If x - 4 2 x 2 x 2 x x - 4 2 x 2 x 2 x x - 4 = A + B x x - A 2 , then the ordered pair ( A , B ) is equal to

x - 4 2 x 2 x 2 x x - 4 2 x 2 x 2 x x - 4 = A + B x x - A 2 Put x = 0 ⇒ - 4 0 0 0 - 4 0 0 0 - 4 = A 3 ⇒ A = - 4 x - 4 2 x 2 x 2 x x - 4 2 x 2 x 2 x x - 4 = B x - 4 x + 4 2 1 - 4 x 2 2 2 1 - 4 x 2 2 2 1 - 4 x = B - 4 x 1 + 4 x 2 Put x → ∞ ⇒ 1 2 2 2 1 2 2 2 1 = B On expanding the determinant along the first row, we get B = 5 .

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Mathematics - Determinant Question with Solution | TestHub

MathematicsDeterminantProperties of det.Medium2 minPYQ_2018

MathematicsMediumsingle choice

If $|\begin{matrix} x - 4 & 2 x & 2 x \\ 2 x & x - 4 & 2 x \\ 2 x & 2 x & x - 4 \end{matrix}| = (A + B x) {(x - A)}^{2},$ then the ordered pair $(A, B)$ is equal to

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Solution:

$|\begin{matrix} x - 4 & 2 x & 2 x \\ 2 x & x - 4 & 2 x \\ 2 x & 2 x & x - 4 \end{matrix}| = (A + B x) {(x - A)}^{2}$

Put $x = 0 \Rightarrow |\begin{matrix} - 4 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & - 4 \end{matrix}| = A^{3} \Rightarrow A = - 4$

$|\begin{matrix} x - 4 & 2 x & 2 x \\ 2 x & x - 4 & 2 x \\ 2 x & 2 x & x - 4 \end{matrix}| = (B x - 4) {(x + 4)}^{2}$

$|\begin{matrix} 1 - \frac{4}{x} & 2 & 2 \\ 2 & 1 - \frac{4}{x} & 2 \\ 2 & 2 & 1 - \frac{4}{x} \end{matrix}| = (B - \frac{4}{x}) {(1 + \frac{4}{x})}^{2}$

Put $x \to \infty \Rightarrow |\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}| = B$

On expanding the determinant along the first row, we get $B = 5$ .

Stream:JEESubject:MathematicsTopic:DeterminantSubtopic:Properties of det.

⏱ 2mℹ️ Source: PYQ_2018

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