নির্ণয়কের সাহায্যে AX = D এর সমাধান কর।
উত্তর: দেওয়া আছে, [গণিত] Ax = D হলে, [গণিত] এখন, [গণিত]
HSC Higher Math 1st Paper Chapter 1: ম্যাট্রিকস ও নির্ণায়ক CQ — Prosthuti প্রশ্নব্যাংক
HSC · Higher Math 1st Paper · Chapter 1: ম্যাট্রিকস ও নির্ণায়ক · সৃজনশীল
নির্ণয়কের সাহায্যে AX = D এর সমাধান কর।
উদ্দীপক
\begin{aligned} & A=\left[\begin{array}{rrr}3 & 2 & 3 \\ 4 & 2 & 1 \\ 2 & 1 & -2\end{aligned}\right] \\ & \mathrm{B}=\left[\[\begin{array}{rrr}2 & 10 & 3 \\ 3 & 8 & 2 \\ 1 & 8 & 1\end{array}\]\right] \text{, } \\ & \mathrm{C}=\left[\[\begin{array}{ccr}\mathrm{a} & 12 & 6 \\ \mathrm{~b} & 10 & 3 \\ 3 & 9 & -1\end{array}\]\right], X=\left[\[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\]\right], \mathrm{D}=\left[\[\begin{array}{r}5 \\ 10 \\ 15\end{array}\]\right] \\ & \end{aligned}
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Comilla · 2022
উত্তর
দেওয়া আছে,
\mathrm{A}=\left|\[\begin{array}{ccc}3 & 2 & 3 \\4 & 2 & 1 \\2 & 1 & -2\end{array}\]\right| \quad \mathrm{B}=\left|\[\begin{array}{ccc}2 & 10 & 3 \\3 & 8 & 2 \\1 & 8 & 1\end{array}\]\right| \quad \mathrm{C}=\left|\[\begin{array}{ccc}a & 12 & 6 \b & 10 & 3 \\3 & 9 & -1\end{array}\]\right|
এখন,
A+B=\left[\[\begin{array}{lll}3 & 2 & 3 \\ 4 & 2 & 1 \\ 2 & 1 & -2\end{array}\]\right]+\left[\[\begin{array}{lll}2 & 10 & 3 \\ 3 & 8 & 2 \\ 1 & 8 & 1\end{array}\]\right]
=\left[\[\begin{array}{ccc}5 & 12 & 6 \\7 & 10 & 3 \\3 & 9 & -1\end{array}\]\right]
\Rightarrow\left[\[\begin{array}{ccc}5 & 12 & 6 \\-7 & 10 & 3 \\3 & 9 & -1\end{array}\]\right]=\left[\[\begin{array}{ccc}a & 12 & 6 \b & 10 & 3 \\3 & 9 & -1\end{array}\]\right]
শর্তমতে,
a = 5
b = 7
∴ a, b = 5, 7
উত্তর
দেওয়া আছে,
B=\left[\[\begin{array}{ccc}2 & 10 & 3 \\ 3 & 8 & 2 \\ 1 & 8 & 1\end{array}\]\right]
B এর সহগুনকসমুহঃ
B_{11}=\left|\[\begin{array}{ll}8 & 2 \\8 & 1\end{array}\]\right|
B_{12}=\left|\[\begin{array}{ll}3 & 2 \\1 & 1\end{array}\]\right|
B_{13}=\left|\[\begin{array}{ll}3 & 8 \\1 & 8\end{array}\]\right|
B_{21}=\left|\[\begin{array}{ll}10 & 3 \\8 & 1\end{array}\]\right|
B_{22}=\left|\[\begin{array}{ll}2 & 3 \\1 & 1\end{array}\]\right|
B_{23}=\left|\[\begin{array}{ll}2 & 10 \\1 & 8\end{array}\]\right|
B_{31}=\left|\[\begin{array}{cc}10 & 3 \\8 & 2 \end{array}\]\right|
= –4
\mathrm{B}_{32}=-\left|\[\begin{array}{ll}2 & 3 \\3 & 2\end{array}\]\right|
B_{33}=\left|\[\begin{array}{cc}2 & 10 \\3 & 8\end{array}\]\right|
\operatorname{Adj}(B) =\left|\[\begin{array}{ccc}-8 & -1 & 16 \\14 & -1 & -6 \\-4 & 5 & -14\end{array}\]\right|=\left|\[\begin{array}{ccc}-8 & 14 & -4 \\-1 & -1 & 5 \\16 & -6 & -14\end{array}\]\right|
=\frac{1}{22}\left|\[\begin{array}{ccc}-8 & 14 & -4 \\-1 & -1 & 5 \\16 & -6 & -14\end{array}\]\right|
উত্তর
দেওয়া আছে,
A=\left[\[\begin{array}{ccc}3 & 2 & 3 \\ 4 & 2 & 1 \\ 2 & 1 & -2\end{array}\]\right], \quad x=\left[\[\begin{array}{l}x \y \z\end{array}\]\right], D=\left[\[\begin{array}{c}5 \\ 10 \\ 15\end{array}\]\right]
\mathrm{AX}=\left[\[\begin{array}{ccr}3 & 2 & 3 \\4 & 2 & 1 \\2 & 1 & -2\end{array}\]\right]\left[\[\begin{array}{l}\mathrm{x} \mathrm{y} \mathrm{z}\end{array}\]\right]
=\left[\[\begin{array}{c}3 \mathrm{x}+2 \mathrm{y}+3 \mathrm{z} \\4 \mathrm{x}+2 \mathrm{y}+\mathrm{z} \\2 \mathrm{x}+\mathrm{y}-2 \mathrm{z}\end{array}\]\right]
Ax = D হলে,
{\left[\[\begin{array}{c}3 \mathrm{x}+2 \mathrm{y}+3 \mathrm{z} \\4 \mathrm{x}+2 \mathrm{y}+\mathrm{z} \\2 \mathrm{x}+\mathrm{y}-2 \mathrm{z}\end{array}\]\right]=\left[\[\begin{array}{c}5 \\10 \\15\end{array}\]\right]}
এখন,
\begin{aligned} & D=\left|\begin{array}{ccc}3 & 2 & 3 \\ 4 & 2 & 1 \\ 2 & 1 & -2\end{aligned}\right| \\ & =3(-4-1)-2(-8-2)+3(4-4) \\ & D x=\left[\[\begin{array}{ccc}5 & 2 & 3 \\ 10 & 2 & 1 \\ 15 & 1 & -2\end{array}\]\right] \\ & =5(-4-1)-2(-20-15)+3(10-30) \\ & =-25+70-60 \\ & =-15 \\ & D y=\left[\[\begin{array}{ccc}3 & 5 & 3 \\ 4 & 10 & 1 \\ 2 & 15 & -2\end{array}\]\right] \\ & =3(-20-15)-5(-8-2)+3(60-20) \\ & =-105+50+120 \\ & =65 \\ & \left.D z=\mid \[\begin{array}{lll}3 & 2 & 5 \\ 4 & 2 & 10 \\ 2 & 1 & 15\end{array}\]\right] \\ & =3(30-10)-2(60-20)+5(9-4) \\ & =60-80 \\ & =-20\end{aligned}
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