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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