KC Sinha Mathematics Solution Class 12 Chapter 7 वर्ग आव्यूह का सहखंडज और प्रतिलोम (adjoint and inverse of matrix) Exercise 7.1 (Q1-Q3)

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KC Sinha Mathematics Solution Class 12 Chapter 7 वर्ग आव्यूह का सहखंडज और प्रतिलोम (adjoint and inverse of matrix) Exercise 7.1 (Q1-Q3)

Exercise 7.1

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 :

(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 :

3 comments:

UnknownMarch 5, 2021 at 4:45 PM
Question no 9 ka 3 sol
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MANOJ KUMARMarch 7, 2021 at 7:38 AM
Manoj mathematics
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UnknownJuly 7, 2021 at 9:22 AM
Roshan
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