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Math Help - Determinant of a matrix

  1. #1
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    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}
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  2. #2
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    Quote Originally Posted by ulysses123 View Post
    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.
    Last edited by NonCommAlg; April 19th 2010 at 08:29 AM.
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  3. #3
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    Quote Originally Posted by ulysses123 View Post
    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)
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