 dikcijom2k

2022-07-08

Determine the value of a if the system
$\begin{array}{rl}{x}_{1}+4{x}_{2}-3{x}_{3}+2{x}_{4}& =0\\ 2{x}_{1}+7{x}_{2}-4{x}_{3}+4{x}_{4}& =0\\ -{x}_{1}+a{x}_{2}+5{x}_{3}-2{x}_{4}& =0\\ 3{x}_{1}+10{x}_{2}-5{x}_{3}+\left(a²+4a+1\right){x}_{4}& =0\end{array}$
has more then 1 solution Tanner Hamilton

Expert

We know that a homogeneous linear system has more than one solution if and only if the rank of the matrix of the system is less than the number of unknowns. See Rank-nullity theorem.
One of the possibilities how to find rank is to use row operations. You could start like this:
$\left(\begin{array}{cccc}1& 4& -3& 2\\ 2& 7& -4& 4\\ -1& a& 5& -2\\ 3& 10& -5& {a}^{2}+4a+1\end{array}\right)\sim \left(\begin{array}{cccc}1& 4& -3& 2\\ 0& -1& 2& 0\\ 0& a+4& 2& 0\\ 3& 10& -5& {a}^{2}+4a+1\end{array}\right)\sim \left(\begin{array}{cccc}1& 4& -3& 2\\ 0& 1& -2& 0\\ 0& a+4& 2& 0\\ 3& 10& -5& {a}^{2}+4a+1\end{array}\right)\sim \left(\begin{array}{cccc}1& 4& -3& 2\\ 0& 1& -2& 0\\ 0& a+5& 0& 0\\ 3& 10& -5& {a}^{2}+4a+1\end{array}\right)$
If $a+5=0$, i.e. if $a=5$, then you have a matrix which does not have full rank. So in this case you have that there is more than one solution.
If $a+5\ne 0$, you can divide the third row by $\left(a+5\right)$
$\left(\begin{array}{cccc}1& 4& -3& 2\\ 0& 1& -2& 0\\ 0& a+5& 0& 0\\ 3& 10& -5& {a}^{2}+4a+1\end{array}\right)\sim \left(\begin{array}{cccc}1& 4& -3& 2\\ 0& 1& -2& 0\\ 0& 1& 0& 0\\ 3& 10& -5& {a}^{2}+4a+1\end{array}\right)\sim \left(\begin{array}{cccc}1& 0& -3& 2\\ 0& 0& -2& 0\\ 0& 1& 0& 0\\ 3& 0& -5& {a}^{2}+4a+1\end{array}\right)\sim \left(\begin{array}{cccc}1& 0& -3& 2\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 3& 0& -5& {a}^{2}+4a+1\end{array}\right)\sim \left(\begin{array}{cccc}1& 0& 0& 2\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 3& 0& 0& {a}^{2}+4a+1\end{array}\right)$

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