Question 37
यदि (if)
A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] , साबित करे कि (prove that)
A^{n}=\left[\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right] सभी
n \in N के लिए (for all
n \in \mathbf{N})
Sol :
माना
P(n): A^{n}=\left[\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right]
\begin{array}{rl}n=1 & A^{1}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\end{array}..(i)
P(1)=सत्य है ।
माना P(k) सत्य है
P(k): A^{k}=\left[\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right]
तो साबित करना है कि P(k+1) भी सत्य होगा
P(k+1): \quad A^{k+1}=\left[\begin{array}{cc}1 & k+1 \\ 0 & 1\end{array}\right]
(i) मे दोनो तरफ A से गुना करने पर
A^{k} \cdot A^{1}=\left[\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right] \cdot A
A^{k+1}=\left[\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]
=\left[\begin{array}{cc}1+0 & 1+k \\ 0+0 & 0-1\end{array}\right]
=\left[\begin{array}{ll}1 & k+1 \\ 0 & 1\end{array}\right]
n \in N के लिए
P(n): A^{n}=\left[\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right] सत्य है ।
Question 38
यदि (if)A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right] दिखाएँ कि (show that) A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]
जहाँ n एक धन पूर्णाक है (where n is a positive integer)
Sol :
माना P(n): A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]
n=1 , A^{\prime}=\left[\begin{array}{ccc}1+2(1) & -4(1) \\ 1 & 1-2(1)\end{array}\right]
=\left[\begin{array}{rr}3 & -4 \\ 1 & -1\end{array}\right]
P(1) सत्य हैं
माना P(k) सत्य हैं
P(k) : A^{k}=\left[\begin{array}{cc}1+2 k & -4 k \\ k & 1-2 k\end{array}\right]..(i)
तो , साबित करना है कि P(k+1) भी सत्य होगा ।
A^{k+1}=\left[\begin{array}{cc}1+2(k-n) & -4(k+1] \\ k+1 & 1-2(k+1)\end{array}\right]
(i) मे दोनो तरफ A से गुना करने पर ,
A^{k} \cdot A=\left[\begin{array}{cc}1+2 k & -4 k \\ k & 1-2 k\end{array}\right]\left[\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}\right]
A^{k+1}=\left[\begin{array}{cc}3+6 k-4 k & -4-8 k+4 k \\ 3 k+1-2 k & -4 k-1+2 k\end{array}\right]
A^{k-1}=\left[\begin{array}{cc}3+2 k & -4-4 k \\ k+1 & -2 k-1\end{array}\right]
A^{k+1}=\left[\begin{array}{cc}1+2 k+2 & -4 k-4 \\ k+1 & 1-2 k-2\end{array}\right]
=\left[\begin{array}{cc}1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1)\end{array}\right]
\therefore n \in z^{+} के लिए A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right] सत्य है ।
Question 39
यदि (if) A=diag.[a b c] दिखाएँ कि (show that) \mathbf{A}^{a}=\operatorname{diag}\left[a^{n} \quad b^{n} \quad c^{n}\right] सभी (for all) n \in \mathbf{N} के लिए
Sol :
माना A^{n}=\left[\begin{array}{ccc}a^{n} & 0 & 0 \\ 0 & b^{n} & 0 \\ 0 & 0 & c^{n}\end{array}\right]
n=1 , A^{\prime}=\left[\begin{array}{lll}a^{1} & 0 & 0 \\ 0 & b^{1} & 0 \\ 0 & 0 & c^{1}\end{array}\right]
=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]
P(1) सत्य हैं
माना P(k) सत्य हैं
P(k) : A^{k}=\left[\begin{array}{lll}a^{k} & 0 & 0 \\ 0 & b^{k} & 0 \\ 0 & 0 & c^{k}\end{array}\right]
तो , साबित करना है कि P(k+1) भी सत्य होगा ।
