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

IMPOSSIBLE

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.