Prove that : vmatrix 1+a & 1 & 1 \\ 1 & 1+a & 1 \\ 1 & 1 & 1+a vmatrix = a^2(a+3)

To prove that $$\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+a& 1\\ 1& 1& 1+a\end{array}\right|=a^2(a+3),$$ we evaluate the determinant of the given $3\times 3$ matrix step-by-step.

Step 1: Write the determinant formula

For a general $3\times 3$ matrix $$\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|,$$ the determinant is calculated as: $$a(ei-fh)-b(di-fg)+c(dh-eg).$$

Step 2: Apply the formula to the given matrix

Let the given matrix be: $$\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+a& 1\\ 1& 1& 1+a\end{array}\right|.$$ Here,

• $a=1+a$,
• $b=1$,
• $c=1$,
• $d=1$,
• $e=1+a$,
• $f=1$,
• $g=1$,
• $h=1$,
• $i=1+a$.

Substituting these into the determinant formula: $$\text{Determinant}=(1+a)\left[(1+a)(1+a)-(1)(1)\right]-1\left[(1)(1+a)-(1)(1)\right]+1\left[(1)(1)-(1+a)(1)\right].$$

Step 3: Simplify each term

1. First term: $$(1+a)\left[(1+a{)}^2-1\right]=(1+a)\left[1+2a+a^2-1\right]=(1+a)(2a+a^2)=(1+a)a(2+a).$$

2. Second term: $$-1\left[(1)(1+a)-(1)(1)\right]=-1\left[1+a-1\right]=-1(a)=-a.$$

3. Third term: $$1\left[(1)(1)-(1+a)(1)\right]=1\left[1-1-a\right]=-a.$$

Step 4: Combine the terms

$$\text{Determinant}=(1+a)a(2+a)-a-a.$$ Simplify further: $$=a(1+a)(2+a)-2a.$$

Step 5: Expand and simplify

Expand $(1+a)(2+a)$: $$(1+a)(2+a)=2+a+2a+a^2=2+3a+a^2.$$ Now multiply by $a$: $$a(2+3a+a^2)=2a+3a^2+a^3.$$ Subtract $2a$: $$2a+3a^2+a^3-2a=3a^2+a^3.$$ Factorize: $$a^3+3a^2=a^2(a+3).$$

Final Answer

Thus, we have proven that: $$\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+a& 1\\ 1& 1& 1+a\end{array}\right|=\boxed{a^2(a+3)}.$$