3e543840-b91d-4fda-9adc-8a7ab284977c21f4991f-d11b-47f3-93fa-9ec974c35cd4bf9c8cfb-3123-438a-a941-8945554badac1.3.3.4<p>Determine the value of $$A$$ which makes the system incompatible $$$\left\{ \begin{array} {rcl} x+Ay+z & = & 1 \\ Ax+y+(A-1)z &=& A\\ x+y+z &=& A+1\end{array}\right.$$$</p><p>The first thing is to re-write the system in matricial form: $$$\begin{pmatrix} 1 & A & 1 & | & 1 \\ A & 1 & (A-1) & | & A \\ 1 & 1 & 1 & | & A+1 \end{pmatrix}$$$ We then use the Gauss method $$$\begin{pmatrix} 1 & A & 1 & | & 1 \\ A & 1 & (A-1) & | & A \\ 1 & 1 & 1 & | & A+1 \end{pmatrix} \rightarrow \left\{ \begin{array}{c} r2-Ar1 \\ r3-r1 \end{array} \right. \rightarrow \begin{pmatrix} 1 & A & 1 & | & 1 \\ 0 & 1-A^2 & -1 & | & 0 \\ 0 & 1-A & 0 & | & A \end{pmatrix} \rightarrow $$$ $$$ \rightarrow (c3 \leftrightarrow c2) \rightarrow \begin{pmatrix} 1 & 1 & A & | & 1 \\ 0 & -1 & 1-A^2 & | & 0 \\ 0 & 0 & 1-A & | & A \end{pmatrix}$$$ We obtained: $$(1-A)y=A$$ *, then:</p> <p>*See that, as we changed the columns, we obtain the solution for $$y$$ and not for $$z$$.</p> <p>If $$A=1$$, we have an incompatible system.</p> <p>Otherwise the system is compatible, and it has the following solution:</p> <p>$$y=\dfrac{A}{1-A}; \ z=(1+A)A; \ x=\dfrac{A^3-A^2-2A+1}{1-A}$$</p><p>$$A=1$$</p>7/18/16, 11:44 AM7/18/16, 11:44 AM