(3a^4 b^5)^2 . (10a^3b^2)^3
(2a^5 b^2)^4
Answers
Step-by-step explanation:
After adding we get:
(3a−4b+4c)+(2a+3b−8c)+(a−6b+c)
=3a+2a+a−4b+3b−6b+4c−8c+c
=6a−7b−3c
Answer:
:⟼2A+3B=[
2
4
3
0
]−−(1)
and
\begin{gathered}\rm :\longmapsto\:3A + 2B = \bigg[ \begin{matrix} - 2&2 \\ 1& - 5 \end{matrix} \bigg] - - - (2)\end{gathered}
:⟼3A+2B=[
−2
1
2
−5
]−−−(2)
On Adding equation (2) and equation (1), we get
\begin{gathered}\rm :\longmapsto\:5A + 5B = \bigg[ \begin{matrix}0&5 \\ 5& - 5 \end{matrix} \bigg]\end{gathered}
:⟼5A+5B=[
0
5
5
−5
]
\begin{gathered}\rm :\longmapsto\:A + B = \bigg[ \begin{matrix}0&1 \\ 1& - 1 \end{matrix} \bigg] - - - (3)\end{gathered}
:⟼A+B=[
0
1
1
−1
]−−−(3)
On Subtracting equation (1) from equation (2), we get
\begin{gathered}\rm :\longmapsto\:A - B = \bigg[ \begin{matrix} - 4& - 1\\ - 3& - 5 \end{matrix} \bigg] - - - (4)\end{gathered}
:⟼A−B=[
−4
−3
−1
−5
]−−−(4)
On Adding equation (3) and (4), we get
\begin{gathered}\rm :\longmapsto\:2A = \bigg[ \begin{matrix} - 4&0 \\ - 2& - 6 \end{matrix} \bigg]\end{gathered}
:⟼2A=[
−4
−2
0
−6
]
\begin{gathered}\rm :\longmapsto\:A = \bigg[ \begin{matrix} - 2&0 \\ - 1& - 3 \end{matrix} \bigg]\end{gathered}
:⟼A=[
−2
−1
0
−3
]
On Subtracting equation (4) from equation (3), we get
\begin{gathered}\rm :\longmapsto\:2B = \bigg[ \begin{matrix}4&2 \\ 4&4 \end{matrix} \bigg]\end{gathered}
:⟼2B=[
4
4
2
4
]
\begin{gathered}\rm :\longmapsto\:B = \bigg[ \begin{matrix}2&1 \\ 2&2 \end{matrix} \bigg]\end{gathered}
:⟼B=[
2
2
1
2
]
Hence,
\begin{gathered}\rm :\longmapsto\:A = \bigg[ \begin{matrix} - 2&0 \\ - 1& - 3 \end{matrix} \bigg]\end{gathered}
:⟼A=[
−2
−1
0
−3
]
and
\begin{gathered}\bf :\longmapsto\:B = \bigg[ \begin{matrix}2&1 \\ 2&2 \end{matrix} \bigg]\end{gathered}
:⟼B=[
2
2
1
2
]