1.कक्षा 12 में सारणिक (Determinants in Class 12),सारणिक कक्षा 12 (Determinants Class 12):

Determinants in Class 12

कक्षा 12 में सारणिक (Determinants in Class 12) के इस आर्टिकल में सारणिकों के गुणधर्म,उपसारणिक और सहखण्ड,आव्यूह के सहखण्डों और व्युत्क्रम,सारणिकों और आव्यूहों के अनुप्रयोग पर आधारित सवालों को हल करेंगे।
आपको यह जानकारी रोचक व ज्ञानवर्धक लगे तो अपने मित्रों के साथ इस गणित के आर्टिकल को शेयर करें।यदि आप इस वेबसाइट पर पहली बार आए हैं तो वेबसाइट को फॉलो करें और ईमेल सब्सक्रिप्शन को भी फॉलो करें।जिससे नए आर्टिकल का नोटिफिकेशन आपको मिल सके । यदि आर्टिकल पसन्द आए तो अपने मित्रों के साथ शेयर और लाईक करें जिससे वे भी लाभ उठाए । आपकी कोई समस्या हो या कोई सुझाव देना चाहते हैं तो कमेंट करके बताएं।इस आर्टिकल को पूरा पढ़ें।

Also Read This Article:- Applications of Determinants Class 12

2.कक्षा 12 में सारणिक के उदाहरण (Determinants in Class 12 Examples):

Example:1.सिद्ध कीजिए कि सारणिक \left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|, \theta से स्वतन्त्र है।
Solution: \left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|

प्रथम पंक्ति के अनुसार प्रसरण करने पर:

x\left|\begin{array}{cc} -x & 1 \\ 1 & x \end{array}\right|-\sin \theta\left|\begin{array}{cc} -\sin \theta & 1 \\ \cos \theta & x \end{array}\right|+\cos \theta \left|\begin{array}{cc} \sin \theta & -x \\ \cos \theta & 1 \end{array}\right| \\ =x\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta (-\sin \theta+x \cos \theta) \\ =-x^3-x+x \sin ^2 \theta+\sin \theta \cos \theta-\cos \theta \sin \theta+x \cos ^2 \theta \\ =-x^3-x+x\left(\sin ^2 \theta+\cos ^2 \theta\right) \\ =-x^3-x+x(1) \\ =-x^3-x+x \\ =-x^3
जो कि \theta से स्वतन्त्र है।
Example:2.सारणिक का प्रसरण किए बिना सिद्ध कीजिए कि \left|\begin{array}{lll}a & a^2 & bc \\ b & b^2 & c a \\ c & c^2 & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3\end{array}\right|
Solution: \left|\begin{array}{lll}a & a^2 & bc \\ b & b^2 & c a \\ c & c^2 & a b\end{array} \right|=\left|\begin{array}{lll}1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3\end{array}\right| \\ \text{L.H.S.} \quad \left|\begin{array}{lll}a & a^2 & bc \\ b & b^2 & c a \\ c & c^2 & a b\end{array}\right| \\ R_1 \rightarrow a R_1, R_2 \rightarrow b R_2 तथा R_3 \rightarrow c R_3  संक्रिया तथा \frac{1}{abc} से सारणिक को गुणा करने पर:
=\frac{1}{a b c}\left|\begin{array}{lll} a^2 & a^3 & a b c \\ b^2 & b^3 & a b c \\ c^2 & c^3 & a b c \end{array}\right| \\ =\frac{a b c}{a b c}\left|\begin{array}{lll} a^2 & a^3 & 1 \\ b^2 & b^3 & 1 \\ c^2 & c^3 & 1 \end{array}\right| \\ C_2 \leftrightarrow C_3 संक्रिया से:
=-\left|\begin{array}{lll} a^2 & 1 & a^3 \\ b^2 & 1 & b^3 \\ c^2 & 1 & c^3 \end{array}\right| \\ C_1 \leftrightarrow C_2 संक्रिया से:

