1.आव्यूह के सहखण्डज और व्युत्क्रम कक्षा 12 (Adjoint and Inverse of Matrix Class 12),आव्यूह के सहखण्डज और व्युत्क्रम (Adjoint and Inverse of Matrix):

आव्यूह के सहखण्डज और व्युत्क्रम कक्षा 12 (Adjoint and Inverse of Matrix Class 12) का अध्ययन करने के साथ-साथ इस आर्टिकल में हम आव्यूह के व्युत्क्रम के अस्तित्व के लिए शर्तों की भी व्याख्या 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:- Minor and Co-factor Class 12

2.आव्यूह के सहखण्डज और व्युत्क्रम कक्षा 12 पर आधारित उदाहरण (Examples Based on Adjoint and Inverse of Matrix Class 12):

प्रश्न 1 से 2 में प्रत्येक आव्यूह का सहखण्डज (adjoint) ज्ञात कीजिए।
Example:1. $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
Solution:माना $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
आव्यूह A का सारणिक $|A|=\left|\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right| \\ \Rightarrow|A|=1 \times 4-2 \times 3=4-6=-2 \neq 0 \\ A_{11}=(-1)^{1+1} 4 =4, A_{12}=(-1)^{1+2} 3=-3 \\ A_{21}=(-1)^{2+1} =-2, A_{22}=(-1)^{2+2} 1=1 \\ \text { adjoint } A=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array} \right]^{\prime} \\ =\left[\begin{array}{cc}4 & -3 \\ -2 & 1\end{array}\right]^{\prime} \\ \Rightarrow \text { adjoint } A=\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]$
Example:2. $\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$
Solution:माना $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$
आव्यूह A का सारणिक $|A|=\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{array} \right] \\ |A|=1(3 \times 1-5 \times 0)+1(2 \times 1-5 \times-2)+2(2 \times 0-3 \times-2) \\ =3+(2+10)+2 \times 6 \\ =3+12+12 \\ \Rightarrow|A|=17 \neq 0 \\ A_{11}=(-1)^{1+1}\left|\begin{array}{ll}3 & 5 \\ 0 & 1 \end{array}\right|=3 \times 1-5 \times 0 \\ \Rightarrow A_{11}=3 \\ A_{12}=(-1)^{1+2} \left|\begin{array}{ll}2 & 5 \\ -2 & 1\end{array}\right|=(-1)(2 \times 1-5 \times-2) \\ \Rightarrow A_{12}=-1(2+10) \\ \Rightarrow A_{12}=-12 \\ A_{13}=(-1)^{1+3}\left|\begin{array}{ll}2 & 3 \\ -2 & 0\end{array}\right|=2 \times 0-3 \times-2 \\ \Rightarrow A_{13}=6 \\ A_{24}=(-1)^{2+1}\left|\begin{array}{cc} -1 & 2 \\ 0 & 1 \end{array}\right|=-1(-1 \times 1-2 \times 0) \\ \Rightarrow A_{21}=1 \\ A_{22}=(-1)^{2+2} \left|\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right|=\mid \times 1-2 \times-2 \\ \Rightarrow A_{22}=1+4=5 \\ A_{23}=(-1)^{2+3}\left|\begin{array}{cc} \mid-1 & -1 \end{array}\right|=(-1)(1 \times 0-(-1)(-2)) \\ \Rightarrow A_{23}=+2 \\ A_{31}=(-1)^{3+1}\left|\begin{array}{cc} -1 & 2 \\ 3 & 5 \end{array}\right|=-(-1 \times 5-2 \times 3) \\ \Rightarrow A_{31}=-11 \\ A_{32}=(-1)^{3+2} \left|\begin{array}{cc} 1 & 2 \\ 2 & 5 \end{array}\right|=(-1)(1 \times 5-2 \times 2) \\ \Rightarrow A_{32}=-1 \\ A_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right|=(1 \times 3-2 \times-1) \\ \Rightarrow A_{33}=5 \\ \therefore \text { adjoint } A=\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}\right]^{\prime} \\ =\left[\begin{array}{ccc} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{array}\right]^{\prime} \\ \operatorname{adj} A=\left[\begin{array}{ccc} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5\end{array}\right]$

उपर्युक्त प्रश्नों के उत्तर द्वारा आव्यूह के सहखण्डज और व्युत्क्रम कक्षा 12 (Adjoint and Inverse of Matrix Class 12),आव्यूह के सहखण्डज और व्युत्क्रम (Adjoint and Inverse of Matrix) के बारे में ओर अधिक जानकारी प्राप्त कर सकते हैं।
प्रश्न 3 और 4 में सत्यापित कीजिए कि है।

