lf $A=\left[\begin{array}{ccc}{a}^2& ab& ac\\ ab& b^2& bc\\ ac& bc& c^2\end{array}\right]$ and $a^2+b^2+c^2=1,$ then $A^2=$

The correct option is A $A$
Consider, $A\times A=\left[\begin{array}{ccc}{a}^2& ab& ac\\ ab& b^2& bc\\ ac& bc& c^2\end{array}\right]\left[\begin{array}{ccc}{a}^2& ab& ac\\ ab& b^2& bc\\ ac& bc& c^2\end{array}\right]$

$=\left[\begin{array}{ccc}{a}^4+a^2{b}^2+a^2{c}^2& a^{3}b+a{b}^3+ab{c}^2& a^{3}c+a{b}^{2}c+a{c}^3\\ a^{3}b+a{b}^3+ab{c}^2& a^2{b}^2+b^4+b^2{c}^2& a^{2}bc+b^{3}c+b{c}^3\\ a^{3}c+ac{b}^2+a{c}^3& a^{2}bc+b^{3}c+b{c}^3& a^2{c}^2+b^2{c}^2+c^4\end{array}\right]$

$=\left[\begin{array}{ccc}{a}^2(a^2+b^2+c^2)& ab(a^2+b^2+c^2)& ac(a^2+b^2+c)\\ ab(a^2+b^2+c^2)& b^2(a^2+b^2+c^2)& bc(a^2+b^2+c^2)\\ ac(a^2+b^2+c^2)& bc(a^2+b^2+c^2)& c^2(a^2+b^2+c^2)\end{array}\right]$

Since $a^2+b^2+c^2=1$

$\therefore$ $\left[\begin{array}{ccc}{a}^2& ab& ac\\ ab& b^2& bc\\ ac& bc& c^2\end{array}\right]=A$