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I have to find all $x \in \mathbb{R}$ that matrix $A$ is invertible.

$$ \begin{vmatrix} 1&1&1&1\\ 1&2x&3&4\\ 2&3&4x&5\\ 3&4&5&6x \end{vmatrix} $$

I know that matrix is invertible, when $det(A)$ is not $0$.

So I got to this point $$ \begin{vmatrix} 1&1&1&1\\ 0&2x-1&2&3\\ 0&0&8x^2-8x-2&6x-6\\ 0&0&4-4x&6x-6 \end{vmatrix} $$

I did something wrong in between and if not, what can I do next? I already tried to get $0$ where $4-4x$ is, but in the end I did not get the same result.

Result is $24(x-1)^2(2x+1)$

Niki Sedlarević
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1 Answers1

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You could subtract the fourth row from the third, generating a zero there, then compute the determinant to be $$ \det(A) = (2x-1)(6x-6)(4-4x) = (2x-1)(x-1)^2 \cdot 24. $$ There is an error in your calculation. Here is my try to compute the determinant: $$\begin{split} \det\pmatrix{ 1&1&1&1\\ 1&2x&3&4\\ 2&3&4x&5\\ 3&4&5&6x} &= \det\pmatrix{ 2x-1&2&3\\ 1&4x-2&3\\ 1&2&6x-3}\\ &=6 \det\pmatrix{ 2x-1&1&1\\ 1&2x-1&1\\ 1&1&2x-1}\\& =6 \det\pmatrix{ 0&-2x+2&-4x^2+4x\\ 0&2x-2&-2x+2\\ 1&1&2x-1}\\ &=24 \det\pmatrix{1-x&2x(1-x)\\x-1&1-x}\\& =24 (x-1)^2 \det\pmatrix{-1&-2x\\1&-1}\\& =24 (x-1)^2 (2x+1) \end{split}$$

daw
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