= \left|\begin{array}{lll} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{array}\right|
Example:3. \left|\begin{array}{ccc} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array} \right| का मान ज्ञात कीजिए।
Solution: \left|\begin{array}{ccc} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right| \\ R_1 \rightarrow \sin \alpha R_1 तथा R_3 \rightarrow \cos \alpha R_3 संक्रिया से:
=\frac{1}{\sin \alpha \cos \alpha}\left|\begin{array}{lll} \cos \alpha \sin \alpha \cos \beta & \sin \alpha \cos \alpha \sin \beta & -\sin ^2 \alpha \\ -\sin \beta & \cos \beta & 0 \\ \cos \alpha \sin \alpha \cos \beta & \sin \alpha \cos \alpha \sin \beta & \cos ^2 \alpha \end{array}\right| \\ R_1 \rightarrow R_1-R_3 संक्रिया से
=\frac{1}{\sin \alpha \cos \alpha} \\ \left|\begin{array}{ccc} 0 & 0 & -\left(\sin ^2 \alpha+\cos ^2 \alpha\right) \\ -\sin \beta & \cos \beta & 0 \\ \cos \alpha \sin \alpha \cos \beta & \sin \alpha \cos \alpha \sin \beta & \cos ^2 \alpha \end{array}\right| \\ =\frac{\cos \alpha}{\sin \alpha \cos \alpha}\left| \begin{array}{ccc} 0 & 0 & -1 \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right| \\ R_{1} के अनुसार प्रसरण करने पर:

=\frac{1}{\sin \alpha}[0\left|\begin{array}{cc} \cos \beta & 0 \\ \sin \alpha \sin \beta & \cos \alpha \end{array}\right|-0 \left| \begin{array}{ll} -\sin \beta & 0\\ \sin \alpha \cos \beta & \cos \alpha \end{array}\right| -1\left|\begin{array}{ll} -\sin \beta & \cos \beta \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta \end{array}\right|] \\ =\frac{1}{\sin \alpha}\left[\left(\sin \alpha \sin ^2 \beta+\sin \alpha \cos ^2 \beta\right)\right] \\ =\frac{1}{\sin \alpha} \sin \alpha\left(\sin^2 \beta+\cos ^2 \beta\right) =1
Example:4.यदि a,b और c वास्तविक संख्याएँ हो और सारणिक \Delta=\left|\begin{array}{lll} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|=0 हो तो दर्शाइए कि या तो a+b+c=0 या a=b=c
Solution: \left|\begin{array}{lll}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|=0 \\ R_1 \rightarrow R_1+R_2+R_3  संक्रिया से:
\left|\begin{array}{ccc} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|=0 \\ 2(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|=0 \\ C_1 \rightarrow C_1-C_2 तथा  C_2 \rightarrow C_2-C_3 संक्रिया से:
2(a+b+c)\left|\begin{array}{lll} 0 & 0 & 1 \\ c-b & a-c & b+c \\ a-c & b-a & c+a \end{array}\right|=0 \\ R_1 के अनुसार प्रसरण करने पर:
2(a+b+c)[0\left|\begin{array}{cc} a-c & b+c \\ b-a & c+a \end{array}\right|-0\left|\begin{array}{cc} c-b & b+c \\ a-c & c+a \end{array}\right|+1\left|\begin{array}{cc} c-b & a-c \\ a-c & b-a\end{array} \right| ]=0 \\ \Rightarrow 2(a+b+c)[(c-b)(b-a)-(a-c)(a-c)]=0 \\ \Rightarrow 2(a+b+c)\left(b c-a c-b^2+a b-a^2+a c+a c-c^2\right)=0 \\ \Rightarrow 2(a+b+c)\left(-a^2-b^2-c^2+a b+b c+c a\right)=0 \\ \Rightarrow-(a+b+c)\left(2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a\right)=0 \\ \Rightarrow-(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]=0 \\ \Rightarrow a+b+c=0
या \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0 \\ a-b=0 , b-c=0 ,c-a=0 \\ \Rightarrow a=b, b=c, c=a \\ \Rightarrow a+b+c=0 या a=b=c

