Assertion - - -1-a1b11-a12b12-a1b1-1-a1b1 1-a1b21-a12b22-a1b2-1-a1b2 1-a1b31-a12b32-a1b3-1-a1b3 1-a2b11-a22b12-a2b1-1-a2b1 1-a2b21-a22b22-a2b2-1-a2b2 1-a2b31-a22b32-a2b3-1-a2b3 1-a3b11-a32b12-a3b1-1-a3b1 1-a3b21-a22b22-a3b2-1-a3b2 1-a3b31-a32b32-a3b3-1-a3b3 - -0 Reason- - can be written as product of two determinants-

The correct option is D Assertion is incorrect but Reason is correctWe know that, #160; 1+a_1^3b_1^3=(1+a_1b_1)(1+a_1^2b_1^2 a_1b_1) So the determinant c ...

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

Byju's Answer

Standard XII

Physics

Kepler's Law

Assertion : ...

Question

Assertion :
$Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
$Δ = 0$ Reason: $Δ$ can be written as product of two determinants.

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

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

Assertion is correct but Reason is incorrect

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

Assertion is incorrect but Reason is correct

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

Open in App

Solution

The correct option is D Assertion is incorrect but Reason is correct
We know that,
$1 + a_{1}^{3} b_{1}^{3} = (1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})$
So the determinant can be expressed as,
$Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$

$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1}) & (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2}) & (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1}) & (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2}) & (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1}) & (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2}) & (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3}) \end{matrix} ∣ ∣ ∣ ∣ ∣$

The determinant can be split as product of two determinants as under,
$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a_{1}^{2} & - a_{1} 1 & a_{2}^{2} & - a_{2} 1 & a_{3}^{2} & - a_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ \times ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 b_{1}^{2} & b_{2}^{2} & b_{3}^{2} b_{1} & b_{2} & b_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$∴$ Reason is correct.
However, nothing can be said about $Δ$ is equal to zero as one of the two determinants, in that case, will have to be zero.
That will be possible only when a pair of $a_{1}, a_{2}, a_{3}$ are equal or a pair of $b_{1}, b_{2}, b_{3}$ are equal.
Hence, Assertion is not always true.

Suggest Corrections

Similar questions

Q. The value of the determinant

Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 + a_{1} b_{1} & 1 + a_{1} b_{2} & 1 + a_{1} b_{3} 1 + a_{2} b_{1} & 1 + a_{2} b_{2} & 1 + a_{2} b_{3} 1 + a_{3} b_{1} & 1 + a_{3} b_{2} & 1 + a_{3} b_{3} \end{matrix} ∣ ∣ ∣ ∣

, is

Q. If

a_{1} > a_{2} > a_{3}

b_{1} > b_{2} > b_{3}

and

a_{i} b_{j} \neq 1

for

1 \leq i

j \leq 3

, then the determinant

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1 - a_{1}^{3} b_{1}^{3}}{1 - a_{1} b_{1}} & \frac{1 - a_{1}^{3} b_{2}^{3}}{1 - a_{1} b_{2}} & \frac{1 - a_{1}^{3} b_{3}^{3}}{1 - a_{1} b_{3}} \frac{1 - a_{2}^{3} b_{1}^{3}}{1 - a_{2} b_{1}} & \frac{1 - a_{2}^{3} b_{2}^{3}}{1 - a_{2} b_{2}} & \frac{1 - a_{2}^{3} b_{3}^{3}}{1 - a_{2} b_{3}} \frac{1 - a_{3}^{3} b_{1}^{3}}{1 - a_{1} b_{1}} & \frac{1 - a_{3}^{3} b_{2}^{3}}{1 - a_{3} b_{2}} & \frac{1 - a_{3}^{3} b_{3}^{3}}{1 - a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Q. If

a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} \in R

and are such that

a_{i} b_{j} \neq 1

for

1 \leq i, j \leq 3

, then

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1 - a_{1}^{3} b_{1}^{3}}{1 - a_{1} b_{1}} & \frac{1 - a_{1}^{3} b_{2}^{3}}{1 - a_{1} b_{2}} & \frac{1 - a_{1}^{3} b_{3}^{3}}{1 - a_{1} b_{3}} \frac{1 - a_{2}^{3} b_{1}^{3}}{1 - a_{2} b_{1}} & \frac{1 - a_{2}^{3} b_{2}^{3}}{1 - a_{2} b_{2}} & \frac{1 - a_{2}^{3} b_{3}^{3}}{1 - a_{2} b_{3}} \frac{1 - a_{3}^{3} b_{1}^{3}}{1 - a_{3} b_{1}} & \frac{1 - a_{3}^{3} b_{2}^{3}}{1 - a_{3} b_{2}} & \frac{1 - a_{3}^{3} b_{3}^{3}}{1 - a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

> 0 Provided either

a_{1} < a_{2} < a_{3}

and

b_{1} < b_{2} < b_{3}

, or

a_{1} > a_{2} a_{3}

and

b_{1} > b_{2} > b_{3}

then show

(a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1}) (b_{1} - b_{2}) (b_{2} - b_{3}) (b_{3} - b_{1}) < 0

Q. Assertion: The compressibility factor for

H_{2}

and

H e

(1 + \frac{P b}{R T})

Reason: The compressibility factor for

H_{2}

and

H e

can be derived from Van der Waal's equation.

Q. Assertion: Average kinetic energy in one oscillation during

SHM

of a body is

\frac{1}{4} m ω^{2} A^{2}

.
Reason: Maximum kinetic energy is

\frac{1}{2} m ω^{2} A^{2}

Join BYJU'S Learning Program