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Question Number 116578 by bobhans last updated on 05/Oct/20
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
solveforx|1xx2x3122223133233144243|=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)) :}
|100012−x4−x28−x313−x9−x227−x314−x16−x264−x3|=0|2−x4−x28−x33−x9−x227−x34−x16−x264−x3|=0(2−x)(3−x)(4−x)|12+x4+2x+x213+x9+3x+x214+x16+4x+x2|=0(2−x)(3−x)(4−x){(12+7x)−(16+12x)+(6+5x)}=02(2−x)(3−x)(4−x)=0→{x=2x=3x=4
Answered by Olaf last updated on 05/Oct/20
det(A) = −2x^3 +18x^2 −52x+48 det(A) = −2(x−2)(x−3)(x−4) det(A) = 0 ⇔ x = 2 or 3 or 4
det(A)=−2x3+18x2−52x+48det(A)=−2(x−2)(x−3)(x−4)det(A)=0⇔x=2or3or4
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=|1xx2x3122223133233144243|=0⇔A=|248392741664|−x|148192711664|+x2|12813271464|−x3|1241391416|=48−52x+18x2−2x3=0⇔x3−9x2+26x−24=0⇔(x−2)(x−3)(x−4)=0⇔x∈{2,3,4}

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