$\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -4M=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\qquad(*) \\\\M=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\to4M=\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]$

$(*)\qquad\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\\\\left[\begin{array}{ccc}9-4a&7-4b\\-4-4c&2-4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\.\qquad\qquad\qquad\qquad\ \ \ \Downarrow\\(_{11})\qquad9-4a=-3\ \ \ |-9\\-4a=-12\ \ \ \ |:(-4)\\a=3\\\\(_{12})\qquad7-4b=3\ \ \ \ |-7\\-4b=-4\ \ \ \ |:(-4)\\b=1$

$(_{21})\qquad-4-4c=4\ \ \ \ |+4\\-4c=8\ \ \ \ |:(-4)\\c=-2\\\\(_{22})\qquad2-4d=-18\ \ \ \ |-2\\-4d=-20\ \ \ \ \ |:(-4)\\d=5\\\\Answer:\ A.\ M=\left[\begin{array}{ccc}3&1\\-2&5\end{array}\right]$

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3 years ago