$ ; $2x + 3y + 4z = 0$
For what value of $k$ to the following system of equations possess a non-trivial solution ?
$x + ky + 3z = 0$ ; $3x + ky + 2z = 0$ ; $2x + 3y + 4z = 0$
Evaluate the determinants
$\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|$
Evaluate the determinants
$\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|$
In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$ has the value
In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$ has the value
If the system of equations
$ x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 $
$ x+(\cos \alpha) y+(\sin \alpha) z=0 $
$ x+(\sin \alpha) y-(\cos \alpha) z=0$
has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :
If the system of equations
$ x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 $
$ x+(\cos \alpha) y+(\sin \alpha) z=0 $
$ x+(\sin \alpha) y-(\cos \alpha) z=0$
has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :
- [JEE MAIN 2024]
If $A = \left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]$ and $\det ({A^n} - I) = 1 - {\lambda ^n}\,,\,n \in N$ then $\lambda $ is
If $A = \left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]$ and $\det ({A^n} - I) = 1 - {\lambda ^n}\,,\,n \in N$ then $\lambda $ is