အကယ္၍ $ \displaystyle A=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)$ ၊ $ \displaystyle I=\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)$ နဲ႔ $ \displaystyle xI-A$ ဟာ singular matrix ျဖစ္မယ္ဆိုရင္
$ \displaystyle \begin{array}{l}xI-A=x\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {x-a} & b \\ c & {x-d} \end{array}} \right)\end{array}$
ဒါဆိုရင္ $ \displaystyle xI-A$ ရဲ့ determinant ကို ရွာၾကည့္မယ္။
$ \displaystyle \begin{array}{l} \det (xI-A)=(x-a)(x-d)-bc\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}-(a+d)x+ad-bc\end{array}$
ဒီေနရာမွာ $ \displaystyle a+d$ ကို matrix $ \displaystyle A$ ရဲ့ trace လို႔ ေခၚၿပီး $ \displaystyle \operatorname{tr}(A)$ လို႔ သတ္မွတ္ပါမယ္။ $ \displaystyle ad-bc$ ကေတာ့ matrix $ \displaystyle A$ ရဲ့ determinant ျဖစ္ၿပီး $ \displaystyle \det (A)$ လို႔ သတ္မွတ္ပါမယ္။
$ \displaystyle xI-A$ ဟာ singular matrix ျဖစ္လို႔ $ \displaystyle \det (xI-A)=0$ ျဖစ္မွာေပါ့။ ဒါဆိုရင္
$ \displaystyle \begin{array}{l}\ \ \ \ \ \det (xI-A)=0\\\\\therefore {{x}^{2}}-(a+d)x+ad-bc=0\\\\\therefore {{x}^{2}}-\operatorname{tr}(A)x+\det (A)=0\end{array}$
အျမင္ရွင္းလင္း လြယ္ကူေအာင္ $ \displaystyle \operatorname{tr}(A)=a+d$ ကို $ p$ လို႔ထားၿပီး $ \displaystyle \det (A)=ad-bc$ ကိုေတာ့ $ q$ လို႔ထားလိုက္မယ္။
ဒါ့ေၾကာင့္ $ \displaystyle {{x}^{2}}-px+q=0$ ဆိုတဲ့ polynomial equation တစ္ခု ရပါတယ္။ ၎ equation ကို characteristic equation လို႔ ေခၚပါတယ္။
Chareateristic polynomial ကို $ \displaystyle f(x)$ လို႔ထားလိုက္မယ္ဆိုရင္ characteristic equation က $ \displaystyle f(x)=0$ ေပါ့။ $ \displaystyle x$ ေနရာမွာ matrix $ \displaystyle A$ ကို အစားသြင္းလိုက္ရင္
$ \displaystyle f(A)=0$
$ \displaystyle \therefore {{A}^{2}}-pA+qI=O$
real number မွာေတာ့ အေျမႇာက္ထပ္တူရကိန္းက 1 ျဖစ္ေသာ္လည္း matrix မွာေတာ့ အေျမႇာက္ ထပ္တူရ matrix က identity matrix $ (I)$ ျဖစ္ တယ္ဆိုတာ သတိျပဳဖို႔ လိုပါတယ္။
ဒါ့ေၾကာင့္ မည္သည့္ square matrix မဆို ၎ရဲ့ characteristic equation ကို ေျပလည္ေစပါတယ္။
| $ \displaystyle {{{A}^{2}}-\operatorname{tr}(A)A+\det (A)I=O}$ |
For 2× 2 matrix $ \displaystyle A=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)$,
| $\displaystyle {{A}^{2}}-(a+d)A+(ad-bc)I=O$ |
Proof : If $ \displaystyle A=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)$, then
$ \displaystyle \begin{array}{l}\ \ \ {{A}^{2}}-(a+d)A+(ad-bc)I\\\\=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)-(a+d)\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)+(ad-bc)\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} {{{a}^{2}}+bc} & {ab+bd} \\ {ac+cd} & {bc+{{d}^{2}}} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {{{a}^{2}}+ad} & {ab+bd} \\ {ac+cd} & {ad+{{d}^{2}}} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {ad-bc} & 0 \\ 0 & {ad-bc} \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\=O\end{array}$
Extension of Characteristic Equation
$ \displaystyle \ \ \ \ {{A}^{2}}-pA+qI=O$
ႏွစ္ဘက္လံုးကို $ \displaystyle {{A}^{{-1}}}$ နဲ႔ ေျမႇာက္လိုက္ရင္
$ \displaystyle \begin{array}{l}\ \ \ {{A}^{2}}{{A}^{{-1}}}-pA{{A}^{{-1}}}+qI{{A}^{{-1}}}=O\\\\\ \ \ AA{{A}^{{-1}}}-pA{{A}^{{-1}}}+qI{{A}^{{-1}}}=O\\\\\ \ \ AI-pI+q{{A}^{{-1}}}=O\\\\\therefore A-pI+q{{A}^{{-1}}}=O\\\\\therefore A-(a+d)I+(ad-bc){{A}^{{-1}}}=O\end{array}$
