x = 1, y = 2 হলে f(p) নির্ণয় কর, যেখানে I একটি অভেদক ম্যাট্রিক্স।
উত্তর: দেওয়া আছে, [গণিত] এখন, [গণিত] হলে, [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
x = 1, y = 2 হলে f(p) নির্ণয় কর, যেখানে I একটি অভেদক ম্যাট্রিক্স।
উদ্দীপক
P=\left[\[\begin{array}{ccc}1+x^2-y^2 & 2 x y & 2 y \\ 2 x y & 1-x^2+y^2 & -2 x \\ -2 y & 2 x & 1-x^2-y^2\end{array}\]\right] এবং
Rajshahi · 2021
উত্তর
দেওয়া আছে, \mathrm{A}=\left[\[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\]\right], \mathrm{B}=\left[\[\begin{array}{rr}6 & 2 \\ 1 & 2 \\ 0 & -1\end{array}\]\right]
এখন, \mathrm{BA}=\left[\[\begin{array}{rr}0 & 2 \\ 1 & 2 \\ 0 & -1\end{array}\]\right]\left[\[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\]\right]
\begin{aligned} & =\left[\begin{array}{ccc}0 \cdot 1+2 \cdot 4 & 0 \cdot 2+2 \cdot 5 & 0 \cdot 3+2 \cdot 6 \\ 1 \cdot 1+2 \cdot 4 & 1 \cdot 2+2 \cdot 5 & 1 \cdot 3+2 \cdot 6 \\ 0 \cdot 1+(-1) \cdot 4 & 0 \cdot 2+(-1) \cdot 5 & 0 \cdot 3+(-1) \cdot 6\end{aligned}\right] \\ & =\left[\[\begin{array}{ccc}8 & 10 & 12 \\ 9 & 12 & 15 \\ -4 & -5 & -6\end{array}\]\right] \\ & \therefore \mathrm{BA}=\left[\[\begin{array}{ccc}8 & 10 & 12 \\ 9 & 12 & 15 \\ -4 & -5 & -6\end{array}\]\right]\end{aligned}
\end{array}\end{array}
উত্তর
\begin{aligned}& \operatorname{det}(P)=\left|\begin{array}{ccc}1+x^2-y^2 & 2 x y & 2 y \\2 x y & 1-x^2+y^2 & -2 x \\-2 y & 2 x & 1-x^2-y^2\end{aligned}\right| \\& =\left|\[\begin{array}{ccc}1+\mathrm{x}^2+\mathrm{y}^2 & 2 \mathrm{xy} & 2 \mathrm{y} \\0 & 1-\mathrm{x}^2+\mathrm{y}^2 & -2 \mathrm{x} \\-\mathrm{y}^{\prime}-\mathrm{x}^2 \mathrm{y}-\mathrm{y}^3 & 2 \mathrm{x} & 1-\mathrm{x}^2-\mathrm{y}^2\end{array}\]\right|\left[\mathrm{c}_1^{\prime}=\mathrm{c}_1+\mathrm{yc}_3 \text{ প্রয়োগ করে }\right] \\& =\left(1+x^2+y^2\right)\left|\[\begin{array}{ccc}1 & 2 x y & 2 y \\0 & 1-x^2+y^2 & -2 x \\-y & 2 x & 1-x^2-y^2\end{array}\]\right| \\& =\left(1+x^2+y^2\right)\left|\[\begin{array}{ccc}1 & 0 & 2 y \\0 & 1+x^2+y^2 & -2 x \\-y & x+x^3+x^2 & 1-x^2-y^2\end{array}\]\right|\left[c_2^{\prime}=c_2-xc_3\right. \text{ [প্রয়োগ করে ] } \\& =\left(1+x^2+y^2\right)^2\left|\[\begin{array}{rrc}1 & 0 & 2 y \\0 & 1 & -2 x \\-y & x & 1-x^2-y^2\end{array}\]\right| \\&=\left(1+\mathrm{x}^2+\mathrm{y}^2\right)^2\left\{1\left(1-\mathrm{x}^2-\mathrm{y}^2+2 \mathrm{x}^2\right)-0+2 \mathrm{y}(0+\mathrm{y})\right\} \\& =\left(1+\mathrm{x}^2+\mathrm{y}^2\right)^2\left(1+\mathrm{x}^2-\mathrm{y}^2+2 \mathrm{y}^2\right) \\& =\left(1+\mathrm{x}^2+\mathrm{y}^2\right)^2\left(1+\mathrm{x}^2+\mathrm{y}^2\right) \\& =\left(1+\mathrm{x}^2+\mathrm{y}^2\right)^3 \end{aligned}
\end{array}
প্রশ্নমতে det হলে,
বা,
(প্রমাণিত)\end{array}
উত্তর
দেওয়া আছে, P=\left[\[\begin{array}{ccc}1+\mathrm{x}^2-\mathrm{y}^2 & 2 \mathrm{xy} & 2 \mathrm{y} \\2 \mathrm{xy} & 1-\mathrm{x}^2+\mathrm{y}^2 & -2 \mathrm{x} \\-2 \mathrm{y} & 2 \mathrm{x} & 1-\mathrm{x}^2-\mathrm{y}^2\end{array}\]\right]