Determinants in Class 12

Example:5.यदि a \neq 0 हो तो समीकरण \left|\begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array}\right|=0 को हल कीजिए।
Solution: \left|\begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array}\right|=0 \\ R_1 \rightarrow R_1+R_2+R_3 संक्रिया से:
=\left|\begin{array}{ccc} 3 x+a & 3 x+a & 3 x+a \\ x & x+a & x \\ x & x & x+a \end{array}\right| \\ =(3 x+a)\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & x+a & x \\ x & x & x+a \end{array}\right| \\ C_1 \rightarrow C_1-C_2 तथा  C_2 \rightarrow C_2-C_3 संक्रिया से:
=(3 x+a)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -a & a & x \\ 0 & -a & x+a \end{array}\right| \\ R_{1} के अनुसार प्रसरण करने पर:

=(3 x+a) \left[0\left|\begin{array}{cc} a & x \\ -a & x+a \end{array}\right|-0\left|\begin{array}{cc} -a & x \\ 0 & x+a \end{array}\right|+1\left|\begin{array}{cc} -a & a \\ 0 & -a \end{array}\right| \right] \\ =(3 x+a)\left(a^2-0\right)=0 \\ \Rightarrow 3 x+a=0, a \neq 0 \\ \Rightarrow x=-\frac{a}{3}
Example:6.सिद्ध कीजिए कि \left|\begin{array}{ccc} a^2 & b c & a c+c^2 \\ a^2+a b & b^2 & a c \\ a b & b^2+b c & c^2 \end{array}\right|=4 a^2 b^2 c^2
Solution:\left|\begin{array}{ccc} a^2 & b c & a c+c^2 \\ a^2+a b & b^2 & a c \\ a b & b^2+b c & c^2 \end{array}\right|=4 a^2 b^2 c^2\\ \left|\begin{array}{ccc} a^2 & b c & a c+r^2 \\ a^2+a b & b^2 & a c \\ a b & b^2+b c & c^2 \end{array}\right| \\ \text{L.H.S.} \left|\begin{array}{ccc} a^2 & b c & a c+c^2 \\ a^2+a b & b^2 & a c \\ a b & b^2+b c & c^2 \end{array}\right| \\=abc\left|\begin{array}{ccc} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{array}\right| \\ R_1 \rightarrow R_1+R_3 संक्रिया से:
=a b c\left|\begin{array}{ccc} 2(a+b) & 2(b+c) & 2(a+c) \\ a+b & b & a \\ b & b+c & c\end{array} \right| \\ =2 a b c\left|\begin{array}{ccc} a+b & b+c & a+c \\ a+b & b & a \\ b & b+c & c \end{array} \right| \\ R_1 \rightarrow R_1 - R_2 संक्रिया से:
=2(a b c)\left|\begin{array}{ccc} 0 & c & c \\ a+b & b & a \\ b & b+c & c \end{array}\right| \\ C_2 \rightarrow C_2-C_3  संक्रिया से:
=2(abc)\left|\begin{array}{ccc} 0 & 0 & c \\a+b & b-a & a \\ b & b & c \end{array}\right| \\ R_{1} के अनुसार प्रसरण करने पर:

=2(a b c)\left[0 \left| \begin{array}{cc} b-a & a \\ b & c \end{array}\right|-0\left| \begin{array}{cc} a+b & a \\ b & c \end{array}\right|+c\left| \begin{array}{cc} a+b & b-a \\ b & b \end{array}\right|\right] \\ =2(a b c) c\left(a b+b^2-b^2+a b\right) \\ =2(abc)^2(2 a b) \\ =4a^2 b^2 c^2=R.H.S
Example:7. A^{-1}=\left[\begin{array}{ccc}3 & -1 & 4 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] और B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right] हो तो (A B)^{-1} का मान ज्ञात कीजिए।
Solution: B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]
B का सारणिक

|B|=\left|\begin{array}{ccc} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right| \\ =1\left|\begin{array}{cc} 3 & 0 \\ -2 & 1 \end{array}\right|-2\left|\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right|-2\left|\begin{array}{cc} -1 & 3 \\ 0 & -2 \end{array}\right| \\ =1(3 \times 1-0)-2(-1-0)-2(2-0) \\ =37-4-4=1 \neq 0
अतः B^{-1} का अस्तित्व है।