$A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| \cdot I$
Example:3. $\left[\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right]$
Solution:माना $A=\left[\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right]$
आव्यूह A का सारणिक

$|A|=\left|\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right| \\ \Rightarrow|A|=2 \times -6-3 \times-4=-12+12=0 \\ A_{11}=(-1)^{1+1}(-6)=-6 \\ A_{12}=(-1)^{1+2}(-4)=4 \\ A_{21}=(-1)^{2+1}(3)=-3 \\ A_{22}=(-1)^{2+2}(2)=2 \\ \operatorname{ddj} A=\left[\begin{array}{ll} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right]^{\prime} \\ =\left[\begin{array}{ll} -6 & 4 \\ -3 & 2 \end{array}\right]^{\prime} \\ \operatorname{adj} A=\left[\begin{array}{cc} -6 & -3 \\ 4 & 2 \end{array}\right] \\ A(\operatorname{adj} A) =\left[\begin{array}{cc} 2 & 3 \\ -4 & -6\end{array}\right]\left[\begin{array}{cc} -6 & -3 \\ 4 & 2 \end{array}\right] \\ =\left[\begin{array}{cc}2 \times-6+3 \times 4 & 2 \times-3+3 \times 2 \\-4 \times-6-6 \times 4 & -4 \times-3-6 \times 2 \end{array}\right] \\ =\left[\begin{array}{cc} -12+12 & -6+6 \\ 24-24 & 12-12 \end{array}\right] \\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ =|A|\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \Rightarrow A(\operatorname{adj} A)=|A| I \cdots(2)\\ (\operatorname{adj} A)(A)=\left[\begin{array}{cc} -6 & -3 \\ 4 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ -4 & -6 \end{array}\right] \\ =\left[\begin{array}{cc} -6 \times 2-3 \times-4 & -6 \times 3-3 \times-6 \\ 4 \times 2+2 \times-4 & 4 \times 3+2 \times-6 \end{array}\right] \\ =\left[\begin{array}{cc} -12+12 & -18+18 \\ 8-8 & 12-12 \end{array}\right] \\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \Rightarrow(\operatorname{adj} A) A=|A| I \cdots(2)$
(1) व (2) से:

$A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$
Example:4. $\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$
Solution:माना $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$
आव्यूह A का सारणिक

$|A|=\left|\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right| \\ =1(0 \times 3-0 \times-2)+1(3 \times 3-1 \times-2)+2(3 \times 0-0 \times 1) \\ =0+11+0 \\ \Rightarrow|A|=11 \\ A_{11}=(-1)^{1+1}\left|\begin{array}{cc}0 & -2 \\ 0 & 3\end{array}\right|=0 \\ \Rightarrow A_{11}=0 \\ A_{12}=(-1)^{1+2}\left|\begin{array}{cc}3 & -2 \\ 1 & 3\end{array}\right|=-1(3 \times 3-1 \times-2) \\ \Rightarrow A_{12}=-11 \\ A_{13}=(-1)^{1+3}\left|\begin{array}{ll}3 & 0 \\ 1 & 0\end{array}\right|=0 \\ \Rightarrow A_{13}=0 \\ A_{21}=(-1)^{2+1}\left|\begin{array}{cc}-1 & 2 \\ 0 & 3\end{array}\right|=-1(-1 \times 3-2 \times 0) \\ \Rightarrow A_{21}=3 \\ A_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 2 \\ 1 & 3 \end{array} \right|=1\times 3-2 \times 1 \\ \Rightarrow A_{22}=1 \\ A_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array}\right|=-1(1 \times 0-1 \times-1) \\ \Rightarrow A_{23}=-1 \\ A_{31}=(-1)^{3+1} \left|\begin{array}{cc} -1 & 2 \\ 0 & -2 \end{array}\right|=-1 \times-2-2 \times 0 \\ \Rightarrow A_{31}=2 \\ A_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 2 \\ 3 & -2 \end{array}\right|=-1(1 \times-2-2 \times 3) \\ \Rightarrow A_{32}=8 \\ A_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & -1 \\ 3 & 0 \end{array}\right|=(1 \times 0-3 \times-1) \\ \Rightarrow A_{33}=3 \\ \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} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{array}\right]^{\prime} \\ =\left[\begin{array}{ccc} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{array}\right] \\ A(\operatorname{adj} A) =\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right]\left[\begin{array}{ccc} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{array} \right] \\ =\left[\begin{array}{lll} (1 \times 0-1 \times & (1 \times 3-1 \times & (1 \times 2-1 \times \\ -11+2 \times 0) & 1+2 \times-1) & 8+2 \times 3) \\ (3 \times 0+0 \times & (3 \times 3+0 \times & (3 \times 2+0 \times \\ 11-2 \times 0) & 1-2 \times-1) & 8-2 \times 3) \\ (1 \times 0+0 \times & (1 \times 3+0 \times & (1 \times 2+0 \times \\ -11+3 \times 0) & 1+3 \times-1) & 3+3 \times 3) \end{array}\right]\\=\left[\begin{array}{lll}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}\right] \\ =11\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ \Rightarrow A(\operatorname{adj} A)=|A| I \ldots(1) \\ (\operatorname{adj} A)(A)=\left[\begin{array}{ccc} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{array}\right]\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right] \\ =\left[\begin{array}{ccc} (0 \times 1+3 & (0 \times-1+3 & (0 \times 2+3 \\ \times 3+2 \times 1) & \times 0+2 \times 0) & \times-2+2 \times 3) \\ (-11 \times 1+1 & (-11 \times -1+1 & (1 \times 2+1 \\ \times 3+8 \times 1) & \times 0+8 \times 0) & \times 2+8 \times 3) \\ (0 \times 1-1 & (0 \times -1-1 & (0 \times 2-1 \\ \times 3+3 \times 1) & \times 0+3 \times 0) & \times-2+3 \times 3) \end{array}\right] \\ =\left[\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}\right] \\ =11\left[\begin{array}{lll} 1 & 0 & 0 \\0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ \Rightarrow (\operatorname{adj} A) A=|A| I \cdots(2)$
(1) व (2) से:

