solve for x determinant (((1 x x^2 x^3 )),((1 2 2^2 2^3 )),((1 3 3^2 3^3 )),((1 4 4^2 4^3 )))= 0


Question and Answers Forum

All Questions Topic List
Matrices and Determinants Questions
Previous in All Question Next in All Question
Previous in Matrices and Determinants Next in Matrices and Determinants
Question Number 116578 by bobhans last updated on 05/Oct/20

$solve for x$ $\| \begin{matrix} 1 x x^{2} x^{3} \\ 1 2 2^{2} 2^{3} \\ 1 3 3^{2} 3^{3} \\ 1 4 4^{2} 4^{3} \end{matrix} \| = 0$
	Answered by bemath last updated on 05/Oct/20
	$determinant (((1 0 0 0)),((1 2−x 4−x^2 8−x^3 )),((1 3−x 9−x^2 27−x^3 )),((1 4−x 16−x^2 64−x^3 )))=0 determinant (((2−x 4−x^2 8−x^3 )),((3−x 9−x^2 27−x^3 )),((4−x 16−x^2 64−x^3 )))= 0 (2−x)(3−x)(4−x) determinant (((1 2+x 4+2x+x^2 )),((1 3+x 9+3x+x^2 )),((1 4+x 16+4x+x^2 )))= 0 (2−x)(3−x)(4−x){(12+7x)−(16+12x)+(6+5x)} = 0 2(2−x)(3−x)(4−x)=0 → { ((x=2)),((x=3)),((x=4)) :}$
	$\| \begin{matrix} 1 0 0 0 \\ 1 2 - x 4 - x^{2} 8 - x^{3} \\ 1 3 - x 9 - x^{2} 27 - x^{3} \\ 1 4 - x 16 - x^{2} 64 - x^{3} \end{matrix} \| = 0$ $\| \begin{matrix} 2 - x 4 - x^{2} 8 - x^{3} \\ 3 - x 9 - x^{2} 27 - x^{3} \\ 4 - x 16 - x^{2} 64 - x^{3} \end{matrix} \| = 0$ $(2 - x) (3 - x) (4 - x) \| \begin{matrix} 1 2 + x 4 + 2 x + x^{2} \\ 1 3 + x 9 + 3 x + x^{2} \\ 1 4 + x 16 + 4 x + x^{2} \end{matrix} \| = 0$ $(2 - x) (3 - x) (4 - x) {(12 + 7 x) - (16 + 12 x) + (6 + 5 x)} = 0$ $2 (2 - x) (3 - x) (4 - x) = 0$ $\to {\begin{cases} x = 2 \\ x = 3 \\ x = 4 \end{cases}$
	Answered by Olaf last updated on 05/Oct/20

	$\det (A) = - 2 x^{3} + 18 x^{2} - 52 x + 48$ $\det (A) = - 2 (x - 2) (x - 3) (x - 4)$ $\det (A) = 0 \Leftrightarrow x = 2 or 3 or 4$
	Answered by 1549442205PVT last updated on 05/Oct/20
	$A= determinant (((1 x x^2 x^3 )),((1 2 2^2 2^3 )),((1 3 3^2 3^3 )),((1 4 4^2 4^3 )))= 0 ⇔A= determinant ((2,4,8),(3,9,(27)),(4,(16),(64)))−x determinant ((1,4,8),(1,9,(27)),(1,(16),(64))) +x^2 determinant ((1,2,8),(1,3,(27)),(1,4,(64)))−x^3 determinant ((1,2,4),(1,3,9),(1,4,(16))) =48−52x+18x^2 −2x^3 =0 ⇔x^3 −9x^2 +26x−24=0 ⇔(x−2)(x−3)(x−4)=0 ⇔x∈{2,3,4}$
	$A = \| \begin{matrix} 1 x x^{2} x^{3} \\ 1 2 2^{2} 2^{3} \\ 1 3 3^{2} 3^{3} \\ 1 4 4^{2} 4^{3} \end{matrix} \| = 0$ $\Leftrightarrow A = \| \begin{matrix} 2 & 4 & 8 \\ 3 & 9 & 27 \\ 4 & 16 & 64 \end{matrix} \| - x \| \begin{matrix} 1 & 4 & 8 \\ 1 & 9 & 27 \\ 1 & 16 & 64 \end{matrix} \|$ $+ x^{2} \| \begin{matrix} 1 & 2 & 8 \\ 1 & 3 & 27 \\ 1 & 4 & 64 \end{matrix} \| - x^{3} \| \begin{matrix} 1 & 2 & 4 \\ 1 & 3 & 9 \\ 1 & 4 & 16 \end{matrix} \|$ $= 48 - 52 x + {18 x}^{2} - {2 x}^{3} = 0$ $\Leftrightarrow x^{3} - {9 x}^{2} + 26 x - 24 = 0$ $\Leftrightarrow (x - 2) (x - 3) (x - 4) = 0$ $\Leftrightarrow x \in {2, 3, 4}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum