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Find the Determinant

Problem

det(1)

Solution

  1. Identify the block structure of the matrix. The matrix M is a 10×10 matrix that can be partitioned into a block upper triangular form. Specifically, the first 7 rows and columns contain non-zero entries, but row 8 has only one non-zero entry at (M_8,6)=1

  2. Expand along the rows with the most zeros. Row 8 has only one non-zero entry at column 6. Expanding along row 8:

det(M)=(−1)(8+6)⋅1⋅det((M_8,6))

det(M)=det((M_8,6))

  1. Observe the resulting 9×9 matrix (M_8,6) In this submatrix, the new row 8 (originally row 9) has non-zero entries in columns 6, 7, 8, and 9 (originally columns 7, 8, 9, 10). However, the last column (originally column 10) has only one non-zero entry at row 7. Expanding along the last column:

det((M_8,6))=(−1)(7+9)⋅1⋅det(M′)

where M′ is an 8×8 matrix.

  1. Analyze the remaining block structure. The matrix can be seen as a block diagonal matrix if we reorder or observe the independence of the first 5-6 variables from the last 3. Specifically, the bottom-right 3×3 block involving the original rows 9, 10 and columns 7, 8, 9 is:

B=[[1,1,1],[1,1,0],[0,0,0]]

Since row 10 of the original matrix (which corresponds to the last row of this block) has zeros in the last column (column 9), and the expansion from step 3 removed column 10, we look at the 2×2 block from rows 9 and 10 and columns 7 and 8:

(B_s(u)*b)=[[1,1],[1,1]]

  1. Calculate the determinant of the 2×2 block.

det(1)=(1⋅1)−(1⋅1)=0

  1. Conclude that since a block within the block triangular structure has a determinant of zero, the determinant of the entire matrix must be zero.

Final Answer

det(1)=0

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