Calculate the coordinates of a transformed matrix Algebra I

Let M=\begin{pmatrix}m_{ij}\end{pmatrix} and X=\begin{pmatrix} x_{i1} \end{pmatrix}.

Then the image of X under the transformation whose matrix is M is:

X'=\begin{pmatrix} x'_{i1} \end{pmatrix} =M \cdot X

Thus the entries of X' are given by x'_{i1}=\sum_{a}^{b}m_{ik}x_{k1}

Here we have:

• M=\begin{pmatrix} 11 & 12 \cr\cr 21 & 22 \cr\cr 31 & 32 \end{pmatrix}
• X=\begin{pmatrix} 1 \cr\cr 4 \end{pmatrix}

Therefore:

X'=MX=\begin{pmatrix} 11\times1+12 \times 4 \cr 21 \times 1+ 22\times 4\cr 31\times 1+ 32\times 4\end{pmatrix}= \begin{pmatrix} 59 \cr 109 \cr 159 \end{pmatrix}

Let M=\begin{pmatrix}m_{ij}\end{pmatrix} and X=\begin{pmatrix} x_{i1} \end{pmatrix}.

Then the image of X under the transformation whose matrix is M is:

X'=\begin{pmatrix} x'_{i1} \end{pmatrix} =M \cdot X

Thus the entries of X' are given by x'_{i1}=\sum_{a}^{b}m_{ik}x_{k1}

Here we have:

• M=\begin{pmatrix} 5 & 7 \cr\cr 6 & 3 \cr\cr 1 & 2 \end{pmatrix}
• X=\begin{pmatrix} 2 \cr\cr 3 \end{pmatrix}

Therefore:

X'=MX=\begin{pmatrix} 5\times2+7 \times 3 \cr 6 \times 2+3\times 3\cr 1\times 2+ 2\times 3\end{pmatrix}= \begin{pmatrix} 31 \cr 21 \cr 8 \end{pmatrix}

Let M=\begin{pmatrix}m_{ij}\end{pmatrix} and X=\begin{pmatrix} x_{i1} \end{pmatrix}.

Then the image of X under the transformation whose matrix is M is:

X'=\begin{pmatrix} x'_{i1} \end{pmatrix} =M \cdot X

Thus the entries of X' are given by x'_{i1}=\sum_{a}^{b}m_{ik}x_{k1}

Here we have:

• M=\begin{pmatrix} 2 & 5 \cr\cr 3 & 4 \end{pmatrix}
• X=\begin{pmatrix} 1 \cr\cr 2 \end{pmatrix}

Therefore:

X'=MX=\begin{pmatrix} 2\times1+5 \times 2 \cr 3\times 1+ 4\times 2\end{pmatrix}= \begin{pmatrix} 12 \cr 11 \end{pmatrix}

Let M=\begin{pmatrix}m_{ij}\end{pmatrix} and X=\begin{pmatrix} x_{i1} \end{pmatrix}.

Then the image of X under the transformation whose matrix is M is:

X'=\begin{pmatrix} x'_{i1} \end{pmatrix} =M \cdot X

Thus the entries of X' are given by x'_{i1}=\sum_{a}^{b}m_{ik}x_{k1}

Here we have:

• M=\begin{pmatrix} 8 & 2 \cr\cr 1 & 3 \end{pmatrix}
• X=\begin{pmatrix} 2 \cr\cr 4 \end{pmatrix}

Therefore:

X'=MX=\begin{pmatrix} 8\times2+2 \times 4 \cr 1 \times 2+ 3\times 4\end{pmatrix}= \begin{pmatrix} 24 \cr 14\end{pmatrix}

Let M=\begin{pmatrix}m_{ij}\end{pmatrix} and X=\begin{pmatrix} x_{i1} \end{pmatrix}.

Then the image of X under the transformation whose matrix is M is:

X'=\begin{pmatrix} x'_{i1} \end{pmatrix} =M \cdot X

Thus the entries of X' are given by x'_{i1}=\sum_{a}^{b}m_{ik}x_{k1}

Here we have:

• M=\begin{pmatrix} 1 & 1 & 2 \cr\cr 2 & 2 &1 \cr\cr 1 & 2 &2 \end{pmatrix}
• X=\begin{pmatrix} 2 \cr\cr 3 \cr \cr 1 \end{pmatrix}

Therefore:

X'=MX=\begin{pmatrix} 1\times2+1 \times 3+2\times 1 \cr 2 \times 2+ 2\times 3+1\times 1\cr 1\times 2+ 2\times 3 +2\times 1\end{pmatrix}= \begin{pmatrix} 7 \cr 11 \cr 10 \end{pmatrix}

Let M=\begin{pmatrix}m_{ij}\end{pmatrix} and X=\begin{pmatrix} x_{i1} \end{pmatrix}.

Then the image of X under the transformation whose matrix is M is:

X'=\begin{pmatrix} x'_{i1} \end{pmatrix} =M \cdot X

Thus the entries of X' are given by x'_{i1}=\sum_{a}^{b}m_{ik}x_{k1}

Here we have:

• M=\begin{pmatrix} -2 & 9 \cr\cr 12 & 4 \end{pmatrix}
• X=\begin{pmatrix} -1 \cr\cr 3 \end{pmatrix}

Therefore:

X'=MX=\begin{pmatrix} -2\times-1+9 \times 3 \cr 12 \times -1+ 4\times 3\end{pmatrix}= \begin{pmatrix} 29 \cr 0 \end{pmatrix}

Let M=\begin{pmatrix}m_{ij}\end{pmatrix} and X=\begin{pmatrix} x_{i1} \end{pmatrix}.

Then the image of X under the transformation whose matrix is M is:

X'=\begin{pmatrix} x'_{i1} \end{pmatrix} =M \cdot X

Thus the entries of X' are given by x'_{i1}=\sum_{a}^{b}m_{ik}x_{k1}

Here we have:

• M=\begin{pmatrix} 1 & 2 \cr\cr 2 & 1 \cr\cr 1 & 3 \end{pmatrix}
• X=\begin{pmatrix} -2 \cr\cr -2 \end{pmatrix}

Therefore:

X'=MX=\begin{pmatrix} 1\times-2+2 \times -2 \cr 2 \times -2+ 1\times -2\cr 1\times -2+ 3\times -2\end{pmatrix}= \begin{pmatrix} -6 \cr -6 \cr -8 \end{pmatrix}