$A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| \cdot I$
प्रश्न 5 से 11 में दिए गए प्रत्येक आव्यूहों के व्युत्क्रम (जिनका अस्तित्व हो) ज्ञात कीजिए।
Example:5. $\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$
Solution:माना $A=\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$
आव्यूह A का सारणिक $|A|=\left|\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right| =2 \times 3-4 \times-2 \\ \Rightarrow|A|=6+8=14 \neq 0$
अतः $A^{-1}$ का अस्तित्व है।

$A_{11}=(-1)^{1+1} 3=3 \\ A_{12}=(-1)^{1+2} 4=-4 \\ A_{21}=(-1)^{2+1}(-2)=2 \\A_{22}=(-1)^{2+2}=2 \\ \operatorname{adj} A =\left[\begin{array}{ll} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right]^{\prime} \\ =\left[\begin{array}{cc} 3 & -4 \\ 2 & 2 \end{array}\right]^{\prime} \\ \Rightarrow \operatorname{adj} A=\left[\begin{array}{ll} 3 & 2 \\ -4 & 2 \end{array}\right] \\ A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\ \Rightarrow A^{-1}=\frac{1}{14}\left[\begin{array}{ll} 3 & 2 \\ -4 & 2 \end{array}\right]$
Example:6. $\left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]$
Solution:माना $A=\left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]$
आव्यूह A का सारणिक $|A|=\left|\begin{array}{cc} -1 & 5 \\ -3 & 2 \end{array}\right| \\ \Rightarrow|A|=-1 \times 2-5 \times-3=-2+15=13 \neq 0$
अतः $A^{-1}$ का अस्तित्व है।

$A_{11}=(-1)^{1+1} 2=2 \\ A_{12}=(-1)^{1+2}(-3)=3 \\ A_{21}=(-1)^{2+1} 5=-5 \\ A_{22}=(-1)^{2+2}(-1)=-1 \\ \operatorname{adj} A=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22} \end{array} \right]^{\prime} \\ =\left[\begin{array}{cc} 2 & 3 \\ -5 & -1 \end{array}\right] \\ \Rightarrow \operatorname{adj} A=\left[\begin{array}{ll} 2 & -5 \\ 3 & -1 \end{array}\right] \\ A^{-1} =\frac{1}{|A|} \operatorname{adj} A \\=\frac{1}{13}\left[\begin{array}{ll} 2 & -5 \\ 3 & -1 \end{array}\right]$

उपर्युक्त प्रश्नों के उत्तर द्वारा आव्यूह के सहखण्डज और व्युत्क्रम कक्षा 12 (Adjoint and Inverse of Matrix Class 12),आव्यूह के सहखण्डज और व्युत्क्रम (Adjoint and Inverse of Matrix) के बारे में ओर अधिक जानकारी प्राप्त कर सकते हैं।

Example:7. $\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$
Solution:माना $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$
आव्यूह A का सारणिक $|A|=\left|\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right|$
प्रथम स्तम्भ के अनुसार प्रसरण करने पर:
=$1(2 \times 5-4 \times 0)-0(2 \times 5-3 \times 0)+0(2 \times 4-3 \times 2) \\ \Rightarrow|A|=10 \neq 0$
अतः $A^{-1}$ का अस्तित्व है।

$A_{11}=(-1)^{1+1}\left|\begin{array}{ll} 2 & 4 \\ 0 & 5 \end{array}\right|=2 \times 5-4 \times 0 \\ \Rightarrow A_{11}=10 \\ A_{12}=(-1)^{1+2}\left|\begin{array}{ll} 0 & 4 \\ 0 & 5 \end{array}\right|=-1(0 \times 5-0 \times 44) \\ A_{12}=0 \\ A_{13}=(-1)^{1+3}\left|\begin{array}{ll} 0 & 2 \\ 0 & 0 \end{array} \right| =(0 \times 0.0 \times 2) \\ \Rightarrow A_{13}=0 \\ A_{21}=(-1)^{2+1}\left|\begin{array}{ll} 2 & 3 \\ 0 & 5 \end{array}\right|=-1(2 \times 5-0 \times 3) \\ \Rightarrow A_{21}=-10 \\ A_{22}=(-1)^{2+2} \left|\begin{array}{ll} 1 & 3 \\ 0 & 5 \end{array}\right|=1 \times 5-3 \times 0 \\ \Rightarrow A_{22}=5 \\ A_{23}=(-1)^{2+3}\left|\begin{array}{ll} 1 & 2 \\ 0 & 0 \end{array}\right|=-1(1 \times 0-2 \times 0) \\ \Rightarrow A_{23}=0 \\ A_{31}=(-1)^{3+1}\left|\begin{array}{ll} 2 & 3 \\ 2 & 4 \end{array}\right|=2 \times 4-3 \times 2 \\ \Rightarrow A_{31}=2 \\ A_{32}=(-1)^{3+2}\left|\begin{array}{ll} 1 & 3 \\ 0 & 4 \end{array} \right|=-1(1 \times 4-3 \times 0) \\ \Rightarrow A_{32}=-4 \\ A_{33}=(-1)^{3+3} \left|\begin{array}{ll} 1 & 2 \\ 0 & 2 \end{array}\right|=(1 \times 2-0 \times 2) \\ \Rightarrow A_{33}=2 \\ \operatorname{adj} A=\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}\right]^{\prime} \\ =\left[\begin{array}{ccc} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{array}\right]^{\prime} \\ \Rightarrow \operatorname{adj} A=\left[\begin{array}{ccc} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right] \\ A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\ =\frac{1}{10} \left[\begin{array}{ccc} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right]$ 
Example:8. $\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$
Solution:माना $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$ 
आव्यूह A का सारणिक $|A|=\left|\begin{array}{ccc}1 & 0 & 0 \\3 & 3 & 0 \\5 & 2 & -1\end{array}\right|$ 
प्रथम पंक्ति के अनुसार प्रसरण करने पर:
$0=1(3 \times-1-2 \times 0)+0(3 \times-1-5 \times 0)+0(3 \times 2-3 \times 5) \\ \Rightarrow|A| =-3 \neq 0$ 
अतः $A^{-1}$  का अस्तित्व है।

