Question
A toy rocket is fired, from a platform, vertically into the air, its height above the ground after $t$ seconds is given by $s(t)= a t ^2+ b t + c$, where $a , b , c \in R ; a \neq 0$ and $s(t)$ is measured in
metres. After $1 0$ second, the rocket is $1 6 ~ m$ above the ground; after $2 0$ seconds, $2 2 ~ m$; after 30 seconds, $2 5$ m.
(i) Write down a system of three linear equations in terms of $a , b$ and $c$.
(ii) Hence find the values of $a , b$ and $c$, using matrix method.

Answer

(i) $s(t)=a t^2+b t+c ; t \geq 0$
Clearly, $(10,16),(20,22),(30,25)$ lie on the curve of $s(t)$.
Then, 100a+10b+c=16
$
\begin{aligned}
400 a+20 b+c & =22 \\
900 a+30 b+c & =25
\end{aligned}
$
(ii) Let, $A=\left(\begin{array}{ccc}100 & 10 & 1 \\ 400 & 20 & 1 \\ 900 & 30 & 1\end{array}\right) ; X=\left(\begin{array}{l}a \\ b \\ c\end{array}\right) ; B=\left(\begin{array}{l}16 \\ 22 \\ 25\end{array}\right)$
Then, the system becomes, $A X=B$
$
\begin{aligned}
|A|= & 100(-10)-400(-20)+900(-10) \\
& =-1000+8000-9000 \\
& =-2000 \neq 0
\end{aligned}
$
Now, adj $A=\left(\begin{array}{ccc}-10 & 500 & -6000 \\ 20 & -800 & 6000 \\ -10 & 300 & -2000\end{array}\right)^T=\left(\begin{array}{ccc}-10 & 20 & -10 \\ 500 & -800 & 300 \\ -6000 & 6000 & -2000\end{array}\right)$
Therefore, $A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{-2000}\left(\begin{array}{ccc}-10 & 20 & -10 \\ 500 & -800 & 300 \\ -6000 & 6000 & -2000\end{array}\right)$
Then, $X=A^{-1} B=\frac{1}{-2000}\left(\begin{array}{ccc}-10 & 20 & -10 \\ 500 & -800 & 300 \\ -6000 & 6000 & -2000\end{array}\right)\left(\begin{array}{l}16 \\ 22 \\ 25\end{array}\right)$
$
=\frac{1}{-2000}\left(\begin{array}{c}
30 \\
-2100 \\
-14000
\end{array}\right)
$
$
=\left(\begin{array}{c}
-\frac{3}{200} \\
\frac{21}{20} \\
7
\end{array}\right)
$
Therefore, $a=-\frac{3}{200}, b=\frac{21}{20}, c=7$.

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