Matrices & Determinants Question - None

Question

For Problems 12 and 13
Let for $A = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{bmatrix}$, there be three row matrices $R_1, R_2$ and $R_3$, satisfying the relations, $R_1 A = [1\ 0\ 0]$, $R_2 A = [2\ 3\ 0]$ and $R_3 A = [2\ 3\ 1]$. If $B$ is square matrix of order 3 with rows $R_1, R_2$ and $R_3$ in order, then

The value of det.$(2A^{100}B^3 - A^{99}B^4)$ is

Accepted Answer

Step 1: Verify that B = A⁻¹. Since R₁A = [1 0 0], R₂A = [0 2 3], R₃A = [0 0 1], stacking these rows gives BA = I, so B = A⁻¹.

Step 2: Rewrite the expression: det(2A^100B³ - A^99B⁴) = det(A^99(2A·B³ - B⁴)) = det(A^99) · det(2AB³ - B⁴).

Step 3: Since B = A⁻¹, we have AB = I and B³ = (A⁻¹)³ = A⁻³. Thus: 2AB³ - B⁴ = 2·I·A⁻³ - A⁻⁴ = 2A⁻³ - A⁻⁴ = A⁻⁴(2A - I).

Step 4: Calculate det(A) = 1 (lower triangular: product of diagonal = 1×1×1). So det(A^99) = 1 and det(A⁻⁴) = 1.

Step 5: Find det(2A - I) = det(begin{bmatrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 6 & 4 & 1 end{bmatrix}) = 1.

Step 6: det(2A^100B³ - A^99B⁴) = 1 · 1 · 1 = 1.

∴ Answer: B

Question:

Step-by-Step Solution

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