$A_{11}=(-1)^{1+1}\left|\begin{array}{cc} 3 & 0 \\ 2 & -1 \end{array}\right|=(3 \times-1-0 \times 2) \\ \Rightarrow A_{11}=-3 \\ A_{12}=(-1)^{1+2}\left|\begin{array}{cc}3 & 0 \\5 & -1 \end{array}\right|=-1(3 \times-1-0 \times 5) \\ \Rightarrow A_{12}=3 \\ A_{13}=(-1)^{1+3}\left|\begin{array}{cc} 3 & 3 \\ 5 & 2 \end{array}\right|=(3 \times 2-5 \times 3) \\ \Rightarrow A_{13}=-9 \\ A_{21}=(-1)^{2+1}\left|\begin{array}{ll}0 & 0 \\ 2 & -1 \end{array}\right|=-1(0 \times-1-0 \times 2) \\ \Rightarrow A_{21}=0 \\ A_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & 0 \\ 5 & -1 \end{array}\right|=(1 \times-1-0 \times 5) \\ \Rightarrow A_{22}=-1 \\ A_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & 0 \\ 5 & 2 \end{array}\right|=-1(1 \times 2-0 \times 5) \\ \Rightarrow A_{23}=-2 \\ A_{31}=(-1)^{3+1}\left|\begin{array}{ll} 0 & 0 \\ 3 & 0 \end{array} \right|=0 \times 0-0 \times 3 \\ \Rightarrow A_{31}=0 \\ A_{32}=(-1)^{3+2}\left|\begin{array}{ll} 1 & 0 \\ 3 & 0 \end{array}\right|=1 \times 0-0 \times 3 \\ \Rightarrow A_{32}=0 \\ A_{33}=(-1)^{3+3} \left|\begin{array}{ll} 1 & 0 \\ 3 & 3 \end{array}\right|=1 \times 3-0 \times 3  \\ \Rightarrow A_{33}=3 \\ \\ \operatorname{adj} A=\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}\right]^{\prime} \\ =\left[\begin{array}{ccc} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{array}\right]^{\prime} \\ \Rightarrow \operatorname{adj} A=\left[\begin{array}{ccc} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{array}\right] \\ A^{-1}= \frac{1}{|A|} \operatorname{adj} A \\ \Rightarrow A^{-1}=-\frac{1}{3}\left[\begin{array}{ccc} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{array}\right]$ 
Example:9. $\left[\begin{array}{ccc} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{array}\right]$ 
Solution:माना $A=\left[\begin{array}{ccc} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{array}\right]$ 
आव्यूह A का सारणिक $|A|=\left|\begin{array}{ccc} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{array}\right|$ 
द्वितीय पंक्ति के अनुसार प्रसरण करने पर:
=-4(1×1-3×2)-1(2×1-3×-7)+0(2×2-1×-7)
=-4(1-6)-1(2+21)+0
=$20-23 \\ \Rightarrow |A|=-3 \neq 0$ 
अतः $A^{-1}$ का अस्तित्व है।

$A_{11}=(-1)^{1+1}\left|\begin{array}{rr} -1 & 0 \\ 2 & 1 \end{array}\right|=(-1 \times 1-0 \times 2) \\ \Rightarrow A_{11}=-1 \\ A_{12}=(-1)^{1+2}\left|\begin{array}{cc} 4 & 0 \\ -7 & 1 \end{array}\right|=-1(4 \times 1-0 \times-7) \\ \Rightarrow A_{12}=-4 \\ A_{13}=(-1)^{1+3}\left|\begin{array}{cc} 4 & -1 \\ -7 & 2 \end{array}\right|=(4 \times 2-(-1)(-7)) \\ \Rightarrow A_{13}=8-7=1 \\ A_{21}=(-1)^{2+1} \left| \begin{array}{ll} 1 & 3 \\ 2 & 1 \end{array}\right|=-1(1 \times 1-3 \times 2) \\ \Rightarrow A_{21}=-(1-6)=5 \\ A_{22}=(-1)^{2+2}\left|\begin{array}{cc} 2 & 3 \\ -7 & 1 \end{array}\right|=(2 \times 1-3 \times-7) \\ \Rightarrow A_{22}=23 \\ A_{23}=(-1)^{2+3}\left|\begin{array}{ll} 2 & 1 \\ -7 & 2 \end{array}\right|=-1(2 \times 2-1 \times-7) \\ \Rightarrow A_{23}=-1(4+7)=-11 \\ A_{31}=(-1)^{3+1}\left|\begin{array}{cc} 1 & 3 \\ -1 & 0 \end{array}\right|=1 \times 0-3 \times-1 \\ \Rightarrow A_{31}=3 \\ A_{32}=(-1)^{3+2} \left|\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right|=-1(2 \times 0-3 \times 4) \\ \Rightarrow A_{32}=12 \\ A_{33}=(-1)^{3+3}\left|\begin{array}{cc} 2 & 1 \\ 4 & -1 \end{array}\right|=2 \times-1-1 \times 4 \\ \Rightarrow A_{33}=-6 \\ \operatorname{adj} A=\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}\right]^{\prime} \\ =\left[\begin{array}{ccc} -1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6 \end{array}\right]^{\prime} \\ \Rightarrow \operatorname{adj} A=\left[\begin{array}{ccc} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{array}\right] \\ A^{-1}= \frac{1}{|A|} \operatorname{adj} A \\ \Rightarrow A^{-1}=-\frac{1}{3}\left[\begin{array}{ccc} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{array}\right]$ 
Example:10. $\left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$ 
Solution:माना $A=\left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$ 
आव्यूह A का सारणिक $|A|=\left|\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right|$
द्वितीय पंक्ति के अनुसार प्रसरण करने पर:
= $-0\left|\begin{array}{ll} -1 & 2 \\ -2 & 4 \end{array}\right|+2\left|\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right|+3\left|\begin{array}{cc} 1 & -1 \\ 3 & -2 \end{array}\right| \\ =2(1 \times 4-2 \times 3)+3(1 \times-2-3 \times-1) \\ =2(4-6)+3(-2+3) \\ =-4+3 \\ \Rightarrow|A| =-1 \neq 0$
अतः $A^{-1}$ का अस्तित्व है।

