প্রমান কর যে , [গণিত]
উত্তর: [গণিত] A এর সহগুনক গুলো বের করে পাই , [গণিত] এখন, [গণিত] adj [গণিত] আবার , [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
প্রমান কর যে ,
উদ্দীপক
A=\left[\[\begin{array}{ccc}1 & 2 & 4 \\ 2 & 2 & 3 \\ 4 & 3 & -5\end{array}\]\right], f(x)=x^{2}-x+3.
Barisal · 2021
উত্তর
দেওয়া আছে,
P=\left[\[\begin{array}{rr}3 & 3 \\ -3 & -3\end{array}\]\right]
এখন,
\begin{aligned} \therefore P^2 & =P \cdot P=\left[\begin{array}{rr}3 & 3 \\ -3 & -3\end{aligned}\right] \cdot\left[\[\begin{array}{rr}3 & 3 \\ -3 & -3\end{array}\]\right] \\ & =\left[\[\begin{array}{rr}9-9 & 9-9 \\ -9+9-9+9\end{array}\]\right] \\ & =\left[\[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\]\right] \\ & =0\end{aligned}
\end{array}
সুতরাং, P হলো শূণ্যঘাতি ম্যাট্রিক্স ।\end{array}
উত্তর
এখানে,
বা,
\begin{aligned}& =\left[\begin{array}{ccc}1 & 2 & 4 \\2 & 2 & 3 \\4 & 3 & -5\end{aligned}\right]\left[\[\begin{array}{ccc}1 & 2 & 4 \\2 & 2 & 3 \\4 & 3 & -5\end{array}\]\right] \\& =\left[\[\begin{array}{lll}1+4+16 & 1+4+16 & 1+4+16 \\1+4+16 & 1+4+16 & 1+4+16 \\1+4+16 & 1+4+16 & 1+4+16\end{array}\]\right]\end{aligned}
\end{array}
\begin{aligned} & f(A)=A^2-A+3 I \\ & =\left[\begin{array}{rrr}21 & 18 & -10 \\ 18 & 17 & -1 \\ -10 & -1 & 50\end{aligned}\right]-\left[\[\begin{array}{rrr}1 & 2 & 4 \\ 2 & 2 & 3 \\ 4 & 3 & -5\end{array}\]\right]+3\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right] \\ & =\left[\[\begin{array}{rrr}21 & 18 & -10 \\ 18 & 17 & -1 \\ -10 & -1 & 50\end{array}\]\right]-\left[\[\begin{array}{rrr}1 & 2 & 4 \\ 2 & 2 & 3 \\ 4 & 3 & -5\end{array}\]\right]+\left[\[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\]\right] \\ & =\left[\[\begin{array}{rrr}23 & 16 & -14 \\ 16 & 18 & -4 \\ -14 & -4 & 58\end{array}\]\right] \text{ [Ans. }\end{aligned}
\end{array}\end{array}
উত্তর
A=\left|\[\begin{array}{ccc}1 & 2 & 4 \\ 2 & 2 & 3 \\ 4 & 3 & -5\end{array}\]\right|
\begin{aligned} & |A|=\left|\begin{array}{ccc}1 & 2 & 4 \\ 2 & 2 & 3 \\ 4 & 3 & -5\end{aligned}\right| \\ & =1(-10-9)-2(-10-12)+4(6-8) \\ & =-19+44-8=17 \neq 0\end{aligned}
\end{array}
A এর সহগুনক গুলো বের করে পাই ,
\begin{aligned} & \mathrm{A}_{11}=(-1)^{1+1}\left|\begin{array}{cc}2 & 3 \\ 3 & -5\end{aligned}\right|=-10-9=-19 \\ & \mathrm{~A}_{12}=(-1)^{1+2}\left|\[\begin{array}{cc}2 & 3 \\ 4 & -5\end{array}\]\right|=-(-10-12)=22 \\ & \mathrm{~A}_{13}=(-1)^{1+3}\left|\[\begin{array}{ll}2 & 2 \\ 4 & 3\end{array}\]\right|=6-8=-2 \\ & \mathrm{~A}_{21}=(-1)^{2+1}\left|\[\begin{array}{cc}2 & 4 \\ 3 & -5\end{array}\]\right|=-(-10-12)=22 \\ & \mathrm{~A}_{22}=(-1)^{2+2}\left|\[\begin{array}{cc}1 & 4 \\ 4 & -5\end{array}\]\right|=-5-16=-21 \\ & \mathrm{~A}_{23}=(-1)^{2+3}\left|\[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\]\right|=-(3-8)=5 \\ & \mathrm{~A}_{31}=(-1)^{3+1}\left|\[\begin{array}{ll}2 & 4 \\ 2 & 3\end{array}\]\right|=6-8=-2 \\ & \mathrm{~A}_{32}=(-1)^{3+2}\left|\[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\]\right|=-(3-8)=5 \\ & \mathrm{~A}_{33}=(-1)^{3+3}\left|\[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\]\right|=2-4=-2\end{aligned}
\end{array}
\begin{aligned} & \text{ অতএব, adj } \mathrm{A}=\left[\begin{array}{ccc}-19 & 22 & -2 \\ 22 & -21 & 5 \\ -2 & 5 & -2\end{aligned}\right]^{\mathrm{T}}= {\left[\[\begin{array}{ccc}-19 & 22 & -2 \\ 22 & -21 & 5 \\ -2 & 5 & -2\end{array}\]\right]}\end{aligned}
\end{array}
এখন, adj
=\frac{1}{17}\left[\[\begin{array}{ccc}-19 & 22 & -2 \\ 22 & -21 & 5 \\ -2 & 5 & -2\end{array}\]\right]
=\[\begin{bmatrix}-\frac{19}{17} & \frac{22}{17} & -\frac{2}{17}\\ \frac{22}{17} & -\frac{21}{17} & \frac{5}{17}\\-\frac{2}{17} & \frac{5}{17} & -\frac{2}{17}\end{bmatrix}\]
আবার , A^{-1},A=\[\begin{bmatrix}-\frac{19}{17} & \frac{22}{17} & -\frac{2}{17}\\ \frac{22}{17} & -\frac{21}{17} & \frac{5}{17}\\-\frac{2}{17} & \frac{5}{17} & -\frac{2}{17}\end{bmatrix}\]
\left\lceil\[\begin{array}{ccc}1 & 2 & 4 \\2 & 2 & 3 \\4 & 3 & -5\end{array}\]\right\rceil
=\left[\[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\]\right]=I_3
\end{array}