Answer to Question #233584 in Linear Algebra for rabiatul

Question

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).

(a) By using the above information, construct a system

containing three linear equations.

(b) Express the above system as a matrix equation

AX  B.

(c) Find the inverse of matrix A by using the adjoint matrix

method. Hence, obtain the values of a, b and c.

Answer 1

$y=ax^2+bx+c$

(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)

$AX=B$

$A=\begin{pmatrix} 4 & 2 & 1 \\ 1 & -1 & 1 \\ 9 & 3 & 1 \\ \end{pmatrix}, X=\begin{pmatrix} a \\ b \\ c \end{pmatrix}, B=\begin{pmatrix} 11\\ -16 \\ 28 \end{pmatrix}$

$\begin{pmatrix} 4 & 2 & 1 \\ 1 & -1 & 1 \\ 9 & 3 & 1 \\ \end{pmatrix}\begin{pmatrix} a \\ b \\ c \end{pmatrix}=\begin{pmatrix} 11\\ -16 \\ 28 \end{pmatrix}$

(c)

$AX=B$

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

$X=A^{-1}B$

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

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

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

Find the cofactor matrix:

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

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

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

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

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

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

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

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

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

The cofactor matrix is

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

The transpose of the cofactor matrix is

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

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

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

$\begin{pmatrix} a \\ b \\ c \end{pmatrix}=\begin{pmatrix} -1/3 & 1/12 & 1/4\\ 2/3 & -5/12 & -1/4 \\ 1 & 1 /2& -1/2 \\ \end{pmatrix}\begin{pmatrix} 11\\ -16 \\ 28 \end{pmatrix}$

$=\begin{pmatrix} -11/3-4/3+7\\ 22/3+20/3-7 \\ 11-8-14 \end{pmatrix}=\begin{pmatrix} 2\\ 7 \\ -11 \end{pmatrix}$

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

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