AX = B হলে নির্ণায়কের সাহায্যে X নির্ণয় কর।
উত্তর: দেওয়া আছে, [গণিত] এবং, [গণিত] এখন, [গণিত] হলে, [গণিত] এখানে, [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
AX = B হলে নির্ণায়কের সাহায্যে X নির্ণয় কর।
উদ্দীপক
X=\left[\[\begin{array}{l}x \y \z\end{array}\]\right], A=\left[\[\begin{array}{rrr}2 & -1 & -1 \\ 1 & 3 & 2 \\ 3 & -1 & -5\end{array}\]\right], B=\left[\[\begin{array}{l}6 \\ 1 \\ 1\end{array}\]\right] এবং C=\left[\[\begin{array}{ccc}p & q & r \p^{2} & q^{2} & r^{2} \p^{3}-1 & q^{3}-1 & r^{3}-1\end{array}\]\right]
Dinajpur · 2021
উত্তর
\left[\[\begin{array}{lll}y+z & x & 1 \z+x & y & 1 \x+y & z & 1\end{array}\]\right]
=\left[\[\begin{array}{ccc}x+y+z & x & 1 \x+y+z & y & 1 \x+y+z & z & 1\end{array}\]\right] ;\left[\mathrm{c}^{\prime}=\mathrm{c}_{1}+\mathrm{c}_{2}\right]
=(x+y+z)\left[\[\begin{array}{lll}1 & x & 1 \\ 1 & y & 1 \\ 1 & z & 1\end{array}\]\right]
[ দুইটি সারি/কলাম সমান হলে নির্ণায়কের মান 0]
নির্ণেয় মানঃ 0
উত্তর
দেওয়া আছে, \begin{aligned} & C=\left[\begin{array}{ccc}p & q & r \p^2 & q^2 & r^2 \p^3-1 & q^3-1 & r^2-1\end{aligned}\right] \\ & \therefore|C|=\left|\[\begin{array}{ccc}p & q & r \p^2 & q^2 & r^2 \p^3-1 & q^3-1 & r^2-1\end{array}\]\right| \\ & =\left|\[\begin{array}{ccc}p & q & r \p^2 & q^2 & r^2 \p^3 & q^3 & r^3\end{array}\]\right|-\left|\[\begin{array}{ccc}p & q & r \p^2 & q^2 & r^2 \\ 1 & 1 & 1\end{array}\]\right| \\ & =p q r\left|\[\begin{array}{ccc}1 & 1 & 1 \p & q & r \p^2 & q^2 & r^2\end{array}\]\right|-\left|\[\begin{array}{ccc}1 & 1 & 1 \p & q & r \p^2 & q^2 & r^2\end{array}\]\right| \\ & =(p q r-1)\left|\[\begin{array}{ccc}1 & 1 & 1 \p & q & r \p^2 & q^2 & r^2\end{array}\]\right| \\ & =(p q r-1)\left|\[\begin{array}{ccc}0 & 0 & 1 \p-q & q-r & r \p^2-q^2 & q^2-r^2 & r^2-q^2\end{array}\]\right| \\ & =(p q r-1)\left|\[\begin{array}{cc}p-q & q-r \p^2-q^2 & q^2-r^2\end{array}\]\right| \\ & =(p q r-1)(p-q)(q-r)\left|\[\begin{array}{cc}1 & 1 \p+q & q+r\end{array}\]\right| \\ & =(p q r-1)(p-q)(q-r)(q+r-p-q) \\ & =(p q r-1)(p-q)(q-r)(r-p) \\ & \therefore|C|=(p q r-1)(p-q)(q-r)(r-p) \\ & \end{aligned}
\end{array}
(দেখানো হল)\end{array}
উত্তর
দেওয়া আছে,
A=\left[\[\begin{array}{ccc}2 & -1 & -1 \\ 1 & 3 & 2 \\ 3 & -1 & -5\end{array}\]\right], B=\left[\[\begin{array}{l}6 \\ 1 \\ 1\end{array}\]\right] এবং, X=\left[\[\begin{array}{l}x \y \z\end{array}\]\right]
এখন, হলে,
\begin{aligned} & {\left[\begin{array}{ccc}2 & -1 & -1 \\ 1 & 3 & 2 \\ 3 & -1 & -5\end{aligned}\right]\left[\[\begin{array}{l}x \y \z\end{array}\]\right]=\left[\[\begin{array}{l}6 \\ 1 \\ 1\end{array}\]\right] } \\ \Rightarrow & {\left[\[\begin{array}{c}2 x-y-z \x+3 y+2 z \\ 3 x-y-5 z\end{array}\]\right]=\left[\[\begin{array}{l}6 \\ 1 \\ 1\end{array}\]\right] } \\ \therefore & 2 x-y-z=6 \\ \therefore & x+3 y+2 z=1 \\ \therefore & 3 x-y-5 z=1\end{aligned}
\end{array}
এখানে,
\therefore D=\left|\[\begin{array}{ccc}2 & -1 & -1 \\1 & 3 & 2 \\3 & -1 & -5\end{array}\]\right|
\therefore D_{x}=\left|\[\begin{array}{ccc}6 & -1 & -1 \\1 & 3 & 2 \\1 & -1 & -5 \end{array}\]\right|
\therefore D_{y}=\left|\[\begin{array}{ccc}2 & 6 & -1 \\1 & 1 & 2 \\3 & 1 & -5 \end{array}\]\right|
\therefore D_{z}=\left|\[\begin{array}{ccc}2 & -1 & 6 \\1 & 3 & 1 \\3 & -1 & 1 \end{array}\]\right|
\therefore X=\left[\[\begin{array}{l}x \y \z\end{array}\]\right]=\left[\[\begin{array}{c} 3 \\-2 \\2\end{array}\]\right]\end{array}