$A_{11}=(-1)^{1+1}\left|\begin{array}{cc} 2 & -3 \\ -2 & 4 \end{array}\right|=(2 \times 4-(-3)(-2)) \\ \Rightarrow A_{11}=2 \\ A_{12}=(-1)^{1+2}\left|\begin{array}{cc} 0 & -3 \\ 3 & 4 \end{array}\right|=-1(0 \times 4-3 \times-3) \\ \Rightarrow A_{12}=-9 \\ A_{13}=(-1)^{1+3}\left|\begin{array}{cc} 0 & 2 \\ 3 & -2 \end{array}\right|=(0 \times-2-2 \times 3) \\ \Rightarrow A_{13}=-6 \\ A_{21}=(-1)^{2+1}\left|\begin{array}{cc} -1 & 2 \\ -2 & 4\end{array}\right|=-1(-1 \times 4-2 \times-2) \\ \Rightarrow A_{21}=-1(-4+4)=0 \\ A_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right|=1 \times 4-2 \times 3 \\ \Rightarrow A_{22}=-2 \\ A_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & -1 \\ 3 & -2 \end{array}\right|=-1(1 \times-2-3 \times-1) \\ \Rightarrow A_{23}=-1(-2+3)=-1 \\ A_{31}=(-1)^{3+1}\left|\begin{array}{cc} -1 & 2 \\ 2 & -3 \end{array}\right|=-1 \times-3-2 \times 2 \\ \Rightarrow A_{31}=-1 \\ A_{32}=(-1)^{3+2} \left|\begin{array}{cc} 1 & 2 \\ 0 & -3 \end{array}\right|=-1(1 \times-3-2 \times 0) \\ \Rightarrow A_{32}=3 \\ A_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right|=1 \times 2-0 \times-1 \\ \Rightarrow A_{33}=2 \\ \operatorname{adj} A=\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}\right]^{\prime} \\ =\left[\begin{array}{ccc} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{array}\right] \\ \Rightarrow \operatorname{adj} A= \left[\begin{array}{ccc} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{array}\right] \\ A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\ =-1\left[\begin{array}{ccc} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{array} \right] \\ \Rightarrow A^{-1}=\left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array} \right] \\ A^{-1}=\frac{1}{|A|} \operatorname{ady} A \\ =-1\left[\begin{array}{ccc} 2 & 0 & -1 \\ -5 & -2 & 3 \\ -6 & -1 & 2 \end{array}\right]$
Example:11. $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right]$
Solution:माना $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right]$
आव्यूह A का सारणिक $|A|=\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right|$
प्रथम पंक्ति के अनुसार प्रसरण करने पर:

=$1\left(-\cos ^2 \alpha-\sin ^2 \alpha\right)-0(0 \times -\cos \alpha-0 \times \sin \alpha)+0(0 \times \sin \alpha-0 \times \cos \alpha) \\=-1\left(\sin ^2 \alpha+\cos ^2 \alpha\right) \\ \Rightarrow |A|=-1 \neq 0$
अतः $A^{-1}$ का अस्तित्व है।

