Class 12Mathematics3-4. Matrices and DeterminantsExpansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line
EasyEvaluate the determinant $\Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$.
- A$-50$
- B$-51$
- C$-52$
- D$-53$
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Similar Questions
The value of the determinant $\left| \begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array} \right|$ is equal to
Difficult
View SolutionThe value of the determinant $\left| \begin{array}{ccc} 0 & b^3 - a^3 & c^3 - a^3 \\ a^3 - b^3 & 0 & c^3 - b^3 \\ a^3 - c^3 & b^3 - c^3 & 0 \end{array} \right|$ is equal to:
Easy
View SolutionIf $a, b, c$ are real numbers such that $a-b=1$ and $b-c=3$, then the number of matrices of the form $A=\begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}$ such that $|A|=-12$ is:
If $a, b, c$ are respectively the $5^{\text{th}}, 8^{\text{th}}, 13^{\text{th}}$ terms of an arithmetic progression,then $\left|\begin{array}{ccc}a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1\end{array}\right|=$
The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is
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