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Sagot :
Sure, let's solve the problem step-by-step.
You are given two vectors [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u = \begin{pmatrix} 3 \\ -2 \end{pmatrix} \][/tex]
[tex]\[ v = \begin{pmatrix} -12 \\ 5 \end{pmatrix} \][/tex]
We need to find [tex]\( u - 2v \)[/tex].
Step 1: Compute [tex]\( 2v \)[/tex].
First, multiply the vector [tex]\( v \)[/tex] by 2:
[tex]\[ 2v = 2 \cdot \begin{pmatrix} -12 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \cdot -12 \\ 2 \cdot 5 \end{pmatrix} = \begin{pmatrix} -24 \\ 10 \end{pmatrix} \][/tex]
Step 2: Compute [tex]\( u - 2v \)[/tex].
Now, subtract [tex]\( 2v \)[/tex] from [tex]\( u \)[/tex]:
[tex]\[ u - 2v = \begin{pmatrix} 3 \\ -2 \end{pmatrix} - \begin{pmatrix} -24 \\ 10 \end{pmatrix} \][/tex]
Subtract the corresponding components of the vectors:
[tex]\[ u - 2v = \begin{pmatrix} 3 - (-24) \\ -2 - 10 \end{pmatrix} = \begin{pmatrix} 3 + 24 \\ -2 - 10 \end{pmatrix} = \begin{pmatrix} 27 \\ -12 \end{pmatrix} \][/tex]
So, the result is:
[tex]\[ u - 2v = \begin{pmatrix} 27 \\ -12 \end{pmatrix} \][/tex]
Thus, the vector [tex]\( u - 2v \)[/tex] is:
[tex]\[ \begin{pmatrix} 27 \\ -12 \end{pmatrix} \][/tex]
You are given two vectors [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u = \begin{pmatrix} 3 \\ -2 \end{pmatrix} \][/tex]
[tex]\[ v = \begin{pmatrix} -12 \\ 5 \end{pmatrix} \][/tex]
We need to find [tex]\( u - 2v \)[/tex].
Step 1: Compute [tex]\( 2v \)[/tex].
First, multiply the vector [tex]\( v \)[/tex] by 2:
[tex]\[ 2v = 2 \cdot \begin{pmatrix} -12 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \cdot -12 \\ 2 \cdot 5 \end{pmatrix} = \begin{pmatrix} -24 \\ 10 \end{pmatrix} \][/tex]
Step 2: Compute [tex]\( u - 2v \)[/tex].
Now, subtract [tex]\( 2v \)[/tex] from [tex]\( u \)[/tex]:
[tex]\[ u - 2v = \begin{pmatrix} 3 \\ -2 \end{pmatrix} - \begin{pmatrix} -24 \\ 10 \end{pmatrix} \][/tex]
Subtract the corresponding components of the vectors:
[tex]\[ u - 2v = \begin{pmatrix} 3 - (-24) \\ -2 - 10 \end{pmatrix} = \begin{pmatrix} 3 + 24 \\ -2 - 10 \end{pmatrix} = \begin{pmatrix} 27 \\ -12 \end{pmatrix} \][/tex]
So, the result is:
[tex]\[ u - 2v = \begin{pmatrix} 27 \\ -12 \end{pmatrix} \][/tex]
Thus, the vector [tex]\( u - 2v \)[/tex] is:
[tex]\[ \begin{pmatrix} 27 \\ -12 \end{pmatrix} \][/tex]
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