The value of 2-i 2-i1-i 1-i is- where i2-1

We have, 2+i amp;2 i 1+i amp;1 i = (2+i)(1 i) (2 i)(1+i) =2 2i+i i^2 2 2i+i+i^2 = 2i ...

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

Adjoint of a Matrix

The value of ...

Question

The value of $∣ ∣ ∣ \begin{matrix} 2 + i & 2 - i 1 + i & 1 - i \end{matrix} ∣ ∣ ∣$ is, where $(i^{2} = - 1)$

1 - i

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

purely imaginary

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

1 + i

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

a real quantity

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

Open in App

Solution

Suggest Corrections

Similar questions

Find the value of $∣ ∣ ∣ \begin{matrix} 2 + i & 2 - i 1 + i & 1 - i \end{matrix} ∣ ∣ ∣$ if $i^{2} = - 1$ .

∣ ∣ ∣ \begin{matrix} 2 + i & 2 - i 1 + i & 1 - i \end{matrix} ∣ ∣ ∣

f (x)

I_{1} = \int_{0}^{\infty} f (x) d x,

I_{2} = \int_{0}^{\infty} f (3 x - \frac{1}{2 x}) d x,

\frac{I_{1}}{I_{2}}

I_{1} & I_{2}

z = \frac{(1 + i) (1 + 2 i) (1 + 3 i) \dots \dots \dots (1 + n i)}{(1 - i) (2 - i) (3 - i) \dots \dots \dots (n - i)}

i = \sqrt{- 1}, n \in N

z

∣ ∣ ∣ ∣ \begin{matrix} 1 + i & 1 - i & i 1 - i & i & 1 + i i & 1 + i & 1 - i \end{matrix} ∣ ∣ ∣ ∣

i = \sqrt{- 1}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program