# Determinant of a matrix

• Apr 19th 2010, 02:49 AM
ulysses123
Determinant of a matrix
I have to find the determinant of this matrix using different properties of the determinant. We can not simply give the solution by finding it with mathematics sotware.

I have tried puting it into reduced row echelon form, but this takes ages.Is there an easier way?

Q1)Show that the determinant of the following matrix is $\displaystyle (a-b)^4$$\displaystyle (a+4b) \displaystyle \begin{array}{ccccc}a&b&b&b&b\\b&a&b&b&b\\b&b&a&b& b\\b&b&b&a&b\\b&b&b&b&a\end{array} • Apr 19th 2010, 07:18 AM NonCommAlg Quote: Originally Posted by ulysses123 I have to find the determinant of this matrix using different properties of the determinant. We can not simply give the solution by finding it with mathematics sotware. I have tried puting it into reduced row echelon form, but this takes ages.Is there an easier way? Q1)Show that the determinant of the following matrix is \displaystyle (a-b)^4$$\displaystyle (a+4b)$

$\displaystyle \begin{array}{ccccc}a&b&b&b&b\\b&a&b&b&b\\b&b&a&b& b\\b&b&b&a&b\\b&b&b&b&a\end{array}$

first do these column operations: (keep the order as it is)

$\displaystyle C_1 - C_2 \to C_1, \ \ C_2 - C_3 \to C_2, \ \ C_3 - C_4 \to C_3, \ \ C_4 - C_5 \to C_4.$

then do these row operations:

$\displaystyle R_1 + R_2 \to R_2, \ \ R_2 + R_3 \to R_3, \ \ R_3 + R_4 \to R_4, \ \ R_4 +R_5 \to R_5.$

then you'll have an upper-triangular matrix and so the determinant would be the product of the entries on the main diagonal.
• Apr 19th 2010, 10:24 AM
Defunkt
Quote:

Originally Posted by ulysses123
I have to find the determinant of this matrix using different properties of the determinant. We can not simply give the solution by finding it with mathematics sotware.

I have tried puting it into reduced row echelon form, but this takes ages.Is there an easier way?

Q1)Show that the determinant of the following matrix is $\displaystyle (a-b)^4$$\displaystyle (a+4b)$

$\displaystyle \begin{array}{ccccc}a&b&b&b&b\\b&a&b&b&b\\b&b&a&b& b\\b&b&b&a&b\\b&b&b&b&a\end{array}$

Denote the mentioned matrix by $\displaystyle A$. Now, note that $\displaystyle rank(A-(a-b)I) = 1$ , meaning $\displaystyle (a-b)$ is an eigenvalue of geometric multiplicity 4 -> algebraic multiplicity $\displaystyle \geq 4$.

Now, use the fact that for any matrix A, $\displaystyle tr(A) = \sum_{i=1}^n \lambda_i$ where $\displaystyle \lambda_1,...,\lambda_n$ are the eigenvalues of A:

In this case, we have that $\displaystyle \lambda_1=...=\lambda_4 = a-b \Rightarrow \lambda_5 = 5a - 4(a-b) = 5a-4a+4b = a+4b$.

Finally, using the fact that $\displaystyle det(A) = \prod_{i=1}^n \lambda_i$, we get $\displaystyle det(A) = (a-b)^4(a+4b)$