# Thread: Determinant of a matrix

1. ## 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 $(a-b)^4$ $(a+4b)$

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

2. 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 $(a-b)^4$ $(a+4b)$

$\begin{array}{ccccc}a&b&b&b&b\\b&a&b&b&b\\b&b&a&b& b\\b&b&b&a&b\\b&b&b&b&a\end{array}$
first do these column operations: (keep the order as it is)

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

then do these row operations:

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

then you'll have an upper-triangular matrix and so the determinant would be the product of the entries on the main diagonal.

3. 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 $(a-b)^4$ $(a+4b)$

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

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

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

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

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