B_{11}=(-1)^{1+1}\left|\begin{array}{cc} 3 & 0 \\ -2 & 1 \end{array}\right|=3 \times 1-0 \times-2 \\ \Rightarrow B_{11}=3-0=3 \\ B_{12}=(-1)^{1+2}\left|\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right|=-(-1 \times 1-0 \times 0) \\ \Rightarrow B_{12}=1 \\ B_{13}=(-1)^{1+3}\left|\begin{array}{cc} -1 & 3 \\ 0 & -2 \end{array}\right|=(-1 \times-2-3 \times 0) \\ \Rightarrow B_{13}=2 \\ B_{21}=(-1)^{2+1}\left|\begin{array}{cc} 2 & -2 \\ -2 & 1 \end{array}\right|=-(2 \times 1-(-2)(-2)) \\ \Rightarrow B_{21}=-(2-4)=2 \\ B_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right|=(1 \times 1-0 \times-2) \\ \Rightarrow B_{22}=1 \\ B_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & 2 \\ 0 & -2 \end{array}\right|=-(1 \times-2-2 \times 0) \\ \Rightarrow B_{23}=2 \\ B_{31}=(-1)^{3+1}\left|\begin{array}{cc} 2 & -2 \\ 3 & 0 \end{array} \right| =(2 \times 0-3 \times-2) \\ \Rightarrow B_{31}=6 \\ B_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & -2 \\ -1 & 0 \end{array}\right|=-(1 \times 0-(-1)(-2)) \\ \Rightarrow B_{32}=2 \\ B_{33}=(-1)^{3+3} \left|\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right|=(1 \times 3-2 \times-1) \\ \Rightarrow B_{33}=3+2=5 \\ \operatorname{adj} B=\left[\begin{array}{lll} B_{11} & B_{12} & B_{13} \\ B_{21} & B_{22} & B_{23} \\ B_{31} & B_{32} & B_{33} \end{array}\right]^{\prime} \\=\left[\begin{array}{lll} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \end{array}\right]^{\prime} \\ \Rightarrow \operatorname{adj} B=\left[\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] \\ B^{-1}=\frac{1}{|B|} \operatorname{adj} B \\ =\frac{1}{1}\left[\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] \\ \Rightarrow B^{-1} =\left[\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] \\ (A B)^{-1}=B^{-1} A^{-1} \\ =\left[\begin{array}{ccc} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] \left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] \\ \left[\begin{array}{ccc} (3 \times 3+2 \times  & (3 \times-1+2 \times  & (3 \times 1+2 \times \\ -15+6 \times 5) & 6+6 \times-2) & -5+6 \times 2) \\ (1 \times 3+1 \times & (1 \times-1+1 \times  & (1 \times 1+1 \times\\ -15+2 \times 5) & 6+2 \times 2) & -5+2 \times 2) \\ (2 \times 3+2 \times & (2 \times-1+2 \times & (2 \times 1+2 \times  \\  -15+5 \times 5)  & 6+5 \times 2) & 5+5 \times 2) \end{array}\right] \\ =\left[\begin{array}{ccc} 9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10 \end{array}\right] \\ = \left[\begin{array}{ccc} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{array}\right] \\ \Rightarrow (A B)^{-1}=\left[\begin{array}{ccc} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{array}\right]
Example:8.मान लीजिए A=\left[\begin{array}{ccc} 1 & -2 & 1 \\-2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right] हो तो सत्यापित कीजिए कि

(i) [\operatorname{adj} A]^{-1}=\operatorname{adj}\left(A^{-1}\right) (ii)\left(A^{-1}\right)^{-1}=A
Solution:(i) A=\left[\begin{array}{ccc} 1 & -2 & 1 \\-2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right] 
आव्यूह A का सारणिक
|A|=\left|\begin{array}{ccc} 1 & -2 & 1 \\-2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right| \\ =1\left|\begin{array}{cc} 3 & 1 \\ 1 & 5 \end{array}\right|+2 \left|\begin{array}{cc} -2 & 1 \\ 1 & 5 \end{array}\right|+1 \left|\begin{array}{cc} -2 & 3 \\ 1 & 1 \end{array}\right|  \\=1(3 \times 5-1 \times 1)+2(-2 \times 5-1 \times 1)+1(-2 \times 1-3 \times 1) \\ =15-1+2(-10-1)+(-2-3) \\=14-22-5 \\ \Rightarrow |A|=-13 \neq 0
अतः A^{-1} का अस्तित्व है।

