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Multiplication of Matrices

x2+3 x-1 x...

Question

$∣ ∣ ∣ ∣ \begin{matrix} x^{2} + 3 & x - 1 & x + 3 x + 3 & - 2 x & x - 4 x - 3 & x + 4 & 3 x \end{matrix} ∣ ∣ ∣ ∣$ $= p x^{4} + q x^{3} + r x^{2} + s x + t,$ then $t =$

A

72

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B

0

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C

24

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D

- 48

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Solution

The correct option is C $72$
On expanding the determinant we get,
$(x^{2} + 3) (- 6 x^{2} - x^{2} + 16) - (x - 1) (3 x^{2} + 9 x - x^{2} + 7 x - 12) + (x + 3) (x^{2} + 7 x + 12 - 2 x^{2} + 6 x)$
Let us consider only the constant terms,
since $t$ is constant term
$∴ (16) (3) - (- 1) (- 12) + 3 (12) = 72$

Hence, $t = 72$

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Similar questions

p x^{4} + q x^{3} + r x^{2} + s x + t = ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + 3 x & x - 1 & x + 3 x + 1 & 2 - x & x - 3 x - 3 & x + 4 & 3 x \end{matrix} ∣ ∣ ∣ ∣

t

x^{2} - 1

p x^{4} + q x^{3} + r x^{2} + s x + u

, show that p + r + u = q + s = 0.

a x^{4} + b x^{3} + c x^{2} + d x + e = ∣ ∣ ∣ ∣ \begin{matrix} x^{3} + 3 x & x - 1 & x + 3 x + 1 & - 2 x & x - 4 x - 3 & x + 4 & 3 x \end{matrix} ∣ ∣ ∣ ∣

a x^{4} + b x^{3} + c x^{2} + d x + e

∣ ∣ ∣ ∣ \begin{matrix} x^{3} + 3 x & x - 1 & x + 3 x + 1 & - 2 x & x - 4 x - 3 & x + 4 & 3 x \end{matrix} ∣ ∣ ∣ ∣

e = 0

|\begin{matrix} x^{2} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{matrix}| = a x^{4} + b x^{3} + c x^{2} + d x + e

be an identity in x, where a, b, c, d, e are independent of x. Then the value of e is
(a) 4
(b) 0
(c) 1
(d) none of these

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