I want to find all values of $x$ for which the following matrix is invertible.
$$\begin{pmatrix} x – \frac12 & 0 & 0\\ x & x – \frac13 & 0\\ x^2 & x^3 & x + \frac14\end{pmatrix}$$
I know that using lower triangular determinant (at $x$ values where determinant not equal to zero), I can find easily, but I want to find using reduced echelon form. How can I do that?
Best Answer
If $x=\frac12$, than all entries of the first row are zeros, and therefore the matrix is not invertible.
Assume now that $x\ne\frac12$. Then divide the first row by $x-\frac12$, and you will get$$\begin{bmatrix}1&0&0\\x&x-\frac13&0\\x^2&x^3&x+\frac14\end{bmatrix}.$$Now you add to the second row the first row times $-x$ and you add to the third row the first row times $-x^2$; you will get$$\begin{bmatrix}1&0&0\\0&x-\frac13&0\\0&x^3&x+\frac14\end{bmatrix}.$$If $x=\frac13$, the the second row only has zeros, and therefore the matrix is not invertible.
Assume now that $x\ne\frac13$. Then divide the second row by $x-\frac13$. You will get$$\begin{bmatrix}1&0&0\\0&1&0\\0&x^3&x+\frac14\end{bmatrix}.$$Now, add to the third row the second row times $-x^3$. You will get$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&x+\frac14\end{bmatrix}.$$
If $x=-\frac14$, the third row only has zeros, and therefore the matrix is not invertible. Otherwise, divide the third row by $x+\frac14$. Then you will get the identity matrix, which is invertible. So, the original matrix will also be invertible.