A_{11}=(-1)^{1+1}\left|\begin{array}{ll} 3 & 1 \\ 1 & 5 \end{array}\right|=(3 \times 5-1 \times 1) \\ \Rightarrow A_{11}=15-1=14 \\ A_{12}=(-1)^{1+2}\left|\begin{array}{cc} -2 & 1 \\ 1 & 5 \end{array}\right| =-(-2 \times 5-1 \times 1) \\ \Rightarrow A_{12}=-(-10-1)=11 \\ A_{13}=(-1)^{1+3} \left|\begin{array}{cc} -2 & 3 \\ 1 & 1 \end{array}\right|=(-2 \times 1-3 \times 1) \\ \Rightarrow A_{13}=(-2-3)=-5 \\ A_{21}=(-1)^{2+1}\left|\begin{array}{cc} -2 & 1 \\ 1 & 5 \end{array}\right|=-(-2 \times 5-1 \times 1)\\ \Rightarrow A_{21}=-1(-10-1)=11 \\ A_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & 1 \\ 1 & 5 \end{array}\right|=(1 \times 5-1 \times 1) \\ \Rightarrow A_{22}=5-1=4\\ A_{23}=(-1)^{2+3} \left|\begin{array}{cc} 1 & -2 \\ 1 & 1 \end{array}\right|=-(1 \times 1-1 \times -2) \\ \Rightarrow A_{23}=-(1+2)=-3 \\ A_{31}=(-1)^{3+1}\left| \begin{array}{cc} -2 & 1 \\ 3 & 1 \end{array}\right|=(-2 \times 1-1 \times 3) \\ \Rightarrow A_{31}=(-2-3)=-5 \\ A_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 1 \\ -2 & 1 \end{array}\right|=-(1 \times 1-1 \times-2) \\ \Rightarrow A_{32}=-(1+2)=-3 \\ A_{33}=(-1)^{3+3} \left|\begin{array}{cc} 1 & -2 \\-2 & 3 \end{array} \right|=(1\times 3-(-2)(-2)) \\ \Rightarrow A_{33}=(3-4)=-1 \\  \operatorname{adj} A=\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array} \right]^{\prime} \\ =\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array} \right]^{\prime} \\ \Rightarrow \operatorname{adj} A= \left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right] \\ A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\ =\frac{1}{-13}\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right] \\ \operatorname{adj} A=B=\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right] \\ \operatorname{adj}|A|=14(-4-9)-11(-11-15)-5(-33+20)=169 \\ B_{11}=(-1)^{1+1} \left|\begin{array}{cc} 4 & -3 \\ -3 & -1 \end{array}\right|=(4 \times-1-(-3)(-3)) \\ \Rightarrow B_{11}=(-4-9)=-13 \\ B_{12}=(-1)^{1+2}\left|\begin{array}{cc} 11 & -3 \\ -5 & -1\end{array}\right|=-(11 \times -1-(-5)(-3)) \\ \Rightarrow B_{12}=-(-11-15)=26 \\ B_{13}=(-1)^{1+3}\left|\begin{array}{cc} 11 & 4 \\ -5 & -3 \end{array}\right|=(11 \times -3-4 \times -5) \\ \Rightarrow B_{13}=(-33+20)=-13 \\ B_{21}=(-1)^{2+1} \left|\begin{array}{cc} 11 & -5 \\-3 & -1 \end{array}\right|=-(11 \times -1-(-5)(-3)) \\ \Rightarrow B_{21} =-(-11-15)=26 \\ B_{22} =(-1)^{2+2}\left|\begin{array}{cc} 14 & -5 \\ -5 & -1 \end{array}\right|=(14 \times -1-(-5)(-5)) \\ \Rightarrow B_{22} =(-14-25)=-39 \\ B_{23} =(-1)^{2+3}\left|\begin{array}{cc} 14 & 11 \\ -5 & -3 \end{array}\right|=-(14 \times-3-11 \times -5) \\ =-(-42+55) \\ B_{23}=-13 \\ B_{31}=(-1)^{3+1} \left|\begin{array}{ll} 11 & -5 \\ 4 & -3 \end{array}\right|=(11 \times -3-4 \times -5)\\ \Rightarrow B_{31}=-33+20=-13 \\ B_{32}=(-1)^{3+2}\left|\begin{array}{ll} 14 & -5 \\ 11 & -3 \end{array}\right|=-(14 \times-3+1 \times-5) \\ \Rightarrow B_{32} =-(-42+55)=-13 \\ B_{33} =(-1)^{3+3}\left|\begin{array}{ll} 14 & 11 \\ 11 & -4 \end{array}\right|=(14 \times 4-11 \times 11) \\ \Rightarrow B_{33}=(56-121)=-65 \\ \operatorname{adj}(\operatorname{adj} A) =\left[\begin{array}{lll} B_{11} & B_{12} & B_{13} \\ B_{21} & B_{22} & B_{23} \\ B_{31} & B_{32} & B_{33} \end{array}\right]^{\prime} \\ =\left[\begin{array}{ccc} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right]^{\prime} \\ \Rightarrow \operatorname{adj} (\operatorname{adj} A) =\left[\begin{array}{ccc} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right] \\ (\operatorname{adj} A)^{-1} =\frac{1}{|\operatorname{adj} A|} \operatorname{adj}(\operatorname{adj} A) \\ =\frac{1}{169}\left[\begin{array}{ccc} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right] \\ =\frac{13}{169}\left[\begin{array}{ccc} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right] \\ \Rightarrow (\operatorname{adj} A)^{-1}=\frac{1}{13} \left[\begin{array}{ccc} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right] \cdots(1)  
माना \Rightarrow A^{-1}=C =-\frac{1}{13}\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right] \\ \Rightarrow c=\left[\begin{array}{ccc} -\frac{14}{13} & \frac{-11}{13} & \frac{5}{13} \\ -\frac{11}{13} & \frac{-4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{array}\right] 
माना A^{-1} का सारणिक

