দেখাও যে, [গণিত] .
উত্তর: বামপক্ষ = [গণিত] ডানপক্ষ [গণিত] . (দেখানো হলো)
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
M=\left(\[\begin{array}{lll}q+r & 1 & q^2+r^2 \r+p & 1 & r^2+p^2 \p+q & 1 & p^2+q^2\end{array}\]\right).
Rajshahi · 2025
উত্তর
বামপক্ষ = =\left|\[\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\]\right|
=\left|\[\begin{array}{ccc}0 & a-b & a^2-b^2 \\ 0 & b-c & b^2-c^2 \\ 1 & c & c^2\end{array}\]\right|
\left[\[\begin{array}{l}r_1{ }^{\prime}=r_1-r_2 \r_2{ }^{\prime}=r_2-r_3\end{array}\]\right]
=\left|\[\begin{array}{cc}a-b & a^2-b^2 \b-c & b^2-c^2\end{array}\]\right|
=(a-b)(b-c)\left|\[\begin{array}{cc}1 & a+b \\ 1 & b+c\end{array}\]\right|
ডানপক্ষ
\therefore\left|\[\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\]\right|
. (দেখানো হলো)
উত্তর
দেওয়া আছে, \mathbf{M}=\left(\[\begin{array}{ccc}\mathrm{q}+\mathrm{r} & 1 & \mathrm{q}^2+\mathrm{r}^2 \\ \mathrm{r}+\mathrm{p} & 1 & \mathrm{r}^2+\mathrm{p}^2 \\ \mathrm{p}+\mathrm{q} & 1 & \dot{\mathrm{p}}^2+\mathrm{q}^2\end{array}\]\right)
\therefore |M| =\left|\[\begin{array}{ccc}q+r & 1 & q^2+r^2 \r+p & 1 & r^2+p^2 \p+q & 1 & p^2+q^2\end{array}\]\right|
=\left|\[\begin{array}{ccc}q+r & 1 & q^2+r^2 \\ (r+p)-(q+r) & 1-1 & \left(r^2+p^2\right)-\left(q^2+r^2\right) \\ (p+q)-(q+r) & 1-1 & \left(p^2+q^2\right)-\left(q^2+r^2\right)\end{array}\]\right|
\left[\[\begin{array}{l}r_2^{\prime}=r_2-r_1 \r_3^{\prime}=r_3-r_1\end{array}\]\right]
=(-1)^{1+2} \cdot 1\left|\[\begin{array}{ccc}q+r & 1 & q^2+r^2 \r+p-q-r & 0 & r^2+p^2-q^2-r^2 \p+q-q-r & 0 & p^2+q^2-q^2-r^2\end{array}\]\right|
=-\left|\[\begin{array}{cc}(p-q) & (p-q)(p+q) \\ (p-r) & (p-r)(p+r)\end{array}\]\right|
=-(p-q)(p-r)\left|\[\begin{array}{ll}1 & p+q \\ 1 & p+r\end{array}\]\right|
M ম্যাট্রিক্সটি ব্যতিক্রমী হলে, হবে।
অর্থাৎ,
বা,
বা,
অথবা,
বা,
অথবা,
বা,
. (দেখানো হলো)
উত্তর
দেওয়া আছে, M=\left(\[\begin{array}{lll}q+r & 1 & q^2+r^2 \r+p & 1 & r^2+p^2 \p+q & 1 & p^2+q^2\end{array}\]\right)
এবং হলে,
M=\left(\[\begin{array}{rrr}1-2 & 1 & 1+4 \\ -2+0 & 1 & 4+0 \\ 0+1 & 1 & 0+1\end{array}\]\right)
\therefore M=\left(\[\begin{array}{rrr}-1 & 1 & 5 \\ -2 & 1 & 4 \\ 1 & 1 & 1\end{array}\]\right)
এখন,
=\left(\[\begin{array}{rrr}-1 & 1 & 5 \\ -2 & 1 & 4 \\ 1 & 1 & 1\end{array}\]\right)\left(\[\begin{array}{rrr}-1 & 1 & 5 \\ -2 & 1 & 4 \\ 1 & 1 & 1\end{array}\]\right)
=\left(\[\begin{array}{rrrr}1 & -2+5 & -1+1+5 & -5+4+5 \\ 2 & -2+4 & -2+1+4 & -10+4+4 \\ -1 & -2+1 & 1+1+1 & 5+4+1\end{array}\]\right)
=\left(\[\begin{array}{rrr}4 & 5 & 4 \\ 4 & 3 & -2 \\ -2 & 3 & 10\end{array}\]\right)
প্রদত্ত রাশি
=2\left(\[\begin{array}{rrr}4 & 5 & 4 \\ 4 & 3 & -2 \\ -2 & 3 & 10\end{array}\]\right)-\left(\[\begin{array}{rrr}-1 & 1 & 5 \\ -2 & 1 & 4 \\ 1 & 1 & 1\end{array}\]\right)+3\left(\[\begin{array}{lll}1 & 0 & 9 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right)
=\left(\[\begin{array}{rrr}8 & 10 & 8 \\ 8 & 6 & -4 \\ -4 & 6 & 20\end{array}\]\right)-\left(\[\begin{array}{rrr}-1 & 1 & 5 \\ -2 & 1 & 4 \\ 1 & 1 & 1\end{array}\]\right)+\left(\[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\]\right)
=\left(\[\begin{array}{rrr}8+1+3 & 10-1+0 & 8-5+0 \\ 8+2+0 & 6-1+3 & -4-4+0 \\ -4-1+0 & 6-1+0 & 20-1+3\end{array}\]\right)
=\left(\[\begin{array}{rrr}12 & 9 & 3 \\ 10 & 8 & -8 \\ -5 & 5 & 22\end{array}\]\right)
\therefore \quad 2 M^2-M+3 I=\left(\[\begin{array}{rrr}12 & 9 & 3 \\ 10 & 8 & -8 \\ -5 & 5 & 22\end{array}\]\right).