2020-12-03T01:14:45-05:00
Denote by a, b and c the column vectors a = (1 2 3), b = (-2 1 -3), c = (-2 -1 1) Calculate 2a - 5b, 2a- 5b +c, a'.b,
1
2020-12-04T12:52:08-0500
a = ( 1 2 3 ) , b = ( − 2 1 − 3 ) , c = ( − 2 − 1 1 ) a=\begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}, b=\begin{pmatrix}
-2 \\
1 \\
-3
\end{pmatrix}, c=\begin{pmatrix}
-2 \\
-1 \\
1
\end{pmatrix} a = ⎝ ⎛ 1 2 3 ⎠ ⎞ , b = ⎝ ⎛ − 2 1 − 3 ⎠ ⎞ , c = ⎝ ⎛ − 2 − 1 1 ⎠ ⎞
2 a − 5 b = 2 ( 1 2 3 ) − 5 ( − 2 1 − 3 ) = ( 2 ( 1 ) − 5 ( − 2 ) 2 ( 2 ) − 5 ( 1 ) 2 ( 3 ) − 5 ( − 3 ) ) 2a-5b=2\begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}-5\begin{pmatrix}
-2 \\
1 \\
-3
\end{pmatrix}=\begin{pmatrix}
2(1)-5(-2) \\
2(2)-5(1) \\
2(3)-5(-3)
\end{pmatrix} 2 a − 5 b = 2 ⎝ ⎛ 1 2 3 ⎠ ⎞ − 5 ⎝ ⎛ − 2 1 − 3 ⎠ ⎞ = ⎝ ⎛ 2 ( 1 ) − 5 ( − 2 ) 2 ( 2 ) − 5 ( 1 ) 2 ( 3 ) − 5 ( − 3 ) ⎠ ⎞
= ( 12 − 1 21 ) =\begin{pmatrix}
12 \\
-1 \\
21
\end{pmatrix} = ⎝ ⎛ 12 − 1 21 ⎠ ⎞
2 a − 5 b + c = 2 ( 1 2 3 ) − 5 ( − 2 1 − 3 ) + ( − 2 − 1 1 ) 2a-5b+c=2\begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}-5\begin{pmatrix}
-2 \\
1 \\
-3
\end{pmatrix}+\begin{pmatrix}
-2 \\
-1 \\
1
\end{pmatrix} 2 a − 5 b + c = 2 ⎝ ⎛ 1 2 3 ⎠ ⎞ − 5 ⎝ ⎛ − 2 1 − 3 ⎠ ⎞ + ⎝ ⎛ − 2 − 1 1 ⎠ ⎞
= ( 12 − 1 21 ) + ( − 2 − 1 1 ) = ( 12 − 2 − 1 − 1 21 + 1 ) = ( 10 − 2 22 ) =\begin{pmatrix}
12 \\
-1 \\
21
\end{pmatrix}+\begin{pmatrix}
-2 \\
-1 \\
1
\end{pmatrix}=\begin{pmatrix}
12-2 \\
-1 -1 \\
21+1
\end{pmatrix}=\begin{pmatrix}
10 \\
-2 \\
22
\end{pmatrix} = ⎝ ⎛ 12 − 1 21 ⎠ ⎞ + ⎝ ⎛ − 2 − 1 1 ⎠ ⎞ = ⎝ ⎛ 12 − 2 − 1 − 1 21 + 1 ⎠ ⎞ = ⎝ ⎛ 10 − 2 22 ⎠ ⎞
a T ⋅ b = ( 1 2 3 ) ⋅ ( − 2 1 − 3 ) a^T\cdot b=\begin{pmatrix}
1 & 2 & 3
\end{pmatrix}\cdot\begin{pmatrix}
-2 \\
1 \\
-3
\end{pmatrix} a T ⋅ b = ( 1 2 3 ) ⋅ ⎝ ⎛ − 2 1 − 3 ⎠ ⎞
= 1 ( − 2 ) + 2 ( 1 ) + 3 ( − 3 ) = − 9 =1(-2)+2(1)+3(-3)=-9 = 1 ( − 2 ) + 2 ( 1 ) + 3 ( − 3 ) = − 9
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