If a,b,c and d are complex numbers, then determinant Delta = left| {,begin{array}{*{20}{c}}2&{a

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3 and 4 .Determinants and Matrices

hard

If $a,b,c$ and $d $ are complex numbers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}2&{a + b + c + d}&{ab + cd}\\{a + b + c + d}&{2(a + b)(c + d)}&{ab(c + d) + cd(a + b)}\\{ab + cd}&{ab(c + d) + cd(a + d)}&{2abcd}\end{array}} \right|$is

Dependent on $a, b, c$ and $ d$

Independent of $a,b,c$and $d$

Dependent on $a,c$and independent of $b,d$

None of these

Solution

(b) We can write the given determinant as a product of two determinants as follows $\Delta = 0\,.\,0 = 0$ (on simplification), which is independent of $a, b, c $ and $d.$

Standard 12

Mathematics

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{10!}&{11!}&{12!}\\{11!}&{12!}&{13!}\\{12!}&{13!}&{14!}\end{array}\,} \right|$ is

easy

If the system of equations

$x+y+z=2$

$2 x+4 y-z=6$

$3 x+2 y+\lambda z=\mu$ has infinitely many solutions, then

medium

(JEE MAIN-2020)

A root of the equation $\left| {\,\begin{array}{*{20}{c}}{3 – x}&{ – 6}&3\\{ – 6}&{3 – x}&3\\3&3&{ – 6 – x}\end{array}\,} \right| = 0$ is

easy

The value of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6,3 x+5 y+5 z=26, x+2 y+\lambda z=\mu$ has no solution, are :

medium

(JEE MAIN-2021)

If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x – 2}\\{2{x^2} + 3x – 1}&{3x}&{3x – 3}\\{{x^2} + 2x + 3}&{2x – 1}&{2x – 1}\end{array}\,} \right| = Ax – 12$, then the value of $A $ is

medium

(IIT-1982)

If $a,b,c$ and $d $ are complex numbers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}2&{a + b + c + d}&{ab + cd}\\{a + b + c + d}&{2(a + b)(c + d)}&{ab(c + d) + cd(a + b)}\\{ab + cd}&{ab(c + d) + cd(a + d)}&{2abcd}\end{array}} \right|$is

Solution

Similar Questions

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{10!}&{11!}&{12!}\\{11!}&{12!}&{13!}\\{12!}&{13!}&{14!}\end{array}\,} \right|$ is

If the system of equations $x+y+z=2$ $2 x+4 y-z=6$ $3 x+2 y+\lambda z=\mu$ has infinitely many solutions, then

A root of the equation $\left| {\,\begin{array}{*{20}{c}}{3 – x}&{ – 6}&3\\{ – 6}&{3 – x}&3\\3&3&{ – 6 – x}\end{array}\,} \right| = 0$ is

The value of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6,3 x+5 y+5 z=26, x+2 y+\lambda z=\mu$ has no solution, are :

If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x – 2}\\{2{x^2} + 3x – 1}&{3x}&{3x – 3}\\{{x^2} + 2x + 3}&{2x – 1}&{2x – 1}\end{array}\,} \right| = Ax – 12$, then the value of $A $ is

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If the system of equations

$x+y+z=2$

$2 x+4 y-z=6$

$3 x+2 y+\lambda z=\mu$ has infinitely many solutions, then