If left| {begin{array}{*{20}{c}}{a, + ,1}&{a, + ,2}&{a, + ,p}{a, + ,2}&{a, +,3}&{a,

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

normal

If $\left| {\begin{array}{*{20}{c}}{a\, + \,1}&{a\, + \,2}&{a\, + \,p}\\{a\, + \,2}&{a\, +\,3}&{a\, + \,q}\\{a\, + \,3}&{a\, + \,4}&{a\, + \,r}\end{array}} \right|$ $= 0$ , then $p, q, r$ are in :

$AP$

$GP$

$HP$

none

Solution

Use $R_2 \rightarrow R_2 – R_1 \, \& \, R_3 \rightarrow R_3 – R_2 \, \& $ then $c_1 \rightarrow c_1 – c_2$ to get $\left| {\,\begin{array}{*{20}{c}}{ – 1} & {a +2}&{a + p}\\0 &1&{q – p}\\0&1&{r – q}\end{array}\,} \right|$ open by $c_1$ to get $p + r = 2q$

Standard 12

Mathematics

$\left| {\,\begin{array}{*{20}{c}}{a – 1}&a&{bc}\\{b – 1}&b&{ca}\\{c – 1}&c&{ab}\end{array}\,} \right| = $

easy

Find values of $\mathrm{k}$ if area of triangle is $4$ square units and vertices are $(\mathrm{k}, 0),(4,0),(0,2)$

medium

If $A = \left| {\,\begin{array}{{20}{c}}{ – 1}&2&4\\3&1&0\\{ – 2}&4&2\end{array}\,} \right|$and $B = \left| {\,\begin{array}{{20}{c}}{ – 2}&4&2\\6&2&0\\{ – 2}&4&8\end{array}\,} \right|$, then $B$ is given by

easy

If the system of equations

$ 2 x+7 y+\lambda z=3 $

$ 3 x+2 y+5 z=4 $

$ x+\mu y+32 z=-1$

has infinitely many solutions, then $(\lambda-\mu)$ is equal to $\qquad$

hard

(JEE MAIN-2024)

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 :

hard

(JEE MAIN-2024)

If $\left| {\begin{array}{*{20}{c}}{a\, + \,1}&{a\, + \,2}&{a\, + \,p}\\{a\, + \,2}&{a\, +\,3}&{a\, + \,q}\\{a\, + \,3}&{a\, + \,4}&{a\, + \,r}\end{array}} \right|$ $= 0$ , then $p, q, r$ are in :

Solution

Similar Questions

$\left| {\,\begin{array}{*{20}{c}}{a – 1}&a&{bc}\\{b – 1}&b&{ca}\\{c – 1}&c&{ab}\end{array}\,} \right| = $

Find values of $\mathrm{k}$ if area of triangle is $4$ square units and vertices are $(\mathrm{k}, 0),(4,0),(0,2)$

If $A = \left| {\,\begin{array}{*{20}{c}}{ – 1}&2&4\\3&1&0\\{ – 2}&4&2\end{array}\,} \right|$and $B = \left| {\,\begin{array}{*{20}{c}}{ – 2}&4&2\\6&2&0\\{ – 2}&4&8\end{array}\,} \right|$, then $B$ is given by

If the system of equations $ 2 x+7 y+\lambda z=3 $ $ 3 x+2 y+5 z=4 $ $ x+\mu y+32 z=-1$ has infinitely many solutions, then $(\lambda-\mu)$ is equal to $\qquad$

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 :

Start a Free Trial Now

If $A = \left| {\,\begin{array}{{20}{c}}{ – 1}&2&4\\3&1&0\\{ – 2}&4&2\end{array}\,} \right|$and $B = \left| {\,\begin{array}{{20}{c}}{ – 2}&4&2\\6&2&0\\{ – 2}&4&8\end{array}\,} \right|$, then $B$ is given by

If the system of equations

$ 2 x+7 y+\lambda z=3 $

$ 3 x+2 y+5 z=4 $

$ x+\mu y+32 z=-1$

has infinitely many solutions, then $(\lambda-\mu)$ is equal to $\qquad$

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 :