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  • Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 57 math

Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 57 math

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Answer: Hence proved, A^{n}=\left[\begin{array}{cc}a^{n} & b\left(\frac{a^{n}-1}{a-1}\right) \\ 0 & 1\end{array}\right] for all positive integers n.

Hint: We use the principle of mathematical induction.

Given:

A=\left[\begin{array}{ll}a & b \\ 0 & 1\end{array}\right]

Prove:

A^{n}=\left[\begin{array}{cc}a^{n} & b\left(\frac{a^{n}-1}{a-1}\right) \\ 0 & 1\end{array}\right] for every positive integer n                  …(i)

Solution:

step 1:

Put n=1 in eqn (i)

\\A^{1}=\left[\begin{array}{cc}a^{1} & b\left(\frac{a^{1}-1}{a-1}\right) \\ 0 & 1\end{array}\right]\\\\ A=\left[\begin{array}{cc}a & b\left(\frac{a-1}{a-1}\right) \\ 0 & 1\end{array}\right] \quad \therefore\left\{a^{1}=a\right\} \\\\ A=\left[\begin{array}{ll}a & b \\ 0 & 1\end{array}\right]

So,  A^n is true for n=1

Step 2: let A^n be true for n=k, so

A^{k}=\left[\begin{array}{cc} a^{k} & b\left(\frac{a^{k}-1}{a-1}\right) \\ 0 & 1 \end{array}\right] ... (ii)

Step 3: we have to show that

\\A^{k+1}=A^{k} \times A\\\\ =\left[\begin{array}{cc} a^{k} & b\left(\frac{a^{k}-1}{a-1}\right) \\ 0 & 1 \end{array}\right] \times\left[\begin{array}{ll} a & b \\ 0 & 1 \end{array}\right]\\\\\\ =\left[\begin{array}{cc} a^{k+1}+0 & a^{k} b+b\left(\frac{a^{k}-1}{a-1}\right) \\ 0+0 & 0+1 \end{array}\right]\\\\\\ =\left[\begin{array}{cc} a^{k+1} & \frac{a^{k+1} b-a^{k} b+a^{k} b-b}{a-1} \\ 0 & 1 \end{array}\right]\\\\\\ A^{k+1}=\left[\begin{array}{cc} a^{k+1} & \frac{b\left(a^{k+1}-1\right)}{a-1} \\ 0 & 1 \end{array}\right]

So,

A^n is true for n=k+1 whenever it is true n=k

Hence, by principle of mathematical induction A^n is true for all positive integer n.

Posted by

infoexpert22

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