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Question Number 55374 by gunawan last updated on 23/Feb/19

Given matrices A(x)= [((x−1),3),(4,(x+3)) ]∈ M_2 (R). The smallest det(A(x)) is..

$$\mathrm{Given}\:\mathrm{matrices}\:{A}\left({x}\right)=\begin{bmatrix}{{x}−\mathrm{1}}&{\mathrm{3}}\\{\mathrm{4}}&{{x}+\mathrm{3}}\end{bmatrix}\in\:{M}_{\mathrm{2}} \left(\mathbb{R}\right). \\ $$$$\mathrm{The}\:\mathrm{smallest}\:\mathrm{det}\left({A}\left({x}\right)\right)\:\mathrm{is}.. \\ $$

Answered by Joel578 last updated on 23/Feb/19

∣A(x)∣ = (x − 1)(x + 3) − 12 = x^2 + 2x − 15 = (x + 1)^2 − 16 So the smallest ∣A(x)∣ is −16, when x = −1

$$\mid{A}\left({x}\right)\mid\:=\:\left({x}\:−\:\mathrm{1}\right)\left({x}\:+\:\mathrm{3}\right)\:−\:\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{15} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({x}\:+\:\mathrm{1}\right)^{\mathrm{2}} \:−\:\mathrm{16}\:\: \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{smallest}\:\mid{A}\left({x}\right)\mid\:\mathrm{is}\:−\mathrm{16},\:\mathrm{when}\:{x}\:=\:−\mathrm{1} \\ $$

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