If A- - 0 -1- 1 0 - and B- - 0 i- i 0 - -then

The correct option is A A^2=B^2= I A^2=[ 0 amp; 1; #160; #160;1 amp; 0 ][ 0 amp; 1; #160; #160;1 amp; 0 ] #160; #160; ...

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

Multiplication of Matrices

If A= [ 0 -...

Question

If A= $[\begin{matrix} 0 & - 1 1 & 0 \end{matrix}]$ and B= $[\begin{matrix} 0 & i i & 0 \end{matrix}]$ ,then

A^{2} = B^{2} = I

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

A^{2} = B^{2} = - I

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

A^{2} = I, B^{2} = - I

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

A^{2} = - I, B^{2} = I

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

Open in App

Solution

Suggest Corrections

Similar questions

A = [\begin{matrix} i & 0 0 & - i \end{matrix}], B = [\begin{matrix} 0 & - 1 1 & 0 \end{matrix}], C = [\begin{matrix} 0 & i i & 0 \end{matrix}]

A^{2} + B^{2} + C^{2} =

a^{2}

b^{2}

16 b^{2} + 9 a^{2} = a^{2} b^{2}

36 b^{2} + 4 a^{2} = a^{2} b^{2}

Prove the following identities :

$(i) (a + b)^{2} = a^{2} + b^{2} + 2 a b$

$(i i) (a - b)^{2} = a^{2} + b^{2} - 2 a b$ [2 MARKS]

(1 + i) (1 + 2 i) (1 + 3 i) . . . (1 + n i) = a + i b, then 2 \times 5 \times 10 \times . . . . \times (1 + n^{2})

\sqrt{a^{2} + b^{2}}

\sqrt{a^{2} - b^{2}}

a^{2} + b^{2}

a^{2} - b^{2}

a + b

Join BYJU'S Learning Program