Finding the det:4x4 Matrix using inspection, not by direct evaluation?

Question

I have the determinant of a 4x4 matrix I need to solve for uni. I understand that if a row (or column) is the same then det of a matrix will equal zero, however the rows = the columns in this example. So this rule does not apply. I can not see a way to multiply a row or column to get zeros. And I'm not allowed to use gaussian elimination as that would be solving by evaluation.

I need to use properties of determinants, but I can not seem to figure it out. A hint I did get was C=the product of A and A transposed. The below matrix is C \begin{pmatrix}39&3&18&x\\ 3&86&-50&-6x\\ 18&-50&68&8x\\ x&-6x&8x&x^2\end{pmatrix} I know the solution is 1764x^2, I am just unsure how to get there without solving on paper.

Any help would be appreciated.

Answer 1

To start, $\begin{pmatrix}39&3&18&x\\ 3&86&-50&-6x\\ 18&-50&68&8x\\ x&-6x&8x&x^2\end{pmatrix} =x\begin{pmatrix}39&3&18&1\\ 3&86&-50&-6\\ 18&-50&68&8\\ x&-6x&8x&x\end{pmatrix} =x^2\begin{pmatrix}39&3&18&1\\ 3&86&-50&-6\\ 18&-50&68&8\\ 1&-6&8&1\end{pmatrix} $.

Then you can start row and column operations.

I'll do a few.

$\begin{array}\\ \begin{pmatrix}39&3&18&1\\ 3&86&-50&-6\\ 18&-50&68&8\\ 1&-6&8&1\end{pmatrix} &\to \begin{pmatrix}38&9&10&0\\ 3&86&-50&-6\\ 18&-50&68&8\\ 1&-6&8&1\end{pmatrix} \quad \text{r1 - r4}\\ &\to \begin{pmatrix}38&9&10&0\\ 9&68&-26&0\\ 18&-50&68&8\\ 1&-6&8&1\end{pmatrix} \quad \text{r2 + 3r4}\\ &\to \begin{pmatrix}38&9&10&0\\ 9&68&-26&0\\ 10&-2&2&0\\ 1&-6&8&1\end{pmatrix} \quad \text{r3 - 8r4}\\ \end{array} $

I'll leave the rest for you.

I know I'm not taking advantage of the symmetry.

Answer 2

COMMENT.-Can you see why you do have

$$2^2x^2\det\begin{pmatrix}39&3&9&1\\ 3&86&-25&-6\\ 9&-25&17&4\\ 1&-6&4&1\end{pmatrix}?$$ There are several ways of reducing this numerical matrix. According with your post the determinant to be calculated must be equal to $(21)^2$.

Answer 3

From the final comments below the answer, it would appear the OP was given the following information, where the first two matrices are transposes of each other. Perhaps just one of the (transposed) pair was displayed and given a name; probably the matrix on the left was shown and called $A.$

$$ \left( \begin{array}{rrrr} 3& 2& 5& 1 \\ 0& 7& -1& -6 \\ 0& 0& 2& 8 \\ 0& 0& 0& x \\ \end{array} \right) \left( \begin{array}{rrrr} 3& 0& 0& 0 \\ 2& 7& 0& 0 \\ 5& -1& 2& 0 \\ 1& -6& 8& x \\ \end{array} \right) = \left( \begin{array}{rrrr} 39& 3& 18& x \\ 3& 86& -50& -6x \\ 18& -50& 68& 8x \\ x& -6x& 8x& x^2 \\ \end{array} \right) $$