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Solve for $$ \alpha\epsilon\mathbb{R} $$ so that rank of matrix $A$ won't be full. Find all results of the system $Ax = b$, where $$ b = [2\hspace{2 mm} 6\hspace{2 mm} 7]^T $$

$$A= \left( \begin{array}{ccc} 1 & 1 & 2 \\ -1 & 2 & \alpha \\ 2 & 1 & 3 \end{array} \right) $$

We had this two part question on our exam today but I can't catch any of my professors to explain this to me.

  1. I understand if the determinant of matrix is not $0$ than it has a full rank. So I solved the matrix of the determinant: $\text{det}(A) = 0$ so I was looking for $\alpha$ that makes $\text{det}(A) = 0$; Is my assumption here correct?

  2. I had no clue how to solve the second part so I calculated $$ x = A^{-1}\cdot b $$

Were my steps here correct or did I completely misfire?

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Notice that if you add together the first two columns, you almost get the third column. So alpha can be 1. The answer to the second part will probably be a line.

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