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On using elementary column operations $C_2\times C_2-2C_1$ in the following matrix equation\[\begin{bmatrix}1 & 3\\2 & 4\end{bmatrix}=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}3 & 1\\2 & 4\end{bmatrix}, \text{we have}\]

\begin{array}{1 1}(A)\quad\begin{bmatrix}1 & -5\\0 & 4\end{bmatrix}=\begin{bmatrix}1 & 1\\2 & 2\end{bmatrix}\begin{bmatrix}3 & -5\\2 & 0\end{bmatrix}\\(B)\quad\begin{bmatrix}1 & -5\\0 & 4\end{bmatrix}=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}3 & -5\\-0 & 2\end{bmatrix}\\(C)\quad\begin{bmatrix}1 & 5\\0 & 2\end{bmatrix}=\begin{bmatrix}1 & 3\\0 & 1\end{bmatrix}\begin{bmatrix}3 & 1\\2 & 4\end{bmatrix}\\(D)\quad\begin{bmatrix}1 & 5\\2 & 0\end{bmatrix}=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}3 & -5\\2 & 0\end{bmatrix}\\\end{array}

1 Answer

On using elementary column operations $C_2\rightarrow C_2-2C_1$ in the following matrix equation.
 
$\begin{bmatrix}1 & -3\\2 & 4\end{bmatrix}=\begin{bmatrix}1 &-1\\0 & 1\end{bmatrix}\begin{bmatrix}3 & 1\\2 & 4\end{bmatrix}$ We have,
 
Apply $C_2\rightarrow C_2-2C_1$
 
$\Rightarrow \begin{bmatrix}1 & -5\\2 & 0\end{bmatrix}=\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}\begin{bmatrix}3 & -5\\2 & 0\end{bmatrix}$
 
Hence the option (D) is the right answer.

 

answered Mar 11, 2013 by sreemathi.v
 

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