দেখাও যে, [গণিত] .
উত্তর: দেওয়া আছে, [গণিত] এবং [গণিত] . (দেখানো হলো) [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
দেখাও যে, .
উদ্দীপক
A=\left[\[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 0 \\ 3 & 2 & 7\end{array}\]\right] ; f(x)=x^2-3 x+2 I
এবং
B=\left[\[\begin{array}{ccc}(y+z)^2 & x^2 & 1 \\ (z+x)^2 & y^2 & 1 \\ (x+y)^2 & z^2 & 1\end{array}\]\right]
রংপুর ক্যান্টনমেন্ট পাবলিক স্কুল অ্যান্ড কলেজ · 2025
উত্তর
ধরি, P=\left[\[\begin{array}{ll}2 & -1 \\ 3 & -2\end{array}\]\right]
P ম্যাট্রিক্সটি অভেদঘাতি হবে যদি হয়।
এখন,
\begin{aligned} P^2 & =P \cdot P=\left[\begin{array}{ll} 2 & -1 \\ 3 & -2 \end{aligned}\right]\left[\[\begin{array}{ll} 2 & -1 \\ 3 & -2 \end{array}\]\right] \\ & =\left[\[\begin{array}{ll} 4-3 & -2+2 \\ 6-6 & -3+4 \end{array}\]\right]=\left[\[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\]\right]=I \end{aligned}
\end{array}
যেহেতু
সুতরাং, P=\left[\[\begin{array}{ll}2 & -1 \\ 3 & -2\end{array}\]\right] একটি অভেদঘাতি ম্যাট্রিক্স। (দেখানো হলো)\end{array}
উত্তর
দেওয়া আছে, A=\left[\[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 0 \\ 3 & 2 & 7\end{array}\]\right]
এখन,
=\left[\[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 0 \\ 3 & 2 & 7\end{array}\]\right] \cdot\left[\[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 0 \\ 3 & 2 & 7\end{array}\]\right]
=\left[\[\begin{array}{lll}1+8+9 & 2+10+6 & 3+0+21 \\ 4+20+0 & 8+25+0 & 12+0+0 \\ 3+8+21 & 6+10+14 & 9+0+49\end{array}\]\right]
=\left[\[\begin{array}{lll}18 & 18 & 24 \\ 24 & 33 & 12 \\ 32 & 30 & 58\end{array}\]\right]
আবার, I=\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right]
\therefore \quad 2 I=2\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right]=\left[\[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\]\right]
(i) নং হতে,
=\left[\[\begin{array}{lll}18 & 18 & 24 \\ 24 & 33 & 12 \\ 32 & 30 & 58\end{array}\]\right]-3\left[\[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 0 \\ 3 & 2 & 7\end{array}\]\right]+\left[\[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\]\right]
=\left[\[\begin{array}{ccc}8 & 18 & 24 \\ 24 & 33 & 12 \\ 32 & 30 & 58\end{array}\]\right]-\left[\[\begin{array}{ccc}3 & 6 & 9 \\ 12 & 15 & 0 \\ 9 & 6 & 21\end{array}\]\right]+\left[\[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\]\right]
=\left[\[\begin{array}{ccc}18-3+2 & 18-6+0 & 24-9+0 \\ 24-12+0 & 33-15+2 & 12-0+0 \\ 32-9+0 & 30-6+0 & 58-21+2\end{array}\]\right]
=\left[\[\begin{array}{lll}17 & 12 & 15 \\ 12 & 20 & 12 \\ 23 & 24 & 39\end{array}\]\right]
নির্ণেয়, f(A)=\left[\[\begin{array}{lll}17 & 12 & 15 \\ 12 & 20 & 12 \\ 23 & 24 & 39\end{array}\]\right]
উত্তর
দেওয়া আছে, B=\left[\[\begin{array}{lll}(y+z)^2 & x^2 & 1 \\ (z+x)^2 & y^2 & 1 \\ (x+y)^2 & z^2 & 1\end{array}\]\right]
\therefore|B|=\left|\[\begin{array}{lll} (y+z)^2 & x^2 & 1 \\ (z+x)^2 & y^2 & 1 \\ (x+y)^2 & z^2 & 1 \end{array}\]\right|
\begin{aligned} & =\left|\begin{array}{lll}(y+z)^2-x^2 & x^2 & 1 \\ (z+x)^2-y^2 & y^2 & 1 \\ (x+y)^2-z^2 & z^2 & 1\end{aligned}\right|\left[\because c_1^{\prime}=c_1-c_2\right] \\ & =\left|\[\begin{array}{lll}(x+y+z)(y+z-x) & x^2 & 1 \\ (x+y+z)(z+x-y) & y^2 & 1 \\ (x+y+z)(x+y-z) & z^2 & 1\end{array}\]\right| \\ & =(x+y+z)\left|\[\begin{array}{lll}y+z-x & x^2 & 1 \z+x-y & y^2 & 1 \x+y-z & z^2 & 1\end{array}\]\right| \\ & =(x+y+z)\left|\[\begin{array}{lll}-2(x-y) & x^2-y^2 & 0 \\ -2(y-z) & y^2-z^2 & 0 \x+y-z & z^2 & 1\end{array}\]\right|\end{aligned}
\end{array}
এবং
\begin{aligned} & =(x+y+z) \cdot\left|\begin{array}{ll}-2(x-y) & x^2-y^2 \\ -2(y-z) & y^2-z^2\end{aligned}\right| \\ & =(x+y+z)\left|\[\begin{array}{ll}-2(x-y) & (x+y)(x-y) \\ -2(y-z) & (y+z)(y-z)\end{array}\]\right| \\ & =(x+y+z).(-2)(x-y)(y-z)\left|\[\begin{array}{ll}1 & x+y \\ 1 & y+z\end{array}\]\right| \\ & =-2.(x+y+z)(x-y)(y-z)(y+z-x-y)\end{aligned}
\end{array}
. (দেখানো হলো)\end{array}