এখন, হলে,
\begin{aligned} & P=\left[\begin{array}{ccc}1+1^2-2^2 & 2 \cdot 1 \cdot 2 & 2 \cdot 2 \\ 2 \cdot 1 \cdot 2 & 1-1^2+2^2 & -2 \cdot 1 \\ -2 \cdot 2 & 2 \cdot 1 & 1-1^2-2^2\end{aligned}\right] \\ & =\left[\[\begin{array}{ccc}-2 & 4 & 4 \\ 4 & 4 & -2 \\ -4 & 2 & -4\end{array}\]\right] \\ & \therefore P^2=P \cdot P=\left[\[\begin{array}{ccc}-2 & 4 & 4 \\ 4 & 4 & -2 \\ -4 & 2 & -4\end{array}\]\right]\left[\[\begin{array}{ccc}-2 & 4 & 4 \\ 4 & 4 & -2 \\ -4 & 2 & -4\end{array}\]\right] \\ & =\left[\[\begin{array}{lll}-2(-2)+4 \cdot 4+4(-4) & -2 \cdot 4+4 \cdot 4+4 \cdot 2 & (-2) \cdot 4+4(-2)+(4)(-4) \\ 4(-2)+4 \cdot 4+(-2)(-4) & 4 \cdot 4+4 \cdot 4+(-2) \cdot 2 & 4 \cdot 4+4(-2)+(-2) \cdot(-4) \\ -4 \cdot(-2)+2 \cdot 4+(-4)(-4) & -4 \cdot 4+2 \cdot 4+(-4) \cdot 2 & -4 \cdot 4+2 \cdot(-2)+(-4) \cdot(-4)\end{array}\]\right]\end{aligned}
\end{array}
\begin{aligned} & =\left[\begin{array}{rrr}4 & 16 & -32 \\ 16 & 28 & 16 \\ 32 & -16 & -4\end{aligned}\right] \\ & \mathrm{P}^3=\mathrm{P}^2 \cdot \mathrm{P} \text{. } \\ & =\left[\[\begin{array}{rrr}4 & 16 & -32 \\ 16 & 28 & 16 \\ 32 & -16 & -4\end{array}\]\right]\left[\[\begin{array}{ccc}-2 & 4 & 4 \\ 4 & 4 & -2 \\ -4 & 2 & -4\end{array}\]\right] \\ & =\left[\[\begin{array}{lll}4(-2)+16 \cdot 4+(-32) \cdot(-4) & 4 \cdot 4+16 \cdot 4+(-32) \cdot 2 & 4 \cdot 4+16(-2)+(-32)(-4) \\ 16 \cdot(-2)+28 \cdot 4+16(-4) & 16+4+28 \cdot 4+16 \cdot 2 & 16 \cdot 4+22(-2)+16 \cdot(-4) \\ 32(-2)+(-16) 4+(-4)(-4) & 32 \cdot 4+1-16) \cdot 4+(-4) \cdot 2 & 32 \cdot 4+(-16) \cdot(-2)+(-4)\end{array}\]\right] \\ & =\left[\[\begin{array}{rrr}184 & 16 & 112 \\ 16 & 208 & -56 \\ -112 & 56 & 176\end{array}\]\right] \\ & \text{ এখन, } f(x)=x^3-3 x+2 I \\ & \therefore \mathrm{f}(\mathrm{P})=\mathrm{P}^3-3 \mathrm{P}+2 \mathrm{I} \\ & =\left[\[\begin{array}{rrr}184 & 16 & 112 \\ 16 & 208 & -56 \\ -112 & 56 & 176\end{array}\]\right]-3\left[\[\begin{array}{ccc}-2 & 4 & 4 \\ 4 & 4 & -2 \\ -4 & 2 & -4\end{array}\]\right]+2\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right] \\ & =\left[\[\begin{array}{rrr}184 & 16 & 112 \\ 16 & 208 & -56 \\ -112 & 56 & 176\end{array}\]\right]-\left[\[\begin{array}{rrr}-6 & 12 & 12 \\ 12 & 12 & -6 \\ -12 & 6 & -12\end{array}\]\right]+\left[\[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\]\right] \\ & =\left[\[\begin{array}{rrr}184+6+2 & 16-12-0 & 112-12-0 \\ 16-12+0 & 208-12+2 & -56+6+0 \\ -112+12+0 & 56-6+0 & 176+12+2\end{array}\]\right] \\ & =\left[\[\begin{array}{rrr}192 & 4 & 100 \\ 4 & 198 & -50 \\ -100 & 50 & 190\end{array}\]\right] \\ & \therefore \mathrm{f}(\mathrm{P})=\left[\[\begin{array}{rrr}192 & 4 & 100 \\ 4 & 198 & -50 \\ -100 & 50 & 190\end{array}\]\right] \\ & \end{aligned}
\end{array}\end{array}