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Identity Matrix

The value of ...

Question

The value of $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + p & 1 + p + q 2 & 3 + 2 p & 4 + 3 p + 2 q 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} ∣ ∣ ∣ ∣$ is:

A

2

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B

- 1

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C

3

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D

1

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Solution

The correct option is D $1$
Given: $Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + p & 1 + p + q 2 & 3 + 2 p & 4 + 3 p + 2 q 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} ∣ ∣ ∣ ∣$
$C_{2} \to C_{2} - p C_{1}$ and $C_{3} \to C_{3} - q C_{1}$
$\Rightarrow Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 + p 2 & 3 & 4 + 3 p 3 & 6 & 10 + 6 p \end{matrix} ∣ ∣ ∣ ∣$
Now, $C_{3} \to C_{3} - p C_{2}$
$\Rightarrow Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 2 & 3 & 4 3 & 6 & 10 \end{matrix} ∣ ∣ ∣ ∣ = 1$

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Similar questions

∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + p & 1 + p + q 2 & 3 + 2 p & 4 + 3 p + 2 q 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} ∣ ∣ ∣ ∣

= 1

Using properties of determinants, prove that:

x = \frac{6 p q}{p + q}

\frac{x + 3 p}{x - 3 p} + \frac{x + 3 q}{x - 3 q}

x = 2^{p}

y = 4^{q}

, then value of

{log}_{2} (x^{3} y)

Q. Find the value of

p

q

using elimination method:

3 p + 4 q = - 1

6 p - 2 q = 8

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Identity Matrix

Standard X Mathematics

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