দেখাও যে, det (B) = (2 – b) (b - c) (c - 2) (x - y).
উত্তর: দেওয়া আছে, [গণিত] = (2-b) (b-c)(x-y) (b+c-2-b) [গণিত] (Showed)
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, det (B) = (2 – b) (b - c) (c - 2) (x - y).
উদ্দীপক
A=\left|\[\begin{array}{ccc}3+\mathrm{x} & 4 & 1 \\4 & 1+\mathrm{x} & 3 \\1 & 3 & 4+\mathrm{x}\end{array}\]\right|, B=\left|\[\begin{array}{ccc}2+\mathrm{x} & \mathrm{b}+\mathrm{x} & \mathrm{c}+\mathrm{x} \\2+\mathrm{y} & \mathrm{b}+\mathrm{y} & \mathrm{c}+\mathrm{y} \\4 & \mathrm{~b}^2 & \mathrm{c}^2\end{array}\]\right|
Rajshahi · 2022
উত্তর
B =\left[\[\begin{array}{ll}2 & 1 \\0 & 2\end{array}\]\right] \mathrm{B}^{\mathrm{t}}=\left[\[\begin{array}{ll}2 & 0 \\1 & 2\end{array}\]\right]
এখন,
\mathrm{B} \cdot \mathrm{B}^{\mathrm{t}} =\left[\[\begin{array}{ll}2 & 1 \\0 & 2\end{array}\]\right] \cdot\left[\[\begin{array}{ll}2 & 0 \\1 & 2\end{array}\]\right]
=\left[\[\begin{array}{ll}4+1 & 0+2 \\0+2 & 0+4\end{array}\]\right] \mid
=\left|\[\begin{array}{ll}5 & 2 \\2 & 4\end{array}\]\right| Ans
উত্তর
দেওয়া আছে, \quad B=\left|\[\begin{array}{ccc}2+\mathrm{x} & \mathrm{b}+\mathrm{x} & \mathrm{c}+\mathrm{x} \\ 2+\mathrm{y} & \mathrm{b}+\mathrm{y} & \mathrm{c}+\mathrm{y} \\ 4 & \mathrm{~b}^2 & \mathrm{c}^2\end{array}\]\right|
\operatorname{det}(B)=|B|=\left|\[\begin{array}{ccc}2+x & b+x & c+x \\2+y & b+y & c+y \\4 & b^2 & c^2\end{array}\]\right|
=\left|\[\begin{array}{ccc}2-b & b-c & c+c \\2-b & b-c & c+y \\4-b^2 & b^2-c^2 & c^2\end{array}\]\right|\left[\[\begin{array}{l}c_1^1=c_1-c_2 \c_2=c_2-c_3\end{array}\]\right]
=\left|\[\begin{array}{ccc}2-b & b-c & c+x \\2+b & b-c & c+y \\(2+b)(2-b) & (b+c)(b-c) & c^2\end{array}\]\right|
=(2-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\[\begin{array}{ccc}1 & 1 & c+x \\1 & 1 & c+y \\2+b & b+c & c^2\end{array}\]\right|
=(2-b)(b-c)\left|\[\begin{array}{ccc}0 & 0 & x-y \\1 & 1 & c+y \\2+b & b+c & c^2\end{array}\]\right| \quad\left[r_1^{\prime}=r_1-r_2\right]
=(2-b)(b-c)(x-y)\left|\[\begin{array}{ccc}0 & 0 & 1 \\1 & 1 & c+y \\2+b & b+c & c^2\end{array}\]\right|
=(2-b)(b-c)(x-y)\left|\[\begin{array}{cc} 1 & 1 \\2+b & b+c\end{array}\]\right|
= (2-b) (b-c)(x-y) (b+c-2-b)
(Showed)
উত্তর
(গ)
দেওয়া আছে,
A=\left|\[\begin{array}{ccc}3+x & 4 & 1 \\4 & 1+x & 3 \\1 & 3 & 4+x\end{array}\]\right|
\operatorname{det}(\mathrm{A})=|\mathrm{A}|=\left|\[\begin{array}{ccc}3+\mathrm{x} & 4 & 1 \\4 & 1+\mathrm{x} & 3 \\1 & 3 & 4+\mathrm{x}\end{array}\]\right|
=\left|\[\begin{array}{ccc}8+\mathrm{x} & 4 & 1 \\8+\mathrm{x} & 1+\mathrm{x} & 3 \\8+\mathrm{x} & 3 & 4+\mathrm{x} \end{array}\]\right|\left[\mathrm{C}_1^{\prime}=\mathrm{c}_1+\mathrm{c}_2+\mathrm{c}_3\right]
=(8+\mathrm{x})\left|\[\begin{array}{ccc}1 & 4 & 1 \\1 & 1+\mathrm{x} & 3 \\1 & 3 & 4+\mathrm{x}\end{array}\]\right|
=(8+x)\left|\[\begin{array}{ccc}0 & -3-\mathrm{x} & -2 \\0 & -2+\mathrm{x} & -1-\mathrm{x} \\1 & 3 & 4+\mathrm{x}\end{array}\]\right| \mathrm{r}_2^{\prime}=\mathrm{r}_1-\mathrm{r}_2-\mathrm{r}_3
=(8+x)\left|\[\begin{array}{cc}-3-x & -2 \\-2+x & -1-x\end{array}\]\right|
=(8+x)(-1)(-1)\left|\[\begin{array}{ll}3+x \\2-x & 1+x\end{array}\]\right|
কিন্তু,
det (A) = 0
8 + x অথবা, + 16x -1 = 0
এখানে, x = 0 এর অবাস্তব মূল রয়েছে x = –8
x = –8
এখন,
A =\left[\[\begin{array}{ccc}3-8 & 4 & 1 \\4 & 1-8 & 3 \\1 & 3 & 4-8\end{array}\]\right]
=\left[\[\begin{array}{ccc}-5 & 4 & 1 \\4 & -7 & 3 \\1 & 3 & -4\end{array}\]\right]
ট্রেস = কর্ন বরাবর ভূক্তি গুলোর যোগফল (Ans)
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