Matrices & Determinants Question - None

Question

If $a, b, c$ and $d$ are the roots of the equation $x^4 - 2x^3 - 4x^2 - 8x + 16 = 0$, the value of the determinant $\begin{vmatrix} 1-a & 1 & 1 & 1 \ 1 & 1-b & 1 & 1 \ 1 & 1 & 1-c & 1 \ 1 & 1 & 1 & 1-d \end{vmatrix}$ is

Accepted Answer

Step 1: Rewrite the determinant in matrix form. Let M be the given matrix. We can write M = J - diag(a, b, c, d), where J is the matrix of all 1s and diag(a, b, c, d) is the diagonal matrix with entries a, b, c, d.

Step 2: Use the Sherman-Morrison formula for determinants. For a rank-1 perturbation, det(I + uv^T) = 1 + v^Tu. However, we recognize that our matrix can be written as:

M = 𝟙𝟙^T - diag(a, b, c, d)

where 𝟙 = (1, 1, 1, 1)^T.

Step 3: Apply the matrix determinant lemma. For a matrix of the form J - D where J = 𝟙𝟙^T and D = diag(a, b, c, d):

det(M) = det(𝟙𝟙^T - D) = (-1)⁴ det(D) · det(I - D^-1𝟙𝟙^T)

Step 4: Simplify using the determinant formula. By the matrix determinant lemma:

det(M) = det(-D) · det(I + D^-1𝟙𝟙^T) = (-1)⁴abcd · (1 + 𝟙^TD^-1𝟙)

where 𝟙^TD^-1𝟙 = 1/a + 1/b + 1/c + 1/d

Step 5: Find Vieta's formulas for the polynomial. From x⁴ - 2x³ - 4x² - 8x + 16 = 0:

• a + b + c + d = 2

• ab + ac + ad + bc + bd + cd = -4

• abc + abd + acd + bcd = 8

• abcd = 16

Step 6: Calculate 1/a + 1/b + 1/c + 1/d.

1/a + 1/b + 1/c + 1/d = (bcd + acd + abd + abc)/(abcd) = 8/16 = 1/2

Step 7: Compute the determinant.

det(M) = abcd · (1 + (1/a + 1/b + 1/c + 1/d))

det(M) = 16 · (1 + 1/2) = 16 · (3/2) = 24

∴ Answer: 24

Question:

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