How to quickly compute the determinant of given matrix

Question

I need to find the determinant of given matrix :

$\begin{bmatrix} 1&0&0&0&0&2\\ 0&1&0&0&2&0\\ 0&0&1&2&0&0\\ 0&0&2&1&0&0\\ 0&2&0&0&1&0\\ 2&0&0&0&0&1\\ \end{bmatrix}$

I know that It can be computed with the help of row operations;by applying

R1 $\to$ R1 + R2+R3+R4+R5+R6, and then subsequently simplifying using further row operations to find the determinant.

However my question is : Is there another way to compute the determinant quickly ?,because the row operations are quite time taking and error prone.

I thought of using eigenvalues here but could not reach the solution

Can anyone help me to find this determinant quickly ?

Thank you

Answer 1

By reordering both rows and columns you can see that the determinant is equal to that of a block diagonal matrix:

$$ \begin{vmatrix} 1&2&0&0&0&0\\ 2&1&0&0&0&0\\ 0&0&1&2&0&0\\ 0&0&2&1&0&0\\ 0&0&0&0&1&2\\ 0&0&0&0&2&1\\ \end{vmatrix} = \left(\begin{vmatrix} 1&2 \\ 2&1 \end{vmatrix}\right)^3 $$

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Or with Schur complements: We have $$ M = \begin{pmatrix} A & B \\ C &D \end{pmatrix} \text{ with } A = D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0 & 1 \end{pmatrix} \text{ and } B = C = \begin{pmatrix} 0 & 0 & 2 \\ 0 & 2 & 0 \\ 2& 0 & 0 \end{pmatrix} $$ and therefore (with $I_3$ denoting the $3\times 3$ identity matrix): $$ \det M = \det(A) \det(D - CA^{-1}B) = \det(I_3 - B^2) = \det(-3I_3) = -27 \, . $$

Answer 2

I think proceeding by finding eigenvalues is much harder than ERO in this case. The determinant of a matrix is the product of its eigenvalues, but to find those you would have to compute a more complicated determinant in the first place (to find $\chi(\lambda)$).

If you do these three simple operations:

1. $2R_1-R_6\to R_6$,
2. $2R_2-R_5\to R_5$,
3. $2R_3-R_4\to R_4$,

then the matrix becomes upper-triangular and the determinant is simply the product of the entries on the diagonal: $-27$.

Answer 3

Looking at the matrix, I think you can solve this problem recursively.

So, let $X_6$ your matrix (the matrix having the principal diagonal with $1$s and the secondary diagonal with only with $2$s. Then, using Laplace on the first row, you can obtain a relation with the 4x4 version of the matrix. $$\det{X_6} = -3\det{X_4}$$.

Then, continue applying this recurrence until you get to the 2x2 version. Then, not only you find the determinant of your matrix, but you can also generalize it to any such matrix with even number of rows.

Answer 4

or

$$ P^T H P = D $$ $$\left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & - 2 & 1 & 0 & 0 \\ 0 & - 2 & 0 & 0 & 1 & 0 \\ - 2 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 & 2 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 & 1 & 0 \\ 2 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & - 2 \\ 0 & 1 & 0 & 0 & - 2 & 0 \\ 0 & 0 & 1 & - 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & - 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & - 3 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 & 1 & 0 \\ 2 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & - 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & - 3 \\ \end{array} \right) \left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 & 2 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 & 2 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 & 1 & 0 \\ 2 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) $$