f(A), নির্ণয় কর।
উত্তর: [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
f(A), নির্ণয় কর।
উদ্দীপক
A =\left|\[\begin{array}{lll}2 & 0 & 1 \\ 3 & 4 & 2 \\ 2 & 1 & 3\end{array}\]\right|, f(x)=x^2+3 x-5 I
Barisal · 2022
উত্তর
\begin{aligned} & \left|\begin{array}{lll}1 & a & a-2 \\ 2 & b & b-4 \\ 3 & c & c-6\end{aligned}\right| \\ = & \left|\[\begin{array}{lll}1 & a & a \\ 2 & b & b \\ 3 & c & c\end{array}\]\right|-\left|\[\begin{array}{lll}1 & a & 2 \\ 2 & b & 4 \\ 3 & c & 6\end{array}\]\right| \\ = & 0-\left|\[\begin{array}{lll}1 & a & 2 \\ 2 & b & 4 \\ 3 & c & 6\end{array}\]\right|\end{aligned}
\end{array} দুটো কলাম একই
= -(2)\left|\[\begin{array}{lll}1 & a & 1 \\2 & b & 2 \\3 & c & 3\end{array}\]\right|
\end{array}
উত্তর
\begin{aligned} & A^2=A \cdot A \\ & =\left|\begin{array}{lll}2 & 0 & 1 \\ 3 & 4 & 2 \\ 2 & 1 & 3\end{aligned}\right|\left|\[\begin{array}{lll}2 & 0 & 1 \\ 3 & 4 & 2 \\ 2 & 1 & 3\end{array}\]\right| \\ & =\left[\[\begin{array}{lll}4+0+2 & 0+0+1 & 2+0+3 \\ 6+12+4 & 0+16+2 & 3+8+6 \\ 4+3+6 & 0+4+3 & 2+2+9\end{array}\]\right] \\ & =\left|\[\begin{array}{ccc}6 & 1 & 5 \\ 22 & 18 & 17 \\ 13 & 7 & 13\end{array}\]\right| \\ & \therefore f(A)=A^2+3 A-5 I \\ & =\left|\[\begin{array}{ccc}6 & 1 & 5 \\ 22 & 18 & 17 \\ 13 & 7 & 13\end{array}\]\right|+3\left|\[\begin{array}{lll}2 & 0 & 1 \\ 3 & 4 & 2 \\ 2 & 1 & 3\end{array}\]\right|-5\left|\[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right| \\ & =\left[\[\begin{array}{ccc}6 & 1 & 5 \\ 22 & 18 & 17 \\ 13 & 7 & 13\end{array}\]\right]+\left[\[\begin{array}{ccc}6 & 0 & 3 \\ 9 & 12 & 6 \\ 6 & 3 & 9\end{array}\]\right]-\left[\[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\]\right] \\ & =\left|\[\begin{array}{lll}6+6-5 & 1+0-0 & 5+3-0 \\ 22+5-0 & 18+12-5 & 17+6-0 \\ 13+6-0 & 7+3-0 & 13+9-5\end{array}\]\right|=\left[\[\begin{array}{ccc}7 & 1 & 8 \\ 31 & 25 & 23 \\ 19 & 10 & 17\end{array}\]\right] \text{ Ans. } \\ & \end{aligned}
\end{array}\end{array}
উত্তর
দেওয়া আছে,
A=\[\begin{vmatrix}2 & 0 & 1 \\ 3 & 4 & 2 \\ 2 & 1 & 3\end{vmatrix}\]
ম্যাট্রিক্সের বীপরিত ম্যাট্রিক্স রয়েছে
এখন,
\therefore A^{-1}=\frac{1}{|A|} \text{ adj } A=\frac{1}{|A|}\left[\[\begin{array}{ccc}A_{11} & A_{12} & A_{13} \A_{21} & A_{22} & A_{23} \A_{31} & A_{32} & A_{33}\end{array}\]\right]^t
=\frac{1}{15}\left[\[\begin{array}{ccc}10 & 1 & -4 \\-5 & 4 & -1 \\-5 & -2 & 8\end{array}\]\right] Ans.
or =\left[\[\begin{array}{ccc}\frac{2}{3} & \frac{1}{15} & \frac{-4}{15} \\-1 / 3 & \frac{4}{15} & -1 / 15 \\-1 / 3 & \frac{-2}{15} & \frac{8}{15}\end{array}\]\right] Ans.