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  • Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 60 math

Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 60 math

Answers (1)

Answer: Hence proved,

A^{n}=\left[\begin{array}{ccc}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right] for every integer n.

Hint: We use the principle of mathematical induction.

Given:

A=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]                                     

Prove:

A^{n}=\left[\begin{array}{ccc}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right] for every positive integer n.                 …(i)

Solution:

step 1: put n=1 in eqn(i)

A^{\mathbf{1}}=\left[\begin{array}{ccc}1 & 1 & \frac{1(1+1)}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\\\\ A^{\mathbf{1}}=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 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}{ccc}1 & k & \frac{k(k+1)}{2} \\ 0 & 1 & k \\ 0 & 0 & 1\end{array}\right]

Step 3:  we will prove  A^n that will be true for n=k+1

Now,\\A^{k+\mathbf{1}}=A^{k} A\) \(=\left[\begin{array}{ccc}1 & k & \frac{k(k+1)}{2} \\ 0 & 1 & k \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}1+0+0 & 1+k+0 & 1+k+\frac{k(k+1)}{2} \\ 0+0+0 & 0+1+0 & 0+1+k \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}1 & k+1 & \frac{(k+1)(k+2)}{2} \\ 0 & 1 & (k+1) \\ 0 & 0 & 1\end{array}\right]

Hence, A^n is true for n=k+1 wherever it is true for n=k

So, by principle of mathematical induction A^n  is true for all positive integers n.

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