Let $A = begin{bmatrix} 0 & 9 \ -1 & 6 end{bmatrix}$ and $B={b_1,b_2}$, where
Let $A = begin{bmatrix}
0 & 9 \
-1 & 6
end{bmatrix}$ and $B={b_1,b_2}$, where $b_1=begin{bmatrix}
3 \
1
end{bmatrix}, b_2 =begin{bmatrix}
2 \
1
end{bmatrix}$. Define $T: mathbb{R}^2 rightarrow mathbb{R}^2$ by $T(x)=Ax$. Find the matrix for $T$ relative to the basis $B$.
Is this the same as the change of coordinates from one matrix to another? If so I have the augmented matrix $begin{bmatrix}
3 & 2 & 0 & 9\
1 & 1 & -1 & 6
end{bmatrix}$ which yields $begin{bmatrix}
1 & 0 & 2 & -3 \
0 & 1 & -3 & 9
end{bmatrix}$, so then would the matrix be $begin{bmatrix}
2 & -3 \
-3 & 9
end{bmatrix}$? This is also a little different from other problems I've done, as some given property usually has to be satisfied.
linear-algebra change-of-basis
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Let $A = begin{bmatrix}
0 & 9 \
-1 & 6
end{bmatrix}$ and $B={b_1,b_2}$, where $b_1=begin{bmatrix}
3 \
1
end{bmatrix}, b_2 =begin{bmatrix}
2 \
1
end{bmatrix}$. Define $T: mathbb{R}^2 rightarrow mathbb{R}^2$ by $T(x)=Ax$. Find the matrix for $T$ relative to the basis $B$.
Is this the same as the change of coordinates from one matrix to another? If so I have the augmented matrix $begin{bmatrix}
3 & 2 & 0 & 9\
1 & 1 & -1 & 6
end{bmatrix}$ which yields $begin{bmatrix}
1 & 0 & 2 & -3 \
0 & 1 & -3 & 9
end{bmatrix}$, so then would the matrix be $begin{bmatrix}
2 & -3 \
-3 & 9
end{bmatrix}$? This is also a little different from other problems I've done, as some given property usually has to be satisfied.
linear-algebra change-of-basis
add a comment |
Let $A = begin{bmatrix}
0 & 9 \
-1 & 6
end{bmatrix}$ and $B={b_1,b_2}$, where $b_1=begin{bmatrix}
3 \
1
end{bmatrix}, b_2 =begin{bmatrix}
2 \
1
end{bmatrix}$. Define $T: mathbb{R}^2 rightarrow mathbb{R}^2$ by $T(x)=Ax$. Find the matrix for $T$ relative to the basis $B$.
Is this the same as the change of coordinates from one matrix to another? If so I have the augmented matrix $begin{bmatrix}
3 & 2 & 0 & 9\
1 & 1 & -1 & 6
end{bmatrix}$ which yields $begin{bmatrix}
1 & 0 & 2 & -3 \
0 & 1 & -3 & 9
end{bmatrix}$, so then would the matrix be $begin{bmatrix}
2 & -3 \
-3 & 9
end{bmatrix}$? This is also a little different from other problems I've done, as some given property usually has to be satisfied.
linear-algebra change-of-basis
Let $A = begin{bmatrix}
0 & 9 \
-1 & 6
end{bmatrix}$ and $B={b_1,b_2}$, where $b_1=begin{bmatrix}
3 \
1
end{bmatrix}, b_2 =begin{bmatrix}
2 \
1
end{bmatrix}$. Define $T: mathbb{R}^2 rightarrow mathbb{R}^2$ by $T(x)=Ax$. Find the matrix for $T$ relative to the basis $B$.
Is this the same as the change of coordinates from one matrix to another? If so I have the augmented matrix $begin{bmatrix}
3 & 2 & 0 & 9\
1 & 1 & -1 & 6
end{bmatrix}$ which yields $begin{bmatrix}
1 & 0 & 2 & -3 \
0 & 1 & -3 & 9
end{bmatrix}$, so then would the matrix be $begin{bmatrix}
2 & -3 \
-3 & 9
end{bmatrix}$? This is also a little different from other problems I've done, as some given property usually has to be satisfied.
linear-algebra change-of-basis
linear-algebra change-of-basis
edited Nov 29 at 9:54
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Nov 29 at 9:13
HighSchool15
1,065517
1,065517
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