Question

if A={ a+ib c+id -c+id a-id } ,a2+b2+c2+d2=1 then find inverse of A in matrix chapter exercise 3e​

Answer 1

$\textbf{Given:}$

$\mathsf{A=\left(\begin{array}{cc}a+ib&c+id\\-c+id&a-ib\end{array}\right)\;\;a^2+b^2+c^2+d^2=1}$

$\textbf{To find:}$

$\textsf{Inverse of A}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{|A|=\left|\begin{array}{cc}a+ib&c+id\\-c+id&a-ib\end{array}\right|}$

$\mathsf{|A|=(a+id)(a-ib)-(id-c)(id+c)}$

$\mathsf{|A|=a^2+b^2-(-d^2-c^2)}$

$\mathsf{|A|=a^2+b^2+c^2+d^2}$

$\mathsf{|A|=1\;{\neq}\;0}$

$\implies\mathsf{A^{-1}\;exists}$

$\mathsf{adjA=\left(\begin{array}{cc}a-ib&-c-id\\c-id&a+ib\end{array}\right)}$

$\mathsf{Now,}$

$\mathsf{A^{-1}=\dfrac{1}{|A|}adjA}$

$\mathsf{A^{-1}=\dfrac{1}{1}\left(\begin{array}{cc}a-ib&-c-id\\c-id&a+ib\end{array}\right)}$

$\implies\mathsf{A^{-1}=\left(\begin{array}{cc}a-ib&-c-id\\c-id&a+ib\end{array}\right)}$

$\textbf{Find more:}$

1.If A is a non-singular square matrix of order 3 such that A^2 = 3A, then

value of det.A is

(A) -3

(B) 3IA|

(C) 9

D) 27​

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2.A is a square matrix with |A|=4 then find the value of |A. (adjA)|

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3.If A is a matrix such that A^3=I , then show that A is invertible

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4.If A and B are invertible matrices of same order such that |(AB)-1 | =8 .if |A|=2 then| B | is

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Answer 2

$\huge\underline \texttt\blue{Given:-}$

$\mathbf{A=\left(\begin{array}{cc}a+ib&c+id\\-c+id&a-ib\end{array}\right)\;\;a^2+b^2+c^2+d^2=1}$

$\huge\underline\texttt\orange{To find:-}$

$\textbf{Inverse of A}$

$\huge\underline\texttt\red{Solution:-}$

$\textbf{Consider,}$

$\mathbf{|A|=\left|\begin{array}{cc}a+ib&c+id\\-c+id&a-ib\end{array}\right|}$

$\mathbf{|A|=(a+id)(a-ib)-(id-c)(id+c)}$

$\mathbf{|A|=a^2+b^2-(-d^2-c^2)}$

$\mathbf{|A|=a^2+b^2+c^2+d^2}$

$\mathbf{|A|=1\;{\neq}\;0}$

$\implies\mathbf{A^{-1}\;exists}$

$\mathbf{adjA=\left(\begin{array}{cc}a-ib&-c-id\\c-id&a+ib\end{array}\right)}$

$\mathbf{Now,}$

$\mathbf{A^{-1}=\dfrac{1}{|A|}adjA}$

$\mathbf{A^{-1}=\dfrac{1}{1}\left(\begin{array}{cc}a-ib&-c-id\\c-id&a+ib\end{array}\right)}$

$\implies\mathbf{A^{-1}=\left(\begin{array}{cc}a-ib&-c-id\\c-id&a+ib\end{array}\right)}$

________________________

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