$A_{11}=(-1)^{1+1}\left|\begin{array}{lc}\cos \alpha & \sin \alpha \\\sin \alpha & -\cos \alpha \end{array} \right| =(\cos \alpha \times -\cos \alpha-\sin \alpha \times \sin \alpha) \\ \Rightarrow A_{11}=-\left(\cos ^2 \alpha+\sin ^2 \alpha\right)=-1 \\A_{12}=(-1)^{1+2}\left|\begin{array}{cc}0 & \sin \alpha \\0 & -\cos \alpha\end{array}\right|=-1(0 \times-\cos \alpha-\sin \alpha \times 0) \\\Rightarrow A_{12}=0 \\ A_{13}=(-1)^{1+3}\left|\begin{array}{cc} 0 & \cos \alpha \\ 0 & \sin \alpha \end{array}\right|=0 \times \sin \alpha-0 \times \cos \alpha \\ \Rightarrow A_{13}=0 \\ A_{21}=(-1)^{2+1}\left|\begin{array}{cc} 0 & 0 \\ \sin \alpha & -\cos \alpha \end{array}\right|=0 \times-\cos \alpha-0 \times \sin \alpha \\ \Rightarrow A_{21}=0 \\ A_{22}=(-1)^{2+2}\left|\begin{array}{cc} 0 & 0 \\ 0 & -\cos \alpha \end{array}\right|=1 \times-\cos \alpha-0 \times 0 \\ \Rightarrow A_{22}=-\cos \alpha \\ A_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & 0 \\ 0 & \sin \alpha \end{array}\right|=-1(1 \times \sin \alpha-0 \times 0) \\ \Rightarrow A_{23}=-\sin \alpha \\ A_{31}=(-1)^{3+1}\left|\begin{array}{cc}0 & 0 \\ \cos \alpha & \sin \alpha \end{array}\right|=0 \times \sin \alpha-0 \times \cos \alpha \\ \Rightarrow A_{31}=0 \\ A_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 0 \\ 0 & \sin \alpha \end{array}\right|=-1(1 \times \sin \alpha-0 \times 0) \\ \Rightarrow A_{32}=-\sin \alpha \\ A_{33}=(-1)^{3+3}\left|\begin{array}{ll} 1 & 0 \\ 0 & \cos \alpha \end{array}\right|=1 \times \cos \alpha-0 \times 0 \\ \Rightarrow A_{33}=\cos \alpha \\ \operatorname{adj} A =\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} 1 & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right]^{\prime} \\=\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right] \\ A^{-1} =\frac{1}{|A|} \operatorname{adj} A \\ = -\frac{1}{1}\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right] \\ \Rightarrow A^{-1} =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]$
Example:12.यदि $A=\left[\begin{array}{ll} 3 & 7 \\2 & 5 \end{array}\right]$ और $B=\left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right]$ है तो सत्यापित कीजिए कि $(A B)^{-1}=B^{-1} A^{-1}$ है।
Solution: $A=\left[\begin{array}{ll} 3 & 7 \\2 & 5 \end{array}\right]$
आव्यूह A का सारणिक $|A|=\left|\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right| \\ \Rightarrow|A|=3 \times 5-2 \times 7= 1 \neq 0$
अतः $A^{-1}$ का अस्तित्व है।

$A_{11}=(-1)^{1+1} 5=5, A_{12}=(-1)^{1+2} 2=-2 \\ A_{21} =(-1)^{2+1} 7=-7, A_{22}=(-1)^{2+2} 3=3 \\ \operatorname{adj} A =\left[\begin{array}{ll} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array} \right]^{\prime} \\ =\left[\begin{array}{cc}5 & -2 \\-7 & 3\end{array}\right]^{\prime}\\ \Rightarrow \operatorname{adj} A=\left[\begin{array}{rr}5 & -7 \\-2 & 3\end{array}\right] \\ A^{-1} =\frac{1}{|A|} \operatorname{adjA} \\ =\frac{1}{1}\left[\begin{array}{cc} 5 & -7 \\ -2 & 3 \end{array}\right] \\ \Rightarrow A^{-1} =\left[\begin{array}{cc} 5 & -7 \\ -2 & 8 \\ -2 & 3 \\ 3 & 8 \end{array}\right]$
आव्यूह B का सारणिक $|B|=\left|\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right| \\ |B|=6 \times 9-8 \times 7=-2 \neq 0$
अतः $B^{-1}$ का अस्तित्व है।