(1) व (2) से

(\operatorname{adj} A)^{-1}=\left(\operatorname{adj} A^{-1}\right)
(ii) \left(A^{-1}\right)^{-1}=A
माना D=A^{-1} =-\frac{1}{13}\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\-5 & -3 & -1 \end{array} \right] \\ D=\left[\begin{array}{ccc} -\frac{14}{13} & \frac{-11}{13} & \frac{5}{13} \\ -\frac{11}{13} & \frac{-4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{array}\right] \\ A^{-1} का सारणिक

|A^{-1}|=|D|=\left|\begin{array}{ccc} \frac{-14}{13}& -\frac{11}{13} & \frac{5}{13} \\ -\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{array}\right| \\ \Rightarrow \left|A^{-1}\right|=-\frac{1}{13} \neq 0 \\ D_{11}=(-1)^{1+1}\left|\begin{array}{cc} -\frac{4}{13} & \frac{3}{13} \\ \frac{3}{13} & \frac{1}{13} \end{array}\right|=\left(\frac{-4}{13} \times \frac{1}{13}-\frac{3}{13} \times \frac{3}{13}\right) \\ D_{11}=-\frac{4}{169}-\frac{9}{169}=\frac{-13}{169}=-\frac{1}{13}
इसी प्रकार D_{12}=C_{12}=\frac{2}{13}, D_{13}=C_{13}=-\frac{1}{13} \\ D_{21}=C_{21}=\frac{2}{13}, D_{22}=C_{22}=\frac{-3}{13}, D_{23}=C_{23}=-\frac{1}{13} \\ D_{31}=C_{31}=-\frac{1}{13}, D_{32}=C_{32}=-\frac{1}{13}, D_{33}=C_{33}=-\frac{5}{13} \\ \operatorname{adj}\left(A^{-1}\right)=\frac{1}{13} \left[\begin{array}{ccc} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -4 & -5 \end{array}\right][(2) से] 

