Suppose D- a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 and D- a 1 -p b 1 b 1 -q c 1 c 1 -r a 1 a 2 -p b 2 b 2 -q c 2 c 2 -r a 2 a 3 -p b 3 b 3 -q c 3 c 2 -r a 3- Then

The correct option is D D'=D(1+pqr) Given, D'= a _ 1 +p b _ 1 amp; b _ 1 +q c _ 1 amp; c _ 1 +r a _ 1 a _ 2 +p b _ 2 amp; b _ 2 +q c ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

Suppose D= ...

Question

Suppose $D = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$ and $D^{'} = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} + p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} + p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} + p b_{3} & b_{3} + q c_{3} & c_{2} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣$ . Then

D^{'} = D

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D^{'} = D (1 - p q r)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D^{'} = D (1 + p + q + r)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D^{'} = D (1 + p q r)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $D^{'} = D (1 + p q r)$
Given, $D^{'} = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} + p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} + p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} + p b_{3} & b_{3} + q c_{3} & c_{2} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} & b_{3} + q c_{3} & c_{2} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} p b_{3} & b_{3} + q c_{3} & c_{2} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣$

In the first determinant apply $C_{3} \to C_{3} - r C_{1}$ and then ; $C_{2} \to C_{2} - q C_{3}$ and then
in second determinant take $p$ common from $C_{1}$ and then apply $C_{2} \to C_{2} - C_{1}$
then take $q$ common from $C_{2}$ and apply $C_{3} \to C_{3} - C_{2}$ .
Finally taking $r$ common from $C_{3}$ ,
$D^{'} = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ + p q r ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$
we have ultimately $D^{'} = (1 + p q r) D$

Suggest Corrections

Similar questions

Q. If D=

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

and D'=

∣ ∣ ∣ ∣ \begin{matrix} a_{1} + p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} + p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} + p b_{3} & b_{3} + q c_{3} & c_{3} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣

, then
prove that D'=D(1+pqr).

Q. If

Δ =

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

and

Δ_{1}

∣ ∣ ∣ ∣ \begin{matrix} a_{1} + p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} + p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} + p b_{3} & b_{3} + q c_{3} & c_{3} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣

then

Δ_{1} =

A body of density d and volume V floats with volume $V_{1}$ of it immersed in a liquid of density $d_{1}$ and rest of the volume $V_{2}$ immersed in another liquid of density $d_{2}$ (< $d_{1}$ ) then $V_{1}$ is

Q. Assertion :If

D = d i a g [d_{1}, d_{2}, . . . ., d_{n}]

, then

D^{- 1} = d i a g [d_{1}^{- 1}, d_{2}^{- 1} . . . . ., d_{n}^{- 1}]

Reason: If

D = d i a g [d_{1}, d_{2}, . . . . . . d_{n}],

then

D^{n} = d i a g [d_{1}^{n}, d_{2}^{n} . . . . . ., d_{n}^{n}] .

A body of density $d_{1}$ is counterpoised by M g of weights of density $d_{2}$ in air of density d. Then the true mass of the body is

Join BYJU'S Learning Program

Suppose D=∣∣ ∣∣a1b1c1a2b2c2a3b3c3∣∣ ∣∣ and D′=∣∣ ∣∣a1+pb1b1+qc1c1+ra1a2+pb2b2+qc2c2+ra2a3+pb3b3+qc3c2+ra3∣∣ ∣∣. Then