Question

Find the matrix X so that \(X\begin{bmatrix}1&2&3\\ 4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\ 2&4&6\end{bmatrix}\)

Answer 1

The correct answer is \(\begin{bmatrix}1&-2\\2&0\end{bmatrix}\)
It is given that \(X\begin{bmatrix}1&2&3\\ 4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\ 2&4&6\end{bmatrix}\)
The matrix given on the R.H.S. of the equation is a \(2 \times 3\) matrix and the one given on the L.H.S. of the equation is a \(2 \times 3\) matrix. Therefore, X has to be a \(2 \times 2\) matrix.
Now, let \(X=\begin{bmatrix}a&c\\b&d\end{bmatrix}\)
Therefore, we have:
\(\begin{bmatrix}a&c\\b&d\end{bmatrix}\begin{bmatrix}1&2&3\\ 4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\ 2&4&6\end{bmatrix}\)
\(\implies\begin{bmatrix}a+4c& 2a+5c& 3a+6c\\ b+4d& 2b+5d& 3b+6d\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\ 2&4&6\end{bmatrix}\)
Equating the corresponding elements of the two matrices, we have:
\(a+4c=-7,2a+5c=-8,3a+6c=-9\)
\(b+4d=2,2b+5d =4,3b+6d=6\)
Now \(a+4c=-7\)
\(=\implies a=-7-4c\)
\(2a+5c=-8\implies 2(-7-4c)+5c=-8\)
\(\implies-14-8c+5c=-8\)
\(\implies c=-2\)
so \(a=-7-4c\implies a=-7-4(-2)=-7+8=1\)
Now \(b+4d=2\)
\(=\implies b=2-4d\)
Now \(2b+5d =4\implies 4-8d+5d=4\)
\(\implies-3d=0\)
\(\implies d=0\)
so \(b=2-4(0)=2\)
Thus, \(a = 1, b = 2, c = −2, d = 0\)
Hence, the required matrix X is \(\begin{bmatrix}1&-2\\2&0\end{bmatrix}\)