#### Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 61 math

Answer: Hence proved, $A^{n+1}=B^{n}(B+(n+1) C$ for every integer $n \in N$.

Hint: We use the principle of mathematical induction.

Given: B, C are n rowed square matrix,

$\\A=B+C, B C=C B, C^{2}=0, \\\\$

$A=B+C$

Squaring both sides, we get

$\\A^{2}=(B+C)^{2}\\\\ A^{2}=(B+C)(B+C)\\\\ A^{2}=B \times B+B C+C B+C$           [using distributive property]

$A^{2}=B^{2}+B C+B C+C^{2}$        [using BC=CB] given and put value of $C^{2}$

$\\A^{2}=B^{2}+2 B C+0\\\\ A^{2}=B^{2}+2 B C \quad \ldots(i)\\\\ A^{2}=B(B+2 C)$

Now consider,$P(n)=A^{n+\mathbf{1}}=B^{n}[B+(n+1) C]$

Step 1:  to prove P(1) is true, put n=1

$\\A^{\mathbf{1 + 1}}=B^{\mathbf{1}}[B+(1+1) C]\\\\ A^{2}=B[B+2 C]\\\\ A^{2}=B^{2}+2 B C$

From equation i, P(1)is true

Step 2:  suppose P(k) is true

$A^{k+1}=B^{k}[B+(k+1) C] ... (ii)$

Step 3 : now we need to show that P(k+1) is true

That is we need to prove that

$A^{k+2}=B^{k+1}[B+(k+2) C]$

Now,

\begin{aligned} A^{k+2} &=A^{k} A^{2} \\ &=B^{(k-1)}[B+k C] \times[B(B+2 C)] \\ &=B^{k}[B+k C] \times[B+2 C] \\ &=B^{k}\left[B \times B+B \times 2 C+k C \times B+2 k C^{2}\right] \\ &=B^{k}\left[B^{2}+2 B C+k B C+2 k \times 0\right] \quad\left[\text { since } B C=C B, C^{2}=0\right] \\ &=B^{k}\left[B^{2}+B C(2+k)\right] \\ &=B^{k} \times B[B+(k+2) C] \\ &=B^{k+1}[B+(k+2) C] \end{aligned}

So, P(n) is true for n=k+1 whenever P(n) is true for n=k.

Therefore, by principle of mathematical induction P(n) is true for all natural number.