ماتریس

$\begin{array}{l}{\left[\begin{array}{ccccc}-1& 2& 0& 3& -5\end{array}\right]}_{1\times 5}\times {\left[\begin{array}{c}-2\\ 3\\ 7\\ -1\\ -2\end{array}\right]}_{5\times 1}\end{array}$

$\begin{array}{l}=\left[\left(-1\right)\left(-2\right)+\left(2\right)\left(3\right)+\left(0\right)\left(7\right)+\left(3\right)\left(-1\right)+\left(-5\right)\left(-2\right)\right]\end{array}$

$\begin{array}{l}=\left[15\right]\\ \\ =15\end{array}$

$\begin{array}{l}A=\left[\begin{array}{cc}2& 1\\ 5& 9\end{array}\right];A_{2\times 2}\\ \\ D=\left[\begin{array}{ccc}3& 7& 4\\ 2& 1& 6\end{array}\right];D_{2\times 3}\end{array}$

$\begin{array}{c}{A}_{2\times 2}\times D_{2\times 3}\end{array}$

$\begin{array}{c}=\left[\begin{array}{cc}2& 1\\ 5& 9\end{array}\right]\times \end{array}\left[\begin{array}{ccc}3& 7& 4\\ 2& 1& 6\end{array}\right]$

$={\left[\begin{array}{ccc}\left[\begin{array}{cc}2& 1\end{array}\right]\left[\begin{array}{c}3\\ 2\end{array}\right]& \left[\begin{array}{cc}2& 1\end{array}\right]\left[\begin{array}{c}7\\ 1\end{array}\right]& \left[\begin{array}{cc}2& 1\end{array}\right]\left[\begin{array}{c}4\\ 6\end{array}\right]\\ \left[\begin{array}{cc}5& 9\end{array}\right]\left[\begin{array}{c}3\\ 2\end{array}\right]& \left[\begin{array}{cc}5& 9\end{array}\right]\left[\begin{array}{c}7\\ 1\end{array}\right]& \left[\begin{array}{cc}5& 9\end{array}\right]\left[\begin{array}{c}4\\ 6\end{array}\right]\end{array}\right]}_{2\times 3}$

$\begin{array}{c}=\end{array}\left[\begin{array}{ccc}\left(2\right)\left(3\right)+\left(1\right)\left(2\right)& \left(2\right)\left(7\right)+\left(1\right)\left(1\right)& \left(2\right)\left(4\right)+\left(1\right)\left(6\right)\\ \left(5\right)\left(3\right)+\left(9\right)\left(2\right)& \left(5\right)\left(7\right)+\left(9\right)\left(1\right)& \left(5\right)\left(4\right)+\left(9\right)\left(6\right)\end{array}\right]$

$=\left[\begin{array}{ccc}8& 15& 14\\ 33& 44& 74\end{array}\right]$

حاصل ‌ضرب $D_{2\times 3}\times A_{2\times 2}$ انجام‌ پذیر نیست، زیرا تعداد ستون های $D$  با تعداد سطرهای $A$ برابر نیستند.

حاصل‌ ضرب دو ماتریس در صورتی امکان ‌پذیر است که تعداد ستون ‌های اولی با تعداد سطرهای دومی برابر باشند.

$\begin{array}{l}={\left[\begin{array}{cc}\left(1\right)\left(2\right)+\left(2\right)\left(-1\right)+\left(-1\right)\left(4\right)& \left(1\right)\left(3\right)+\left(2\right)\left(2\right)+\left(-1\right)\left(5\right)\\ \left(3\right)\left(2\right)+\left(2\right)\left(-1\right)+\left(1\right)\left(4\right)& \left(3\right)\left(3\right)+\left(2\right)\left(2\right)+\left(1\right)\left(5\right)\\ \left(-1\right)\left(2\right)+\left(-2\right)\left(-1\right)+\left(4\right)\left(4\right)& \left(-1\right)\left(3\right)+\left(-2\right)\left(2\right)+\left(4\right)\left(5\right)\end{array}\right]}_{3\times 2}\end{array}$

