project on solar system 6 7 pages in english
Answers
Explanation:
Answer:
System of linear equations has no solution.
Step-by-step explanation:
Given pair of linear equations is
\begin{gathered}\sf \: 3x - 2y = 3 \\ \\ \end{gathered}
3x−2y=3
and
\begin{gathered}\sf \: 6x - 4y = 7 \\ \\ \end{gathered}
6x−4y=7
In matrix form, can be represented as
\begin{gathered}\rm \: \bigg[ \begin{matrix}3& - 2 \\ 6& - 4 \end{matrix} \bigg] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c} 3\\7\end{array}\right] \\ \\ \end{gathered}
[
3
6
−2
−4
][
x
y
]=[
3
7
]
where,
\begin{gathered}\rm \: A = \bigg[ \begin{matrix}3& - 2 \\ 6& - 4 \end{matrix} \bigg] \\ \\ \end{gathered}
A=[
3
6
−2
−4
]
\begin{gathered}\sf \: B = \left[\begin{array}{c} 3\\7\end{array}\right] \\ \\ \end{gathered}
B=[
3
7
]
\begin{gathered}\sf \: X = \left[\begin{array}{c}x\\y\end{array}\right] \\ \\ \end{gathered}
X=[
x
y
]
Now, Consider
\begin{gathered}\sf \: |A| \\ \\ \end{gathered}
∣A∣
\begin{gathered}\qquad\sf \: = \: \bigg | \begin{matrix}3& - 2 \\ 6& - 4 \end{matrix} \bigg | \\ \\ \end{gathered}
=
∣
∣
3
6
−2
−4
∣
∣
\begin{gathered}\qquad\sf \: = \: - 12 - ( - 12) \\ \\ \end{gathered}
=−12−(−12)
\begin{gathered}\qquad\sf \: = \: - 12 + 12 \\ \\ \end{gathered}
=−12+12
\begin{gathered}\qquad\sf \: = \: 0 \\ \\ \end{gathered}
=0
This implies, system of equations is either consistent or inconsistent.
Now, Consider cofactors of matrix A
\begin{gathered}\sf \: A_{11} = {( - 1)}^{1 + 1}(- 4) = - 4 \\ \\ \end{gathered}
A
11
=(−1)
1+1
(−4)=−4
\begin{gathered}\sf \: A_{12} = {( - 1)}^{1 + 2}(6) = - 6 \\ \\ \end{gathered}
A
12
=(−1)
1+2
(6)=−6
\begin{gathered}\sf \: A_{21} = {( - 1)}^{2+ 1}( - 2) = 2 \\ \\ \end{gathered}
A
21
=(−1)
2+1
(−2)=2
\begin{gathered}\sf \: A_{22} = {( - 1)}^{2+ 2}(3) = 3 \\ \\ \end{gathered}
A
22
=(−1)
2+2
(3)=3
Thus,
\begin{gathered}\sf \: adjA = \bigg[ \begin{matrix}A_{11} & A_{21} \\ A_{12} & A_{22} \end{matrix} \bigg] \\ \\ \end{gathered}
adjA=[
A
11
A
12
A
21
A
22
]
\begin{gathered}\sf\implies \sf \: adjA = \bigg[ \begin{matrix}- 4&2 \\ - 6& 3 \end{matrix} \bigg] \\ \\ \end{gathered}
⟹adjA=[
−4
−6
2
3
]
Now, Consider
\begin{gathered}\sf \: (adjA)B \\ \\ \end{gathered}
(adjA)B
\begin{gathered}\sf \: = \: \bigg[ \begin{matrix} - 4& 2 \\ - 6& 3 \end{matrix} \bigg]\left[\begin{array}{c} 3\\7\end{array}\right] \\ \\ \end{gathered}
=[
−4
−6
2
3
][
3
7
]
\begin{gathered}\sf \: = \: \left[\begin{array}{c} - 12 + 14\\ - 18 + 21\end{array}\right] \\ \\ \end{gathered}
=[
−12+14
−18+21
]
\begin{gathered}\sf \: = \: \left[\begin{array}{c} 2\\ 3\end{array}\right] \\ \\ \end{gathered}
=[
2
3
]
So,
\begin{gathered}\sf\implies \sf \: (adjA)B \: \ne \: 0\\ \\ \end{gathered}
⟹(adjA)B
=0
\begin{gathered}\sf\implies \sf \: System\:of\:equations\:has\:no\:solution\\ \\ \end{gathered}
⟹Systemofequationshasnosolution