အကယ္၍ $ \displaystyle A=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)$ ျဖစ္ခဲ့ရင္ $ \displaystyle {\operatorname{tr}(A)=p=a+d}$ and $ \displaystyle \det (A)=q=ad-bc$ ဆိုတာ ေျပာခဲ့ၿပီးပါၿပီ။
$ \displaystyle \begin{array}{l}\ \ \ {{A}^{2}}=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -----(1)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {{{a}^{2}}+bc} & {ab+bd} \\ {ac+cd} & {bc+{{d}^{2}}} \end{array}} \right)\\\\\ \ {{A}^{{-1}}}=\frac{1}{{ad-bc}}\left( {\begin{array}{*{20}{c}} d & {-b} \\ {-c} & a \end{array}} \right)\\\\\ \ pq{{A}^{{-1}}}=(a+d)(ad-bc)\frac{1}{{ad-bc}}\left( {\begin{array}{*{20}{c}} d & {-b} \\ {-c} & a \end{array}} \right)\ \ \\\\\ \ \ \ \ \ \ \ \ \ \ =(a+d)\left( {\begin{array}{*{20}{c}} d & {-b} \\ {-c} & a \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {ad+{{d}^{2}}} & {-ab-bd} \\ {-ac-cd} & {{{a}^{2}}+ad} \end{array}} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -----(2)\\\\\ \ \ (q-{{p}^{2}})I=\left[ {(ad-bc)-{{{(a+d)}}^{2}}} \right]I\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left[ {-bc-ad-{{a}^{2}}-{{d}^{2}}} \right]\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-bc-ad-{{a}^{2}}-{{d}^{2}}} & 0 \\ 0 & {-bc-ad-{{a}^{2}}-{{d}^{2}}} \end{array}} \right)--(3)\end{array}$
ညီမွ်ျခင္း (1), (2), (3) ကို ေပါင္းလိုက္ရင္
$ \displaystyle \begin{array}{l}\ \ \ {{A}^{2}}+pq{{A}^{{-1}}}+(q-{{p}^{2}})I\\\\=\left( {\begin{array}{*{20}{c}} {{{a}^{2}}+bc} & {ab+bd} \\ {ac+cd} & {bc+{{d}^{2}}} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {ad+{{d}^{2}}} & {-ab-bd} \\ {-ac-cd} & {{{a}^{2}}+ad} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {-bc-ad-{{a}^{2}}-{{d}^{2}}} & 0 \\ 0 & {-bc-ad-{{a}^{2}}-{{d}^{2}}} \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\=O\end{array}$
$ \displaystyle \begin{array}{l}\therefore {{A}^{2}}+pq{{A}^{{-1}}}+(q-{{p}^{2}})I=O\\\\\therefore {{A}^{2}}+(a+d)(ad-bc){{A}^{{-1}}}+\left[ {(ad-bc)-{{{(a+d)}}^{2}}} \right]I=O\end{array}$
အားလံုးျပန္ခ်ဳပ္လိုက္ရင္ $ \displaystyle A=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)\ $ ျဖစ္ခဲ့ရင္
| $ \displaystyle \begin{array}{l}(1){{A}^{2}}-(a+d)A+(ad-bc)I=O\\\\(2){A}-(a+d)I+(ad-bc){{A}^{{-1}}}=O\\\\(3){{A}^{2}}+(a+d)(ad-bc){{A}^{{-1}}}+\left[ {(ad-bc)-{{{(a+d)}}^{2}}} \right]I=O\end{array}$ |
Example : If $ \displaystyle A=\left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)\ $ then
$ \displaystyle \begin{array}{l}(1)\ \ {{A}^{2}}-(-2+4)A+(-8+9)I=O\\\\\Rightarrow {{A}^{2}}-2A+I=O\\\\(2){A}-(-2+4)I+(-8+9){{A}^{{-1}}}=O\\\\\Rightarrow {A}+{{A}^{{-1}}}-2I=O\\\\(3){{A}^{2}}+(-2+4)(-8+9){{A}^{{-1}}}+\left[ {(-8+9)-{{{(-2+4)}}^{2}}} \right]I=O\\\\\Rightarrow {{A}^{2}}+2{{A}^{{-1}}}-3I=O\end{array}$.
ေမးခြန္းျပန္လုပ္ေသာ္
$ \displaystyle \begin{array}{l}\text{If}\ A=\left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)\ \ \text{show that}\\\begin{array}{*{20}{l}} {(1){{A}^{2}}-2A+I=O} \\ {(2)A+{{A}^{{-1}}}-2I=O} \\ {(3){{A}^{2}}+2{{A}^{{-1}}}-3I=O} \end{array}\\\text{where}\ I\ \text{is a unit matrix of order 2}\text{.}\end{array}$
ဆရာႀကီး Dr. Shwe Kyaw ပို႔ခ်ခဲ့ေသာ post ကို မွီျငမ္း၍ ျပန္လည္ တင္ျပပါသည္။
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