حل اسئلة مراجعة تراكمية

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50)

$4\left[\begin{array}{cc}\begin{array}{cc}8& -1\\ -3& -4\end{array}& \end{array}\right]-5\left[\begin{array}{cc}\begin{array}{cc}-2& 4\\ 6& 3\end{array}& \end{array}\right]$

$\begin{array}{c}\left[\begin{array}{cc}\begin{array}{cc}4\cdot 8& 4\cdot (-1)\\ 4\cdot (-3)& 4\cdot (-4)\end{array}& \end{array}\right]-\left[\begin{array}{cc}\begin{array}{cc}5\cdot (-2)& 5\cdot 4\\ 5\cdot 6& 5\cdot 6\end{array}& \end{array}\right]=\\ \left[\begin{array}{cc}\begin{array}{cc}32& -4\\ -12& -16\end{array}& \end{array}\right]-\left[\begin{array}{cc}\begin{array}{cc}-10& 20\\ 30& 15\end{array}& \end{array}\right]=\\ \left[\begin{array}{cc}\begin{array}{cc}32-(-10)& (-4)-20\\ (-12)-30& (-16)-15\end{array}& \end{array}\right]=\\ \left[\begin{array}{cc}\begin{array}{cc}42& -24\\ -42& -31\end{array}& \end{array}\right]=\end{array}$

51)

$\begin{array}{r}5\left(\begin{array}{cc}2\left[\begin{array}{cc}\begin{array}{cc}-2& -5\\ -1& 3\end{array}& \end{array}\right]-3& \left[\begin{array}{cc}\begin{array}{cc}-1& -2\\ 6& 4\end{array}& \end{array}\right]\end{array}\right)\\ 5\left(\begin{array}{cc}\left[\begin{array}{cc}\begin{array}{cc}-4& -10\\ -2& 6\end{array}& \end{array}\right]+& \left[\begin{array}{cc}\begin{array}{cc}3& 6\\ -18& -12\end{array}& \end{array}\right]\end{array}\right)=\\ \left[\begin{array}{cc}\begin{array}{cc}-5& -20\\ -100& -30\end{array}& \end{array}\right]=5\left[\begin{array}{cc}\begin{array}{cc}-1& -4\\ -20& -6\end{array}& \end{array}\right]=\end{array}$

52)

$\begin{array}{r}-4\left(\begin{array}{cc}\left[\begin{array}{cc}\begin{array}{cc}8& 9\\ -5& 5\end{array}& \end{array}\right]-2& \left[\begin{array}{cc}\begin{array}{cc}-6& -1\\ 6& 3\end{array}& \end{array}\right]\end{array}\right)\\ -4\left(\begin{array}{cc}\left[\begin{array}{cc}\begin{array}{cc}8& 9\\ -5& 5\end{array}& \end{array}\right]-& \left[\begin{array}{cc}\begin{array}{cc}-12& -2\\ 12& 6\end{array}& \end{array}\right]\end{array}\right)=\\ -4\left[\begin{array}{cc}\begin{array}{cc}20& 11\\ -17& -1\end{array}& \end{array}\right]=-4\left[\begin{array}{cc}\begin{array}{cc}8-(-12)& 9-(-2)\\ -5-12& 5-6\end{array}& \end{array}\right]=\end{array}$

$\begin{array}{r}\left[\begin{array}{cc}\begin{array}{cc}(-4)\cdot 20& (-4)\cdot 11\\ (-4)\cdot (-17)& (-4)\cdot (-1)\end{array}& \end{array}\right]=\\ \left[\begin{array}{cc}\begin{array}{cc}-80& -44\\ 68& 4\end{array}& \end{array}\right]=\end{array}$