The determinant of the matrix [Tex]\begin{bmatrix} 6 & -8 & 1 & 1 \\ 0 & 2 & 4…
1997
The determinant of the matrix [Tex]\begin{bmatrix} 6 & -8 & 1 & 1 \\ 0 & 2 & 4 & 6 \\ 0 & 0 & 4 & 8 \\ 0 & 0 & 0 & -1 \end{bmatrix}[/Tex] is:
Answer: B. -48 — For a triangular matrix — upper or lower — every entry on one side of the main diagonal is zero. Expanding the determinant by cofactors along the first column…
- A.
11
- B.
-48
- C.
0
- D.
-24
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Correct answer: B
For a triangular matrix — upper or lower — every entry on one side of the main diagonal is zero. Expanding the determinant by cofactors along the first column of an upper triangular matrix, every term vanishes except the one for the top-left diagonal entry, because every other entry in that column is zero (a zero entry always contributes a zero term to the expansion, regardless of its own cofactor's value); repeating this reduction row by row shows that the determinant of any triangular matrix equals the product of its diagonal entries.
In the given matrix, every entry below the main diagonal is zero, so it is upper triangular, with diagonal entries 6, 2, 4, and -1. Its determinant is their product, computed step by step:
6 × 2 = 12
12 × 4 = 48
48 × (-1) = -48
As an independent cross-check, expand along the last row, which has only one nonzero entry, -1, in the bottom-right position. Its cofactor is the determinant of the top-left 3 × 3 block, itself upper triangular with diagonal entries 6, 2, and 4, giving 6 × 2 × 4 = 48; the cofactor sign at that position is positive, so the determinant is (-1) × 48 = -48, matching the direct product.
The determinant of the matrix is -48.
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