$={\left[\begin{array}{cc}-4& 2\\ 8& 18\\ 16& 13\end{array}\right]}_{3\times 2}$

$\begin{array}{l}={\left[\begin{array}{cc}3& -2\\ 1& 0\end{array}\right]}_{2\times 2}\times {\left[\begin{array}{c}3\\ 1\end{array}\right]}_{2\times 1}\end{array}$

$\begin{array}{l}={\left[\begin{array}{c}\left[\begin{array}{cc}3& -2\end{array}\right]\left[\begin{array}{c}3\\ 1\end{array}\right]\\ \left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{c}3\\ 1\end{array}\right]\end{array}\right]}_{2\times 1}\end{array}$

$\begin{array}{l}={\left[\begin{array}{c}\left(3\right)\left(3\right)+\left(-2\right)\left(1\right)\\ \left(1\right)\left(3\right)+\left(0\right)\left(1\right)\end{array}\right]}_{2\times 1}\\ \\ ={\left[\begin{array}{c}7\\ 3\end{array}\right]}_{2\times 1}\end{array}$

$\left[\begin{array}{cc}2& 1\\ -1& 2\\ 0& -1\end{array}\right]\begin{array}{l}{}_{3\times 2}\times {\left[\begin{array}{c}3\\ 1\end{array}\right]}_{2\times 1}\end{array}$

$\begin{array}{l}={\left[\begin{array}{c}\left[\begin{array}{cc}2& 1\end{array}\right]\left[\begin{array}{c}3\\ 1\end{array}\right]\\ \left[\begin{array}{cc}-1& 2\end{array}\right]\left[\begin{array}{c}3\\ 1\end{array}\right]\\ \left[\begin{array}{cc}0& -1\end{array}\right]\left[\begin{array}{c}3\\ 1\end{array}\right]\end{array}\right]}_{3\times 1}\end{array}$

$\begin{array}{l}={\left[\begin{array}{c}\left(2\right)\left(3\right)+\left(1\right)\left(1\right)\\ \left(-1\right)\left(3\right)+\left(2\right)\left(1\right)\\ \left(0\right)\left(3\right)+\left(-1\right)\left(1\right)\end{array}\right]}_{3\times 1}\\ \\ ={\left[\begin{array}{c}7\\ -1\\ -1\end{array}\right]}_{3\times 1}\end{array}$

${\left[\begin{array}{ccc}1& 0& -1\\ 2& 1& 0\\ 1& 1& 1\end{array}\right]}_{3\times 3}{\left[\begin{array}{cc}2& 1\\ 1& 1\\ 0& 0\end{array}\right]}_{3\times 1}$

$=\left[\begin{array}{cc}\left[\begin{array}{ccc}1& 0& -1\end{array}\right]\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]& \left[\begin{array}{ccc}1& 0& -1\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]\\ \left[\begin{array}{ccc}2& 1& 0\end{array}\right]\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]& \left[\begin{array}{ccc}2& 1& 0\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]\\ \left[\begin{array}{ccc}1& 1& 1\end{array}\right]\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]& \left[\begin{array}{ccc}1& 1& 1\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]\end{array}\right]$

$=\left[\begin{array}{cc}2& 1\\ 5& 3\\ 3& 2\end{array}\right]$

$=\left[\begin{array}{cc}\left[\begin{array}{cc}2& 1\end{array}\right]\left[\begin{array}{c}\frac{5}{2}\\ -4\end{array}\right]& \left[\begin{array}{cc}2& 1\end{array}\right]\left[\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right]\\ \left[\begin{array}{cc}8& 5\end{array}\right]\left[\begin{array}{c}\frac{5}{2}\\ -4\end{array}\right]& \left[\begin{array}{cc}8& 5\end{array}\right]\left[\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right]\end{array}\right]$

$\begin{array}{l}=\left[\begin{array}{cc}\left(2\right)\left(\frac{5}{2}\right)+\left(1\right)\left(-4\right)& \left(2\right)\left(-\frac{1}{2}\right)+\left(1\right)\left(1\right)\\ \left(8\right)\left(\frac{5}{2}\right)+\left(5\right)\left(-4\right)& \left(8\right)\left(-\frac{1}{2}\right)+\left(5\right)\left(1\right)\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

$=\left[\begin{array}{cc}\frac{5}{2}& -\frac{1}{2}\\ -4& 1\end{array}\right]\times \left[\begin{array}{cc}2& 1\\ 8& 5\end{array}\right]$

$\begin{array}{l}=\left[\begin{array}{cc}\left[\begin{array}{cc}\frac{5}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}2\\ 8\end{array}\right]& \left[\begin{array}{cc}\frac{5}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}1\\ 5\end{array}\right]\\ \left[\begin{array}{cc}-4& 1\end{array}\right]\left[\begin{array}{c}2\\ 8\end{array}\right]& \left[\begin{array}{cc}-4& 1\end{array}\right]\left[\begin{array}{c}1\\ 5\end{array}\right]\end{array}\right]\end{array}$

$\begin{array}{l}=\left[\begin{array}{cc}\left(\frac{5}{2}\right)\left(2\right)+\left(-\frac{1}{2}\right)\left(8\right)& \left(\frac{5}{2}\right)\left(1\right)+\left(-\frac{1}{2}\right)\left(5\right)\\ \left(-4\right)\left(2\right)+\left(1\right)\left(8\right)& \left(-4\right)\left(1\right)+\left(1\right)\left(5\right)\end{array}\right]\end{array}$

$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

${\left[\begin{array}{c}2\\ -5\\ 4\end{array}\right]}_{3\times 1}{\left[-1,2,3\right]}_{1\times 3}={\left[\begin{array}{ccc}-2& 4& 6\\ 5& -10& -15\\ -4& 8& 12\end{array}\right]}_{3\times 3}$

${\left[\begin{array}{c}0\\ 1\\ 0\\ 2\end{array}\right]}_{4\times 1}{\left[\begin{array}{cccc}5& 0& 1& 0\end{array}\right]}_{1\times 4}={\left[\begin{array}{cccc}0& 0& 0& 0\\ 5& 0& 1& 0\\ 0& 0& 0& 0\\ 10& 0& 2& 0\end{array}\right]}_{4\times 4}$

$A\times \left(B+C\right)$

$=\left[\begin{array}{cc}1& 2\\ -1& 0\\ 2& -1\end{array}\right]\times {\left(\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]+\left[\begin{array}{cc}2& 3\\ -1& 1\end{array}\right]\right)}_{3\times 2}$

$=\left[\begin{array}{cc}1& 2\\ -1& 0\\ 2& -1\end{array}\right]\times {\left[\begin{array}{cc}3& 2\\ 1& 4\end{array}\right]}_{2\times 2}$

$=\left[\begin{array}{cc}5& 10\\ -3& -2\\ 5& 0\end{array}\right]$

$\left(A\times B\right)+\left(A\times C\right)$

$=\left(\left[\begin{array}{cc}1& 2\\ -1& 0\\ 2& -1\end{array}\right]\times \left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]\right)+\left(\left[\begin{array}{cc}1& 2\\ -1& 0\\ 2& -1\end{array}\right]\times \left[\begin{array}{cc}2& 3\\ -1& 1\end{array}\right]\right)$

$=\left[\begin{array}{cc}5& 5\\ -1& 1\\ 0& -5\end{array}\right]+\left[\begin{array}{cc}0& 5\\ -2& -3\\ 5& 5\end{array}\right]$

$=\left[\begin{array}{cc}5& 10\\ -3& -2\\ 5& 0\end{array}\right]$

$A\times \left(B\times C\right)$

$=\left[\begin{array}{cc}1& 2\\ -1& 0\\ 2& -1\end{array}\right]\times \left(\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]\times \left[\begin{array}{cc}2& 3\\ -1& 1\end{array}\right]\right)$

$={\left[\begin{array}{cc}1& 2\\ -1& 0\\ 2& -1\end{array}\right]}_{3\times 2}\times {\left[\begin{array}{cc}3& 2\\ 1& 9\end{array}\right]}_{2\times 2}$

$=\left[\begin{array}{cc}5& 20\\ -3& -2\\ 5& -5\end{array}\right]$

$\left(A\times B\right)\times C$

$=\left(\left[\begin{array}{cc}1& 2\\ -1& 0\\ 2& -1\end{array}\right]\times \left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]\right)\times \left[\begin{array}{cc}2& 3\\ -1& 1\end{array}\right]$

$=\left[\begin{array}{cc}5& 5\\ -1& 1\\ 0& -5\end{array}\right]\times \left[\begin{array}{cc}2& 3\\ -1& 1\end{array}\right]$

$=\left[\begin{array}{cc}5& 20\\ -3& -2\\ 5& -5\end{array}\right]$

$\begin{array}{l}A\times B\\ \\ =\left[\begin{array}{cc}-1& 1\\ 0& 1\end{array}\right]\times \left[\begin{array}{cc}2& -1\\ 1& 0\end{array}\right]\end{array}$

$\begin{array}{l}=\left[\begin{array}{cc}\left[\begin{array}{cc}-1& 1\end{array}\right]\left[\begin{array}{c}2\\ 1\end{array}\right]& \left[\begin{array}{cc}-1& 1\end{array}\right]\left[\begin{array}{c}-1\\ 0\end{array}\right]\\ \left[\begin{array}{cc}0& 1\end{array}\right]\left[\begin{array}{c}2\\ 1\end{array}\right]& \left[\begin{array}{cc}0& 1\end{array}\right]\left[\begin{array}{c}-1\\ 0\end{array}\right]\end{array}\right]\end{array}$

$\begin{array}{l}=\left[\begin{array}{cc}\left(-1\right)\left(2\right)+\left(1\right)\left(1\right)& \left(-1\right)\left(-1\right)+\left(1\right)\left(0\right)\\ \left(0\right)\left(2\right)+\left(1\right)\left(1\right)& \left(0\right)\left(-1\right)+\left(1\right)\left(0\right)\end{array}\right]\\ \\ =\left[\begin{array}{cc}-1& 1\\ 1& 0\end{array}\right]\end{array}$

$\begin{array}{l}B\times A\\ \\ =\left[\begin{array}{cc}2& -1\\ 1& 0\end{array}\right]\times \left[\begin{array}{cc}-1& 1\\ 0& 1\end{array}\right]\end{array}$

$\begin{array}{l}=\left[\begin{array}{cc}\left[\begin{array}{cc}2& -1\end{array}\right]\left[\begin{array}{c}-1\\ 0\end{array}\right]& \left[\begin{array}{cc}2& -1\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]\\ \left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{c}-1\\ 0\end{array}\right]& \left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]\end{array}\right]\end{array}$

$\begin{array}{l}=\left[\begin{array}{cc}\left(2\right)\left(-1\right)+\left(-1\right)\left(0\right)& \left(2\right)\left(1\right)+\left(-1\right)\left(1\right)\\ \left(1\right)\left(-1\right)+\left(0\right)\left(0\right)& \left(1\right)\left(1\right)+\left(0\right)\left(1\right)\end{array}\right]\end{array}$

$\begin{array}{l}=\left[\begin{array}{cc}-2& 1\\ -1& 1\end{array}\right]\\ \\ \\ A\times B\ne B\times A\end{array}$

$\left\{\begin{array}{l}AB=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]\left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]=\left[\begin{array}{cc}8& 5\\ 16& 10\end{array}\right]\\ AC=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]\left[\begin{array}{cc}-2& 7\\ 5& -1\end{array}\right]=\left[\begin{array}{cc}8& 5\\ 16& 10\end{array}\right]\end{array}〉\right.\left\{\begin{array}{l}AB=AC\\ B\ne C\end{array}\right.$

$\begin{array}{l}A{A}^2\\ \\ =A\left(AA\right)\end{array}$

$\begin{array}{l}=\left(AA\right)A\\ \\ =A^{2}A\end{array}$

$A^2=\left[\begin{array}{ccc}0& 0& a\\ 0& b& 0\\ c& 0& 0\end{array}\right]\left[\begin{array}{ccc}0& 0& a\\ 0& b& 0\\ c& 0& 0\end{array}\right]=\left[\begin{array}{ccc}ac& 0& 0\\ 0& b^2& 0\\ 0& 0& ac\end{array}\right]$

