Answer to Question #233583 in Linear Algebra for rabiatul

Question #233583

A curve


y  ax  bx  c

2


where a, b and c are constants, passes


through the points (2,11), (-1,-16) and (3,28).


1
Expert's answer
2021-09-06T14:32:31-0400
"y=ax^2+bx+c"

(a)


"a(2)^2+b(2)+c=11"

"a(-1)^2+b(-1)+c=-16"




"a(3)^2+b(3)+c=28"

Construct a system containing three linear equations.


"4a+2b+c=11""a-b+c=-16""9a+3b+c=28"

(b)


"AX=B"


"A=\\begin{pmatrix}\n 4 & 2 & 1 \\\\\n 1 & -1 & 1 \\\\\n 9 & 3 & 1 \\\\\n\\end{pmatrix}, X=\\begin{pmatrix}\n a \\\\\n b \\\\\n c\n\\end{pmatrix}, B=\\begin{pmatrix}\n11\\\\\n -16 \\\\\n 28\n\\end{pmatrix}"

"\\begin{pmatrix}\n 4 & 2 & 1 \\\\\n 1 & -1 & 1 \\\\\n 9 & 3 & 1 \\\\\n\\end{pmatrix}\\begin{pmatrix}\n a \\\\\n b \\\\\n c\n\\end{pmatrix}=\\begin{pmatrix}\n11\\\\\n -16 \\\\\n 28\n\\end{pmatrix}"

(c)


"AX=B"

"A^{-1}AX=A^{-1}B"

"X=A^{-1}B"

"\\det A=\\begin{vmatrix}\n 4 & 2 & 1 \\\\\n 1 & -1 & 1 \\\\\n 9 & 3 & 1 \\\\\n\\end{vmatrix}=4\\begin{vmatrix}\n -1 & 1 \\\\\n 3 & 1\n\\end{vmatrix}-2\\begin{vmatrix}\n 1 & 1 \\\\\n 9 & 1\n\\end{vmatrix}+1\\begin{vmatrix}\n 1 & -1 \\\\\n 9 & 3\n\\end{vmatrix}"

"=4(-1-3)-2(1-9)+(3+9)"

"=-16+16+12=12\\not=0=>A^{-1}\\ exists"


Find the cofactor matrix:


"C_{11}=(-1)^{1+1}\\begin{vmatrix}\n -1 & 1 \\\\\n 3 & 1\n\\end{vmatrix}=-4"

"C_{12}=(-1)^{1+2}\\begin{vmatrix}\n 1 & 1 \\\\\n 9 & 1\n\\end{vmatrix}=8"

"C_{13}=(-1)^{1+3}\\begin{vmatrix}\n 1 & -1 \\\\\n 9 & 3\n\\end{vmatrix}=12"

"C_{21}=(-1)^{2+1}\\begin{vmatrix}\n 2 & 1 \\\\\n 3 & 1\n\\end{vmatrix}=1"

"C_{22}(-1)^{2+2}\\begin{vmatrix}\n 4 & 1 \\\\\n 9 & 1\n\\end{vmatrix}=-5"

"C_{23}=(-1)^{2+3}\\begin{vmatrix}\n 4 & 2 \\\\\n 9 & 3\n\\end{vmatrix}=6"

"C_{31}=(-1)^{3+1}\\begin{vmatrix}\n 2 & 1 \\\\\n -1 & 1\n\\end{vmatrix}=3"

"C_{32}=(-1)^{3+2}\\begin{vmatrix}\n 4 & 1 \\\\\n 1 & 1\n\\end{vmatrix}=-3"

"C_{33}=(-1)^{3+3}\\begin{vmatrix}\n 4 & 2 \\\\\n 1 & -1\n\\end{vmatrix}=-6"

The cofactor matrix is


"C=\\begin{pmatrix}\n -4 &8 & 12 \\\\\n 1 & -5 & 6 \\\\\n 3 & -3 & -6 \\\\\n\\end{pmatrix}"

The transpose of the cofactor matrix is


"C^T=\\begin{pmatrix}\n -4 & 1 & 3\\\\\n 8 & -5 & -3 \\\\\n 12 & 6 & -6 \\\\\n\\end{pmatrix}"

"A^{-1}=\\dfrac{1}{12}\\begin{pmatrix}\n -4 & 1 & 3\\\\\n 8 & -5 & -3 \\\\\n 12 & 6 & -6 \\\\\n\\end{pmatrix}"

"A^{-1}=\\begin{pmatrix}\n -1\/3 & 1\/12 & 1\/4\\\\\n 2\/3 & -5\/12 & -1\/4 \\\\\n 1 & 1 \/2& -1\/2 \\\\\n\\end{pmatrix}"

"\\begin{pmatrix}\n a \\\\\n b \\\\\n c\n\\end{pmatrix}=\\begin{pmatrix}\n -1\/3 & 1\/12 & 1\/4\\\\\n 2\/3 & -5\/12 & -1\/4 \\\\\n 1 & 1 \/2& -1\/2 \\\\\n\\end{pmatrix}\\begin{pmatrix}\n11\\\\\n -16 \\\\\n 28\n\\end{pmatrix}"

"=\\begin{pmatrix}\n-11\/3-4\/3+7\\\\\n22\/3+20\/3-7 \\\\\n 11-8-14\n\\end{pmatrix}=\\begin{pmatrix}\n2\\\\\n7 \\\\\n -11\n\\end{pmatrix}"

"a=2, b=7, c=-11"

"y=2x^2+7x-11"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS