If A -1- bmatrix sin - 2 - 0 - 0 0 -sin - 2 - 0 0 - 0 -sin- 2 lend bmatrix and B -1- begin bmatrix cos- 2 lalpha - 0 - 0 10 -cos - 2 beta - 0 10 - 0 -cos - 2 lendbmatrix where - - and - are real numbers and C-A-5-B-5-5 A-1 B-1A-3-B-3-10 A-2 B-2A-1-B-1- then find - C - -

You visited us 95 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Cofactor

If A ∧-1=. bm...

Question

If $A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} s i n^{2} α & 0 & 0 0 & s i n^{2} β & 0 0 & 0 & s i n^{2} γ \end{matrix} ⎤ ⎥ ⎦$ and $B^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} c o s^{2} α & 0 & 0 0 & c o s^{2} β & 0 0 & 0 & c o s^{2} γ \end{matrix} ⎤ ⎥ ⎦$ where $α, β$ and $γ$ are real numbers and $C = (A^{- 5} + B^{- 5}) + 5 A^{- 1} B^{- 1} (A^{- 3} + B^{- 3}) + 10 A^{- 2} B^{- 2} (A^{- 1} + B^{- 1})$ , then find |C|.___

Open in App

Solution

$C = A^{- 5} + B^{- 5} + 5 A^{- 1} B^{- 1} (A^{- 3} + B^{- 3}) + 10 A^{- 2} B^{- 2} (A^{- 1} + B^{- 1}) C = (A^{- 1} + B^{- 1})^{5} | C | = | A^{- 1} + B^{- 1} |^{5} | C | = 1$

Suggest Corrections

Similar questions

If $A = [\begin{matrix} c o s α & s i n α - s i n α & c o s α \end{matrix}]$ , then $A^{2} =$

Q. Find the inverse of the following matrices:
(a)

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 0 - 1 & 2 & 2 0 & 0 & 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

(b)

⎡ ⎢ ⎢ ⎣ \begin{matrix} c o s θ & - s i n θ & 0 s i n θ & c o s θ & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

(c)

⎡ ⎢ ⎢ ⎣ \begin{matrix} 2 & 2 & 2 2 & 5 & 5 2 & 5 & 11 \end{matrix} ⎤ ⎥ ⎥ ⎦

(d)

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 4 & 5 3 & 5 & 6 \end{matrix} ⎤ ⎥ ⎥ ⎦

(e)

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 0 - 1 & 2 & 2 0 & 0 & 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

Q. If

A = [\begin{matrix} cos α & - sin α sin α & cos α \end{matrix}]

B = [\begin{matrix} cos 2 β & sin 2 β sin 2 β & - cos 2 β \end{matrix}]

, where

0 < β < \frac{π}{2}

, then prove that

B A B = A^{- 1}

. .

If $A = [\begin{matrix} 0 & - t a n \frac{α}{2} t a n \frac{α}{2} & 0 \end{matrix}]$ and I is the identity matrix of order 2, show that I + A $= (I - A) [\begin{matrix} c o s α & - s i n α s i n α & c o s α \end{matrix}]$

Q. If

A = [\begin{matrix} c o s α & s i n α - s i n α & c o s α \end{matrix}]

, then

A^{2} =

Join BYJU'S Learning Program

If A−1=⎡⎢⎣sin2α000sin2β000sin2γ⎤⎥⎦ and B−1=⎡⎢⎣cos2α000cos2β000cos2γ⎤⎥⎦ where α,β and γ are real numbers and C=(A−5+B−5)+5A−1B−1(A−3+B−3)+10A−2B−2(A−1+B−1) , then find |C|.___

C=A−5+B−5+5A−1B−1(A−3+B−3)+10A−2B−2(A−1+B−1)C=(A−1+B−1)5|C|=|A−1+B−1|5|C|=1

$C = A^{- 5} + B^{- 5} + 5 A^{- 1} B^{- 1} (A^{- 3} + B^{- 3}) + 10 A^{- 2} B^{- 2} (A^{- 1} + B^{- 1}) C = (A^{- 1} + B^{- 1})^{5} | C | = | A^{- 1} + B^{- 1} |^{5} | C | = 1$