If A=−1+i32i−1−i32i1+i32i1−i32i,i=−1 and f(x)=x2+2 then f(A) equals to

∵ω=−1+i32 and ω2=−1−i32Also, ω3=1 and ω+ω2=−1Thus A=−iω−iω2iω2iω∴ A2=−iω−iω2iω2iω−iω−iω2iω2iω=−ω2+ω00−ω2+ωNow, f(A)=A2+2I=−ω2+ω00−ω2+ω+2 00 2=−ω2+ω+200−ω2+ω+2=−ω2+ω+21001=(2+i3)1001

If A=−1+i32i−1−i32i1+i32i1−i32i,i=−1 and f(x)=x2+2 then f(A) equals to

∵ω=−1+i32 and ω2=−1−i32Also, ω3=1 and ω+ω2=−1Thus A=−iω−iω2iω2iω∴ A2=−iω−iω2iω2iω−iω−iω2iω2iω=−ω2+ω00−ω2+ωNow, f(A)=A2+2I=−ω2+ω00−ω2+ω+2 00 2=−ω2+ω+200−ω2+ω+2=−ω2+ω+21001=(2+i3)1001

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $A = [\begin{array}{cc} \frac{- 1 + i \sqrt{3}}{2 i} & \frac{- 1 - i \sqrt{3}}{2 i} \\ \frac{1 + i \sqrt{3}}{2 i} & \frac{1 - i \sqrt{3}}{2 i} \end{array}],$ $i = \sqrt{- 1}$ and $f (x) = x^{2} + 2$ then $f (A)$ equals to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$[\begin{array}{ll} 1 0 \\ 0 1 \end{array}]$

$(\frac{3 - i \sqrt{3}}{2}) [\begin{array}{ll} 1 0 \\ 0 1 \end{array}]$

$(2 + i \sqrt{3}) [\begin{array}{ll} 1 0 \\ 0 1 \end{array}]$

$(\frac{5 - i \sqrt{3}}{2}) [\begin{array}{ll} 1 0 \\ 0 1 \end{array}]$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$∵ ω = \frac{- 1 + i \sqrt{3}}{2} and ω^{2} = \frac{- 1 - i \sqrt{3}}{2}$

Also, $ω^{3} = 1 and ω + ω^{2} = - 1$

Thus $A = [\begin{array}{cc} - i ω & - i ω^{2} \\ i ω^{2} & i ω \end{array}]$

$∴ A^{2} = [\begin{array}{cc} - i ω & - i ω^{2} \\ i ω^{2} & i ω \end{array}] [\begin{array}{cc} - i ω & - i ω^{2} \\ i ω^{2} & i ω \end{array}] = [\begin{array}{cc} - ω^{2} + ω & 0 \\ 0 & - ω^{2} + ω \end{array}]$

Now, $f (A) = A^{2} + 2 I = [\begin{array}{cc} - ω^{2} + ω & 0 \\ 0 & - ω^{2} + ω \end{array}] + [\begin{array}{ll} 2 0 \\ 0 2 \end{array}]$

$\begin{aligned} = [\begin{array}{cc} - ω^{2} + ω + 2 & 0 \\ 0 & - ω^{2} + ω + 2 \end{array}] \\ = (- ω^{2} + ω + 2) [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] = (2 + i \sqrt{3}) [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] \end{aligned}$