Question:
Given \(3\begin{bmatrix}x&y\\z&w\end{bmatrix}\)\(=\begin{bmatrix}x&6\\-1&2w\end{bmatrix}+\begin{bmatrix}4& x+y\\ z+w& 3\end{bmatrix}\),find the values of x,y,z and w
Given \(3\begin{bmatrix}x&y\\z&w\end{bmatrix}\)\(=\begin{bmatrix}x&6\\-1&2w\end{bmatrix}+\begin{bmatrix}4& x+y\\ z+w& 3\end{bmatrix}\),find the values of x,y,z and w
Updated On: Sep 4, 2023
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Solution and Explanation
The correct answer is \(x=2,y=4,z=1\space and\space w=3\)
\(3\begin{bmatrix}x&y\\z&w\end{bmatrix}\)\(=\begin{bmatrix}x&6\\-1&2w\end{bmatrix}+\begin{bmatrix}4& x+y\\ z+w& 3\end{bmatrix}\)
\(\implies\begin{bmatrix}3x& 3y\\ 3z& 3w\end{bmatrix}=\begin{bmatrix}x+4& 6+x+y\\ -1+z+w& 2w+3\end{bmatrix}\)
Comparing the corresponding elements of these two matrices, we get:
\(3x=x+4\)
\(\implies2x=4\)
\(\implies x=2\)
\(3y=6+x+y\)
\(\implies2y=6+x\)
\(\implies y=4\)
\(3w=2w+3\)
\(\implies w=3\)
\(3z=-1+z+w\)
\(\implies2z=-1+w\)
\(=-1+3\)
\(\implies z=1\)
\(\therefore x=2,y=4,z=1\space and\space w=3\)
\(3\begin{bmatrix}x&y\\z&w\end{bmatrix}\)\(=\begin{bmatrix}x&6\\-1&2w\end{bmatrix}+\begin{bmatrix}4& x+y\\ z+w& 3\end{bmatrix}\)
\(\implies\begin{bmatrix}3x& 3y\\ 3z& 3w\end{bmatrix}=\begin{bmatrix}x+4& 6+x+y\\ -1+z+w& 2w+3\end{bmatrix}\)
Comparing the corresponding elements of these two matrices, we get:
\(3x=x+4\)
\(\implies2x=4\)
\(\implies x=2\)
\(3y=6+x+y\)
\(\implies2y=6+x\)
\(\implies y=4\)
\(3w=2w+3\)
\(\implies w=3\)
\(3z=-1+z+w\)
\(\implies2z=-1+w\)
\(=-1+3\)
\(\implies z=1\)
\(\therefore x=2,y=4,z=1\space and\space w=3\)
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