(A)^{-1} =\frac{1}{|\bar{A}|} \operatorname{adj}(A^{-1}) \\ =\frac{1}{-\frac{1}{13}} \times \frac{1}{13} \left[\begin{array}{ccc} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right] \\ =-1\left[ \begin{array}{ccc} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right] \\ =\left[\begin{array}{ccc} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right] \\ \Rightarrow(A)^{-1}=A
Example:9. \left|\begin{array}{ccc} x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right| का मान ज्ञात कीजिए।
Solution:माना |A|=\left|\begin{array}{ccc} x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right| \\ R_1 \rightarrow R_1+R_2+R_3 संक्रिया से:
|A|=\left|\begin{array}{ccc}2(x+y) & 2(x+y) & 2(x+y) \\y & x+y & x \\ x+y & x & y \end{array}\right| \\ =2(x+y)\left|\begin{array}{ccc}1 & 1 & 1 \\y & x+y & x \\x+y & x & y\end{array}\right| \\ C_{1} \rightarrow C_1-C_2 तथा C_2 \rightarrow C_2-C_3  संक्रिया से:

=2(x+y)\left|\begin{array}{ccc}0 & 0 & 1 \\-x & y & x \\y & x-y & y\end{array}\right| \\ =2(x+y) \left[0\left|\begin{array}{cc}y & x \\x-y & y\end{array}\right|-0\left|\begin{array}{cc}-x & x \\y & y \end{array}\right|+1\left|\begin{array}{cc}-x & y \\y & x- y\end{array}\right|\right] \\ =2(x+y)\left(-x^2+x y-y^2\right) \\ =-2(x+y)\left(x^2-x y+y^2\right) \\ |A| =-2\left(x^3+y^3\right)
Example:10. |A|=\left|\begin{array}{ccc} 1 & x & y \\1 & x+y & y \\1 & x & x+y\end{array}\right| का मान ज्ञात कीजिए।
Solution:माना |A|=|A|=\left|\begin{array}{ccc} 1 & x & y \\1 & x+y & y \\1 & x & x+y \end{array}\right| \\ R_1 \rightarrow R_1-R_2 तथा R_2 \rightarrow R_2-R_3 संक्रिया से:

=\left|\begin{array}{ccc}0 & -y & 0 \\0 & y & -x \\ 1 & x & x+y \end{array}\right| \\ =0\left|\begin{array}{ll}y & -x \\x & x+y \end{array}\right|+y\left|\begin{array}{cc} 0 & -x \\ 1 & x+y \end{array}\right| +0\left|\begin{array}{ll} 0 &y \\ 1 & x \end{array}\right| \\ =y(x) \\ \Rightarrow |A|=x y
उपर्युक्त उदाहरणों के द्वारा कक्षा 12 में सारणिक (Determinants in Class 12),सारणिक कक्षा 12 (Determinants Class 12) को समझ सकते हैं।

3.कक्षा 12 में सारणिक के सवाल (Determinants in Class 12 Questions):

Determinants in Class 12

(1.)यदि A=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] तो A^{-1} ज्ञात कीजिए।तत्पश्चात इसकी सहायता से सिद्ध कीजिए A^2-4 A-5 I=0
(2.)मैट्रिक्स A ज्ञात कीजिए जबकि:

\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right] A\left[\begin{array}{ll} 4 & 7 \\ 3 & 5 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]
उत्तर (Answers): (1.) A^{-1}=\frac{1}{5}\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\2 & 2 & -3 \end{array}\right] 
(2.) A=\left[\begin{array}{cc} 21 & -29 \\-13 & 18 \end{array}\right]
उपर्युक्त सवालों को हल करने पर कक्षा 12 में सारणिक (Determinants in Class 12),सारणिक कक्षा 12 (Determinants Class 12) को ठीक से समझ सकते हैं।

Also Read This Article:- Adjoint and Inverse of Matrix 12th

4.कक्षा 12 में सारणिक (Frequently Asked Questions Related to Determinants in Class 12),सारणिक कक्षा 12 (Determinants Class 12) से सम्बन्धित अक्सर पूछे जाने वाले प्रश्न:

कक्षा 12 में सारणिक (Determinants in Class 12) के इस आर्टिकल में सारणिकों के गुणधर्म,
उपसारणिक और सहखण्ड,आव्यूह के सहखण्डों और व्युत्क्रम,सारणिकों और आव्यूहों के
अनुप्रयोग पर आधारित सवालों को हल करेंगे।

Satyam Mathematics

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