Question

Find values of a, b, and c (if possible) such that the system of linear equations has a unique solution, no solution, and infinitely many solutions. (If not possible, enter IMPOSSIBLE.) x + y = 4 y + z = 4 x + z = 4 ax + by + cz = 0

Answer 1

Answer:

IMPOSSIBLE

Step-by-step explanation:

First we set the equation system:

$x+y+0z=0\\0x+4y+z=0\\4ax+by+cz=0$

Now we set the matrix in order to have a solution for the system:

$\left[\begin{array}{ccc}1&1&0\\0&4&1\\4a&b&c\end{array}\right]$

Now we are going to apply Gauss-Jordan to find the solution of the system in terms of a, b and c:

$-4aR_{1}+R_{3}\rightarrow R_{3}\\\\{\left[\begin{array}{ccc}1&1&0\\0&4&1\\0&(-4a+b)&c\end{array}\right]$

Next step:

$(4a-b)R_{2}+4R_{3} \rightarrow R_{3}\\\\{\left[\begin{array}{ccc}1&1&0\\0&4&1\\0&0&(4a-b+c)\end{array}\right]$

Next step:

$(4a-b+c)R_{2}-R_{3} \rightarrow R_{2}\\\\{\left[\begin{array}{ccc}1&1&0\\0&4(4a-b+c)&0\\0&0&(4a-b+c)\end{array}\right]$

Next step:

$4(4a-b+c)R_{1}-R_{2} \rightarrow R_{1}\\\\{\left[\begin{array}{ccc}4(4a-b+c)&0&0\\0&4(4a-b+c)&0\\0&0&(4a-b+c)\end{array}\right]$

With this solution, we have a new equation system:

$4(4a-b+c)=0\\4(4a-b+c)=0\\4a-b+c=0$

This system can be solved by Cramer's rule, by finding the matrix determinant:

$\left[\begin{array}{ccc}16&-4&4\\16&-4&4\\4&-1&1\end{array}\right]$

$\Delta s= (-64-64-64)-(-64-64-64)=0$

As the determinant is zero, we can say that the second system is imposible to solve.