P(k+1): A^{k+1}=\left[\begin{array}{ccc}a^{k+1} & 0 & 0 \\ 0 & b^{k+1} & 0 \\ 0 & 0 & c^{k+1}\end{array}\right]
(i) मे दोनो तरफ A से गुना करने पर ,
A^{k} \cdot A=\left[\begin{array}{lll}a^{k} & 0 & 0 \\ 0 & b^{k} & 0 \\ 0 & 0 & c^{k}\end{array}\right]\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]
A^{k+1}=\left[\begin{array}{ccc}a^{k+1} & 0 & 0 \\ 0 & b^{k+1} & 0 \\ 0 & 0 & c^{k-1}\end{array}\right]
\therefore n \in z^{+} के लिए A^{n}=\left[\begin{array}{lll}a^{n} & 0 & 0 \\ 0 & b^{n} & 0 \\ 0 & a & c^{1}\end{array}\right] सत्य है ।
Question 40
यदि (if) A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] ,
तो गणितीय आगमन सिद्धान्त से साबित करे कि (prove by principle of mathematical induction that)
A^{n}=\left[\begin{array}{cc}\cos n \alpha & \sin n \alpha \\ -\sin n \alpha & \cos n \alpha\end{array}\right]
प्रत्येक प्राकृत संख्या n के लिए (for every natural number n)
Sol :
Let P(n): A^{n}=\left[\begin{array}{cc}\cos n \alpha & \sin n \alpha \\ -\sin n \alpha & \cos n \alpha\end{array}\right]
when n=1 ,
A^{1}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]
P(1) is true
Let P(k) be true
P(k): A^{k}=\left[\begin{array}{cc}\cos k \alpha & \sin k \alpha \\ -\sin k \alpha & \cos k \alpha\end{array}\right]
then prove that P(k+1) is true
P(k+1):
A^{k+1}=\left[\begin{array}{cc}\cos (k+1) \alpha & \sin (k+1) \alpha \\ -\sin (k+1) \alpha & \cos (k+1) \alpha\end{array}\right]
<to be added>
A^{k} \cdot A=\left[\begin{array}{cc}\cos k \alpha & \sin k \alpha \\ -\sin k \alpha & \cos k \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]
A^{k+1}=\left[\begin{array}{ll}\cos k \alpha+\cos \alpha-\sin \alpha \sin \alpha & cosk\alpha \sin \alpha+ \sin k \alpha \cos \alpha\\ -\cos \alpha \sin k \alpha-\cos k\alpha\sin \alpha & -\sin k \alpha\sin \alpha+\cos k\alpha \cos \alpha\end{array}\right]
=\left[\begin{array}{ll}\cos (k\alpha+\alpha) & \sin (k \alpha+\alpha) \\ -\sin (k\alpha+\alpha) & \cos (k+\alpha)\end{array}\right]
A^{k+1}=\left[\begin{array}{cc}\cos (k+1) \alpha & \sin (k+1)+\alpha \\ -\sin (k-1) \alpha & \cos (k+1) \alpha \end{array}\right]
∴ nϵN के लिए , A^{n}=\left[\begin{array}{cc}\cos n \alpha & \sin \alpha \\ -\sin n \alpha & \cos n \alpha\end{array}\right] सत्य है ।
यदि (If) A=\left[\begin{array}{ll}\cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta\end{array}\right] , तो गणितीय आगमन सिद्धान्त से साबित करें कि (then prove by principle f mathematical induction that) A^{n}=\left[\begin{array}{cc}\cos n \theta & i \sin n \theta \\ i \sin n \theta & \cos n \theta\end{array}\right] , जहाँ (where) n∊N
Sol :
Question 42
यदि A और B समान कोटिवाले वर्ग आव्यूह इस प्रकार है कि AB=BA , तो गणितीय आगमन सिद्धान्त से साबित करे कि AB^n=B^nA . इसके अतिरिक्त यह भी साबित करें कि (\mathrm{AB})^{n}=\mathrm{A}^{n} \mathrm{~B}^{n} सभी n∈N के लिए।
[If A and B are square matrices of the smae order such that AB=BA then prove by principle of mathematical induction that AB^n=B^nA. Further, prove that (\mathrm{AB})^{n}=\mathrm{A}^{n} \mathrm{~B}^{n} for all n∈N ]
Sol :
Let P(n): A B^{n}=B^{n} A
where n∈N
when n=1 ,
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