$B_{11} =(-1)^{1+1} 9=9, B_{12}=(-1)^{1+2} 7=-7 \\ B_{21} =(-1)^{2+1} 8=-8, B_{2+2}=(-1)^{2+2} 6=6 \\ \operatorname{adj} B =\left[\begin{array}{ll} B_{11} & B_{12} \\ B_{21} & B_{22}\end{array}\right]^{\prime} \\ =\left[\begin{array}{cc} 9 & -7 \\ -8 & 6 \end{array}\right]^{\prime} \\ \Rightarrow \operatorname{adj} B=\left[\begin{array}{cc} 9 & -8 \\ -7 & 6 \end{array}\right] \\ B^{-1}=\frac{1}{|B|} \operatorname{adj} B \\=-\frac{1}{2}\left[\begin{array}{cc} 9 & -8 \\ -7 & 6 \end{array}\right] \\ =\left[\begin{array}{cc} -\frac{9}{2} & \frac{8}{2} \\ \frac{7}{2} & -\frac{6}{2} \end{array}\right]\\ \Rightarrow B^{-1}=\left[\begin{array}{cc} -\frac{9}{2} & 4 \\ -\frac{7}{2} & -3 \end{array}\right] \\ AB=\left[\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right]\left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right] \\ =\left[\begin{array}{ll} 3 \times 6+7 \times 7 & 3 \times 8+7 \times 9 \\ 2 \times 6+5 \times 7 & 2 \times 8+5 \times 9 \end{array}\right] \\=\left[\begin{array}{ll} 67 & 87 \\ 47 & 61 \end{array}\right] \\ |A B| =67 \times 61-87 \times 47=4087-4089 \\ \Rightarrow|A B| =-2 \neq 0 \\ \operatorname{adj}(A B) =\left[\begin{array}{ll} 61 & -47 \\ -87 & 67\end{array}\right] \\ \Rightarrow \operatorname{adj} A B=\left[\begin{array}{cc}61 & -87 \\ -47 & 67 \end{array}\right] \\ (AB)^{-1}=\frac{1}{|A B|} \operatorname{adj}(A B) \\ =-\frac{1}{2}\left[\begin{array}{cc} 61 & -87 \\ -47 & 67 \end{array}\right] \\ \Rightarrow (AB)^{-1}=\left[\begin{array}{cc} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2}\end{array}\right] \cdots(1) \\ B^{-1} A^{-1} =\left[\begin{array}{cc} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{array}\right]\left[\begin{array}{cc} 5 & -7 \\ -2 & 3 \end{array} \right] \\ =\left[\begin{array}{cc} -\frac{45}{2}-8 & \frac{63}{2}+12 \\ \frac{35}{2}+6 & -\frac{49}{2}-9 \end{array}\right] \\ \Rightarrow B^{-1} A^{-1} =\left[\begin{array}{cc} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2}\end{array}\right] \cdots(2)$
(1) व (2) से:

$(A B)^{-1}=B^{-1} A^{-1}$

उपर्युक्त प्रश्नों के उत्तर द्वारा आव्यूह के सहखण्डज और व्युत्क्रम कक्षा 12 (Adjoint and Inverse of Matrix Class 12),आव्यूह के सहखण्डज और व्युत्क्रम (Adjoint and Inverse of Matrix) के बारे में ओर अधिक जानकारी प्राप्त कर सकते हैं।

3.आव्यूह के सहखण्डज और व्युत्क्रम कक्षा 12 के सवाल (Adjoint and Inverse of Matrix Class 12 Questions):

(1.)निम्नलिखित मैट्रिक्स की व्युत्क्रमणीय मैट्रिक्स ज्ञात कीजिए

$\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$
(2.)यदि $A=\left[\begin{array}{lll}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{array}\right]$ तथा $B^{-1}=\left[\begin{array}{ccc}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$ हो तो $(A B)^{-1}$ ज्ञात कीजिए।
उत्तर (Answers): (1.) $\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}\right]$
(2.)$\left[\begin{array}{ccc} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{array}\right]$

उपर्युक्त प्रश्नों के उत्तर द्वारा आव्यूह के सहखण्डज और व्युत्क्रम कक्षा 12 (Adjoint and Inverse of Matrix Class 12),आव्यूह के सहखण्डज और व्युत्क्रम (Adjoint and Inverse of Matrix) के बारे में ओर अधिक जानकारी प्राप्त कर सकते हैं।

Also Read This Article:- Area of Triangle Class 12

4.आव्यूह के सहखण्डज और व्युत्क्रम कक्षा 12 (Frequently Asked Questions Related to Adjoint and Inverse of Matrix Class 12),आव्यूह के सहखण्डज और व्युत्क्रम (Adjoint and Inverse of Matrix) से सम्बन्धित अक्सर पूछे जाने वाले प्रश्न:

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Adjoint and Inverse of Matrix Class 12

आव्यूह के सहखण्डज और व्युत्क्रम कक्षा 12
(Adjoint and Inverse of Matrix Class 12)

Adjoint and Inverse of Matrix Class 12