$\left| {\,\begin{array}{*{20}{c}}1&{1 + ac}&{1 + bc}\\1&{1 + ad}&{1 + bd}\\1&{1 + ae}&{1 + be}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}1&{1 + ac}&{1 + bc}\\1&{1 + ad}&{1 + bd}\\1&{1 + ae}&{1 + be}\end{array}\,} \right| = $
- A
$1$
- B
$0$
- C
$3$
- D
$a + b + c$
Similar Questions
Let $D_1 =$ $\left| {\,\begin{array}{*{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a - b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{*{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is
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Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If
$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$
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Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If
$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$
then value of $\lambda^{2}$ is equal to $.....$
- [JEE MAIN 2021]
The parameter on which the value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\{\cos (p - d)x}&{\cos px}&{\cos (p + d)x}\\{\sin (p - d)x}&{\sin px}&{\sin (p + d)x}\end{array}\,} \right|$ does not depend upon
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- [IIT 1997]
If $p, q, r, s$ are in $A.P.$ and $f (x) =$ $\left| {\,\begin{array}{*{20}{c}}
{p\,\, + \,\,\sin \,x}&{q\,\, + \,\,\sin \,x}&{p\,\, - \,\,r\,\, + \,\,\sin \,x}\\
{q\,\, + \,\,\sin \,x}&{r\,\, + \,\,\sin \,x}&{ - \,1\,\, + \,\,\sin \,x}\\
{r\,\, + \,\,\sin \,x}&{s\,\, + \,\,\sin \,x}&{s\,\, - \,\,q\,\, + \,\,\sin \,x}
\end{array}\,} \right|$ such that $f (x)dx = - 4$ then the common difference of the $A.P.$ can be :
If $a,b,c$ are positive integers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a^2} + x}&{ab}&{ac}\\{ab}&{{b^2} + x}&{bc}\\{ac}&{bc}&{{c^2} + x}\end{array}\,} \right|$ is divisible by
If $a,b,c$ are positive integers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a^2} + x}&{ab}&{ac}\\{ab}&{{b^2} + x}&{bc}\\{ac}&{bc}&{{c^2} + x}\end{array}\,} \right|$ is divisible by