Please help me solve this.
Consider for every real number $a$ the linear system of equations: $$ \begin{align} x +( a + 1 )y + a^2 z &= a^3 \\ (1-a)x +( 1 - 2a )y &= a^3 \\ x +( a + 1 )y + az &= a^2 \\ \end{align} $$
- Find the solution for $a = 2$;
- Find the values of a for which the system has no solution, infinitely many solutions, and a unique solution;
- Find the solution for $a = -1$.
You can write $$ \left( \begin{array}{ccc} 1 & a+1 & a^{2} \\ 1-a & 1-2a & 0 \\ 1 & a+1 & a% \end{array}% \right) \left( \begin{array}{c} x \\ y \\ z% \end{array}% \right) =\mathbf{A\cdot }\left( \begin{array}{c} x \\ y \\ z% \end{array}% \right) =\left( \begin{array}{c} a^{3} \\ a^{3} \\ a^{2}% \end{array}% \right) . $$ In general, if $\det \mathbf{A}\neq 0$, $\mathbf{A}$ is invertible and you have a unique solution. If you subtract the third equation from the first you obtain $$ \mathbf{B}\left( \begin{array}{c} x \\ y \\ z% \end{array}% \right) =\left( \begin{array}{ccc} 0 & 0 & a(a-1) \\ 1-a & 1-2a & 0 \\ 1 & a+1 & a% \end{array}% \right) \left( \begin{array}{c} x \\ y \\ z% \end{array}% \right) =\left( \begin{array}{c} a^{2}(a-1) \\ a^{3} \\ a^{2}% \end{array}% \right) $$ Here $$ \det \mathbf{B}=a(a-1)\{(1-a)(1+a)-1+2a\}=a^{2}(a-1)(2-a), $$ which vanishes for $a=0,1,2$. For $a=1$, $z$ is undetermined and you can have an infinite number of solutions. In this case you find $x$ and $y$ by substituting $a=1$. If $a=0$ then we have $x=y=0$ and $z$ is undetermined. The case $a=2$ you already considered.