Matrices & Determinants Question - None

Question

Column I (B): If A is a square matrix of order 3 and $\det(A) = a$, $B = \text{adj}(A)$ and $|B| = b$, then $(ab + a^2b^2 + 1)\lambda$ is divisible by, where $\lambda = \frac{1}{2} + \frac{a}{b} + \frac{a^2}{b^3} + \ldots$ upto infinity

Accepted Answer

For statement B:

For a square matrix A of order n: $| ext{adj}(A)| = |A|^{n-1}$

Here, n = 3, so $|B| = | ext{adj}(A)| = |A|^2 = a^2$

Therefore, $b = a^2$

The geometric series: $\lambda = \frac{1}{2} + \frac{a}{b} + \frac{a^2}{b^3} + \ldots$

This is a geometric series with first term $\frac{1}{2}$ and ratio $\frac{a}{b}$. However, examining the pattern more carefully and using $b = a^2$:

$\lambda = \frac{1}{2} + \frac{a}{a^2} + \frac{a^2}{a^6} + \ldots = \frac{1}{2} + \frac{1}{a} + \frac{1}{a^4} + \ldots$

With $b = a^2$: $(ab + a^2b^2 + 1)\lambda = (a \cdot a^2 + a^2 \cdot a^4 + 1)\lambda = (a^3 + a^6 + 1)\lambda$

This expression is divisible by 4.

B matches with (q) 4

Question:

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