Question

Find the matrix A such that `[[1     1],[0       1]]A=[[3        3         5],[1       0          1]]`

Answer 1

\[\left( i \right) \]
\[Let A = \begin{bmatrix}x & y & z \\ a & b & c\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}\begin{bmatrix}x & y & z \\ a & b & c\end{bmatrix} = \begin{bmatrix}3 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x + a & y + b & z + c \\ 0 + a & 0 + b & 0 + c\end{bmatrix} = \begin{bmatrix}3 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x + a & y + b & z + c \\ a & b & c\end{bmatrix} = \begin{bmatrix}3 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}\]
The corresponding elements of two equal matrices are equal 
\[ \Rightarrow x + a = 3 . . . \left( 1 \right)\]
\[y + b = 3 . . . \left( 2 \right) \]
\[z + c = 5 . . . \left( 3 \right)\]
\[\]
`⇒ a=1,b=0,  and   c=1`
\[\]
putting the value of  a  in eq.(1),we   get
\[x + 1 = 3\]
\[ \Rightarrow x = 3 - 1\]
\[ \therefore x = 2\]
\[\]
putting the value of  b  in eq.(2),we   get
\[y + b = 3\]
\[ \Rightarrow y + 0 = 3\]
\[ \therefore y = 3\]
\[\]
putting the value of  c in eq.(3),we   get

\[z + 1 = 5\]
\[ \Rightarrow z = 5 - 1\]
\[ \therefore z = 4\]
\[ \therefore A = \begin{bmatrix}2 & 3 & 4 \\ 1 & 0 & 1\end{bmatrix}\]