Solving for unknown value using properties of determinant

Problem : If $ax^4 +bx^3+cx^2+dx+e=$ $$\begin{vmatrix} x^3+3x & x-1 & x+3 \\ x+1 & -2x & x-4 \\ x-3 & x+4 & 3x \\ \end{vmatrix}$$

Then find $e$.

Solution: I know to solve the question by expanding the determinant

But I want to find determinant by using properties of determinant.

• I don't know that you can get completely away from expanding the determinant, but before you do, you can remove all terms involving $x$ from all the entries. This will not affect $e$. Now instead you have $$e = \begin{vmatrix}0 & -1 & 3 \\\\ 1 & 0 & -4 \\\\ -3 & 4 & 0\end{vmatrix}$$ – Arthur Aug 15 '13 at 12:28
• @Arthur Can we solve the question like this? For finding value of d (i.e coefficient of x) make determinant of coefficient of x – rst Aug 15 '13 at 12:34
• As this determinant is written the equation can't be true in general because the determinant has a term in $x^5$ – Mark Bennet Aug 15 '13 at 13:01
• @rst The reason this works is because the determinant essentially is a sum of products of the entries in the matrix. If your matrix entries are polynomials, then only the constant terms of those polynomials will affect the resulting constant term. For the first degree term you need to take into account both first-degree terms and constant terms, so you can remove $x^3$ from the top left entry and that's that. However, as you're expanding the determinant to find $d$, you can at any point in your calculation discard any second-or-higher degree term, since they will have no way of affecting $d$. – Arthur Aug 15 '13 at 13:13

Considering @Arthur's leading comment, if we accept that this equation is being held for every value of $x$, so put for example $x=0$ to find out $e=0$. Note that we can see that for some $x$ the matrix is skew-symmetric matrix.

Using Arthur's helpful comment, we can find $e$ by evaluating: $$e = \begin{vmatrix}0 & -1 & 3 \\\\ 1 & 0 & -4 \\\\ -3 & 4 & 0\end{vmatrix}$$

Now you can expand the determinant to solve for $e$, or use elementary row operations to row reduce to find that $$e = \begin{vmatrix}1 & 0 & -4 \\\\0 & 1 & - 3\\\\ 0 & 0 & 0\end{vmatrix} = 0$$

And this can be confirmed using the fact that "switch row 1 with row 2" negates the sign of the original determinant, "multiply row 2 by -1" negates the sign of the negated determinant, and adding a multiple of a row to another row has no impact on the determinant. I.e.

\begin{align} e & = \begin{vmatrix}0 & -1 & 3 \\ 1 & 0 & -4 \\ -3 & 4 & 0\end{vmatrix}\\ \\ -e & = \begin{vmatrix}1 & 0 & -4 \\ 0 & -1 & 3 \\ -3 & 4 & 0\end{vmatrix}\tag{R_1 \leftrightarrow R_2}\\ \\ -(-e) &= \begin{vmatrix}1 & 0 & -4 \\ 0 & 1 & -3 \\ -3 & 4 & 0\end{vmatrix} \tag{-1\cdot R_2}\\ \\ e &= \begin{vmatrix}1 & 0 & -4 \\ 0 & -1 & 3 \\ 0 & 4 & -12\end{vmatrix}\tag{3R_1 + R_3 \to R_3} \\ \\ e &= \begin{vmatrix}1 & 0 & -4 \\ 0 & -1 & 3 \\ 0 & 0 & 0\end{vmatrix}\tag{4 R_2 + R_3 \to R_3}\end{align}

• Needs another TU! +1 – Amzoti Aug 16 '13 at 12:52

If you set $x = 0$ you will find that the third row is a linear combination of row one and two, the determinat is zero and e is zero.