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If $a, b, c$ and $d$ are real numbers such that $a^2+b^2+c^2+d^2=1 \quad$ and $A=\left[\begin{array}{c}a+i b c+i d \\ -c+i d a-i b\end{array}\right]$, then $A^{-1}$ equals to
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Verified Answer
The correct answer is:
$\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]$
Given,
$$
a^2+b^2+c^2+d^2=1
$$
$$
\text { and } \quad A=\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]
$$
$$
\begin{aligned}
\text { Now }|A| & =(a+i b)(a-i b) \\
& -(c+i d)(-c+i d) \\
& =a^2-(i b)^2-\left[(i d)^2-(c)^2\right] \\
& =a^2+b^2-\left[-d^2-c^2\right] \\
& =a^2+b^2+d^2+c^2 \\
& =1 \\
\therefore \quad A^{-1} & =\frac{1}{|A|}\left[\begin{array}{cc}
a-i b & -(c+i d) \\
-(-c+i d) & a+i b
\end{array}\right]
\end{aligned}
$$
[from Eq. (i)]
$$
\therefore \quad A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}
a-i b & -(c+i d) \\
-(-c+i d) & a+i b
\end{array}\right]
$$
$\begin{aligned} & =\frac{1}{1}\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right] \\ & =\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]\end{aligned}$
$$
a^2+b^2+c^2+d^2=1
$$
$$
\text { and } \quad A=\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]
$$
$$
\begin{aligned}
\text { Now }|A| & =(a+i b)(a-i b) \\
& -(c+i d)(-c+i d) \\
& =a^2-(i b)^2-\left[(i d)^2-(c)^2\right] \\
& =a^2+b^2-\left[-d^2-c^2\right] \\
& =a^2+b^2+d^2+c^2 \\
& =1 \\
\therefore \quad A^{-1} & =\frac{1}{|A|}\left[\begin{array}{cc}
a-i b & -(c+i d) \\
-(-c+i d) & a+i b
\end{array}\right]
\end{aligned}
$$
[from Eq. (i)]
$$
\therefore \quad A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}
a-i b & -(c+i d) \\
-(-c+i d) & a+i b
\end{array}\right]
$$
$\begin{aligned} & =\frac{1}{1}\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right] \\ & =\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]\end{aligned}$
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