Exercise 7.1
Question 1
निम्नलिखित आव्यूहों के परिवर्त तथा सहखण्डज निकालें छ
[Find the transpose and adjoint of the following matrices]
KC Sinha Mathematics Solution Class 12 Chapter 7 वर्ग आव्यूह का सहखंडज और प्रतिलोम (adjoint and inverse of matrix) Exercise 7.1 (Q1-Q3)
Question 1
निम्नलिखित आव्यूहों के परिवर्त तथा सहखण्डज निकालें छ
[Find the transpose and adjoint of the following matrices]
(i) $\left[\begin{array}{ll}2 & 3 \\ 5 & 1\end{array}\right]$
Sol :
$A=\left[\begin{array}{ll}2 & 3 \\ 5 & 1\end{array}\right]$
$A^{\prime}orA^{T}=\left[\begin{array}{ll}2 & 5 \\ 3 & 1\end{array}\right]$
$\left|\begin{array}{ll}+ & - \\ - & +\end{array}\right|$
A11=1 , A12=-5
A21=-3 , A22=2
adjA $=\left[\begin{array}{cc}1 & -5 \\ -3 & 2\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}1 & -3 \\ -5 & 2\end{array}\right]$
(ii) $\left[\begin{array}{ll}\sec \theta & \operatorname{cosec} \theta \\ \sin \theta & \cos \theta\end{array}\right]$
Sol :
$A=\left[\begin{array}{cc}\sec \theta & cosec \theta \\ \sin \theta & \cos \theta\end{array}\right]$
$A^{\prime}=\left[\begin{array}{ll}sec \theta & \sin \theta \\ cosec \theta & \cos \theta\end{array}\right]$
A11=cosθ , A12=-sinθ
A21=-cosecθ , A22=secθ
adjA $=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\operatorname{cosec} \theta & \sec \theta\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}cos\theta & -\operatorname{cosec} \theta \\ -\sin \theta & \sec \theta\end{array}\right]$
(iii) $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
Sol :
Sol :
$A=\left[\begin{array}{ll}2 & 3 \\ 5 & 1\end{array}\right]$
$A^{\prime}orA^{T}=\left[\begin{array}{ll}2 & 5 \\ 3 & 1\end{array}\right]$
$\left|\begin{array}{ll}+ & - \\ - & +\end{array}\right|$
A11=1 , A12=-5
A21=-3 , A22=2
adjA $=\left[\begin{array}{cc}1 & -5 \\ -3 & 2\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}1 & -3 \\ -5 & 2\end{array}\right]$
(ii) $\left[\begin{array}{ll}\sec \theta & \operatorname{cosec} \theta \\ \sin \theta & \cos \theta\end{array}\right]$
Sol :
$A=\left[\begin{array}{cc}\sec \theta & cosec \theta \\ \sin \theta & \cos \theta\end{array}\right]$
$A^{\prime}=\left[\begin{array}{ll}sec \theta & \sin \theta \\ cosec \theta & \cos \theta\end{array}\right]$
A11=cosθ , A12=-sinθ
A21=-cosecθ , A22=secθ
adjA $=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\operatorname{cosec} \theta & \sec \theta\end{array}\right]^{\prime}$
$=\left[\begin{array}{cc}cos\theta & -\operatorname{cosec} \theta \\ -\sin \theta & \sec \theta\end{array}\right]$
(iii) $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
Sol :
(iv) $\left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]$
Sol :
(v) $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
Sol :
Question 2
Find the adjoint of the following matrices:
(i) $\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 2 \\ 3 & 3 & 4\end{array}\right]$
Sol :
$A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 2 \\ 3 & 3 & 4\end{array}\right]$ $\left| \begin{array}{ccc}+ & -&+ \\ -&+ & - \\ + & -&+\end{array}\right|$
$A_{11}=\left|\begin{array}{cc}3 & 2 \\ 3 & 4\end{array}\right|$
=12-6
=6
$A_{12}=-\left|\begin{array}{cc}2 & 2 \\ 3 & 4\end{array}\right|$
=-(-8-6)
=-2
$A_{13}=\left|\begin{array}{ll}2 & 3 \\ 3 & 3\end{array}\right|$
=6-9
=-3
$A_{2 1}=-\left|\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right|$
=-(8-9)
=1
$A_{22}=\left|\begin{array}{ll}1 & 3 \\ 3 & 4\end{array}\right|$
=4-9
=-5
$A_{23}=-\left|\begin{array}{cc}1 & 2 \\ 3 & 3\end{array}\right|$
=-(3-6)
=3
$A_{31}=\left|\begin{array}{cc}2 & 3 \\ 3 & 2\end{array}\right|$
=4-9
=-5
$A_{32}=-\left|\begin{array}{ll}1 & 3 \\ 2 & 2\end{array}\right|$
=-(2-6)
=4
$A_{33}=\left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right|$
=3-4
=-1
$\operatorname{adj} A=\left[\begin{array}{ccc}6 & -2 & -3 \\ 1 & -5 & 3 \\ -5 & 4 & -1\end{array}\right]'$
$=\left[\begin{array}{ccc}6 & 1 & -5 \\ -2 & -5 & 4 \\ -3 & 3 & -1\end{array}\right]$
(ii) $\left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$
Sol :
(iii) $\left[\begin{array}{rrr}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$
Sol :
Question 3
Find the inverse of the following matrices(i) $\left[\begin{array}{rr}2 & 5 \\ -3 & 1\end{array}\right]$
Sol :
$A=\left[\begin{array}{ll}2 & 5 \\ -3 & 1\end{array}\right]$
$|A|=\left|\begin{array}{ll}2 & 5 \\ -3 & 1\end{array}\right|$
=2+15
=17≠0
<to be added>
$A_{11}=1$ ,$A_{12}=-(-3)=3$ ,$A_{21}=-5$
$A_{22}=2$
$a d{j} A-\left[\begin{array}{cc}1 & 3 \\ -5 & 2\end{array}\right]'$
$=\left[\begin{array}{cc}1 & -5 \\ 3 & 2\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|} \cdot$ adj A
$=\frac{1}{17}\left[\begin{array}{ll}1 & -5 \\ 3 & 2\end{array}\right]$
(ii) $\left[\begin{array}{rr}2 & -2 \\ 1 & 3\end{array}\right]$
Sol :
(iii) $\left[\begin{array}{rr}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right]$ where
$a^{2}+b^{2}+c^{2}+d^{2}=1$
Sol :
$A=\left[\begin{array}{rr}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right]$
|A|=(a+ib)(a-ib)-(-c+id)(c+id)
$=a^{2}-1^{2} b^{2}+c^{2}-i^{2} d^{2}$
$=a^{2}+b^{2}+c^{2}+d^{2}$
=1≠0
<to be added>
$A_{11}=a-i b$ , $A_{12}=-(-c+i d)$=c-id
$A_{21}=-(c+i d)$
=-c-id
$A_{22}=a+ib$
$ad{j} A=\left[\begin{array}{cc}a-ib & c-i d \\ -c-i d & a+i b\end{array}\right]$
$=\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]$
$A^{-1}=\frac{1}{(A)} \cdot a d j A$
$=\frac{1}{1}\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+2 b\end{array}\right]$
$A^{-1}=\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]$
(iv) $\left[\begin{array}{cc}1 & \tan x \\ -\tan x & 1\end{array}\right]$
Sol :
(v) $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$. जहाँ (where) ad-bc≠0
Sol :
Question no 9 ka 3 sol
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