#### Please Solve R.D. Sharma class 12 Chapter determinants Exercise  5. 2 Question 41 sub question 1 Maths textbook Solution.

Answer: $a^{3}+3 a^{2}$

Hint Use determinant formula

Given:$\left|\begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+a & 1 \\ 1 & 1 & 1+a \end{array}\right|=a^{3}+3 a^{2}$

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+a & 1 \\ 1 & 1 & 1+a \end{array}\right| \\ &\quad \text { Apply } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} a+3 & 1 & 1 \\ a+3 & 1+a & 1 \\ a+3 & 1 & 1+a \end{array}\right| \end{aligned}

\begin{aligned} &(a+3) \text { common from } \mathrm{C}_{1}\left[C_{1} \rightarrow \frac{C_{1}}{a+3}\right] \\ &=(a+3)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+a & 1 \\ 1 & 1 & 1+a \end{array}\right| \end{aligned}

\begin{aligned} &\text { Now } \mathrm{C}_{3} \rightarrow C_{3}-C_{1} \\ &=(a+3)\left|\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1+a & 0 \\ 1 & 1 & a \end{array}\right| \end{aligned}

Expanding w.r.t $C_{3}$

\begin{aligned} &=(a+3)\left\{a \mid \begin{array}{cc} 1 & 1 \\ 1 & 1+a \end{array}\right\} \\ &=(a+3)\{a(1+a-1)\} \\ &=(a+3) a^{2} \\ &=a^{3}+3 a^{2} \\ &=R H . S \end{aligned}