$\begin{array}{l}{A}^3=A^{2}A=\left(3A\right)A=3A^2=3\left(3A\right)=3^{2}A\end{array}$

$\begin{array}{l}{A}^4=A^{3}A=\left(3^{2}A\right)A=3^2\left(A^2\right)=3^2\left(3A\right)=3^{3}A\end{array}$

$\begin{array}{l} ⋮\\ ⋮\\ A^{100}=3^{99}A\end{array}$

$A^3=A^{2}A$

سطر دوم ماتریس $A^2$ عبارت است از:

سطر دوم ماتریس $A$ در ماتریس $A$

$\left[\begin{array}{ccc}1& -2& 0\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 1& -2& 0\\ 2& 4& 1\end{array}\right]=\left[\begin{array}{ccc}-2& 4& 0\end{array}\right]$

درايه سطر دوم، ستون اول ماتريس $A^3$ عبارت است از:

 سطر دوم ماتریس $A^2$ را در ستون اول ماتریس $A$ 

$=\left[\begin{array}{ccc}-2& 4& 0\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]=4$

$ifAB+BA=\overline{O}\Rightarrow AB=-BA$

$\begin{array}{l}{A}^{2}B\\ \\ =\left(AA\right)B\\ \\ =A\left(AB\right);AB=-BA\end{array}$

$\begin{array}{l}=A\left(-BA\right)\\ \\ =-\left(AB\right)A\end{array}$

$\begin{array}{l}=-\left(-BA\right)A\\ \\ =B{A}^2\end{array}$

اثبات به روش استقراء:

$p\left(1\right):n=1\Rightarrow A{A}^1=A^{1}A$

برقرار است.

فرض استقراء:

$p\left(k\right):n=k\Rightarrow A{A}^k=A^{k}A$

حكم استقراء:

$p\left(k+1\right):n=k+1\Rightarrow A{A}^{k+1}=A^{k+1}A$

$\begin{array}{l}A{A}^{k+1}\\ \\ =A\left(A.A^k\right);A{A}^k=A^{k}A\end{array}$

$\begin{array}{l}=A\left(A^k.A\right)\\ \\ =\left(A{A}^k\right)A\\ \\ =A^{k+1}A\end{array}$

اثبات به روش استقراء:

$p\left(1\right):n=1\Rightarrow A^1=\left[\begin{array}{cc}{2}^2-1& 2^1-1\\ -2^2+2& -2^1+2\end{array}\right]=\left[\begin{array}{cc}3& 1\\ -2& 0\end{array}\right]$

برقرار است.

فرض استقراء:

$p\left(k\right):n=k\Rightarrow A^k=\left[\begin{array}{cc}{2}^{k+1}-1& 2^k-1\\ -2^{k+1}+2& -2^k+2\end{array}\right]$

حكم استقراء:

$p\left(k+1\right):n=k+1\Rightarrow A^{k+1}=\left[\begin{array}{cc}{2}^{k+2}-1& 2^{k+1}-1\\ -2^{k+2}+2& -2^{k+1}+2\end{array}\right]$

$\begin{array}{l}{A}^{k+1}\\ \\ =A{A}^k\end{array}$

$\begin{array}{l}=\left[\begin{array}{cc}3& 1\\ -2& 0\end{array}\right]\end{array}\left[\begin{array}{cc}{2}^{k+1}-1& 2^k-1\\ -2^{k+1}+2& -2^k+2\end{array}\right]$

$\begin{array}{l}=\end{array}\left[\begin{array}{cc}3\times 2^{k+1}-3-2^{k+1}+2& 3\times 2^k-3-2^k+2\\ -2\times 2^{k+1}+2& -2\times 2^k+2\end{array}\right]$

$\begin{array}{l}=\end{array}\left[\begin{array}{cc}2\times 2^{k+1}-1& 2\times 2^k-1\\ -2\times 2^{k+1}+2& -2\times 2^k+2\end{array}\right]$

$=\left[\begin{array}{cc}{2}^{k+2}-1& 2^{k+1}-1\\ -2^{k+2}+2& -2^{k+1}+2\end{array}\right]$

فرض كنيم $x=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$  باشد.

$\begin{array}{l}{x}^2=B\\ \\ \Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\end{array}$