AB = I3 থেকে ক্রেমারের সূত্রের সাহায্যে (x, y, z) নির্ণয় কর।
উত্তর: [গণিত] ম্যাট্রিক্সর সমতা অনুসারে, [গণিত] x, y, z এর সহগ নিয়ে গঠিত নির্ণায়ক [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
AB = I3 থেকে ক্রেমারের সূত্রের সাহায্যে (x, y, z) নির্ণয় কর।
উদ্দীপক
A=\left|\[\begin{array}{ccc}3 & -4 & 2 \\ -2 & 1 & 0 \\ -1 & -1 & 1\end{array}\]\right|, \quad B=\left|\[\begin{array}{ccc}x & 2 & -2 \y & 5 & -4 \z & 7 & -5\end{array}\]\right|
Sylhet · 2022
উত্তর
দেওয়া আছে, \quad \mathrm{A}=\left(\[\begin{array}{ll}3 & -1 \\ 2 & -4\end{array}\]\right)
\mathrm{A}^{\top} =\left(\[\begin{array}{cc}3 & -1 \\2 & -4\end{array}\]\right)^{\top}
=\left(\[\begin{array}{cc}3 & 2 \\-1 & -4\end{array}\]\right)
\mathrm{A}+\mathrm{A}^{\top} =\left(\[\begin{array}{ll}3 & -1 \\2 & -4\end{array}\]\right)+\left(\[\begin{array}{cc}3 & 2 \\-1 & -4\end{array}\]\right)
=\left(\[\begin{array}{cc}3+3 & 2-1 \\-1+2 & -4-4\end{array}\]\right)
=\left(\[\begin{array}{cc}6 & 1 \\1 & -8\end{array}\]\right)
উত্তর
দেওয়া আছে,
A=\left|\[\begin{array}{ccc}3 & -4 & 2 \\-2 & 1 & 0 \\-1 & -1 & 1\end{array}\] \mid\right|
|A|=\left|\[\begin{array}{ccc}-1 & -1 & 1 \\-2 & 1 & 0 \\-1 & -1 & 1\end{array}\]\right|
= 3(1+0) + 4(-2+0) + 2(2+1)
= 3 - 8 + 8
= 1
নির্ণয়যোগ্য
A_{11}=\left|\[\begin{array}{cc}1 & 0 \\-1 & 1\end{array}\]\right|
A_{12}=-\left|\[\begin{array}{ll}-2 & 0 \\-1 & 1\end{array}\]\right|
A_{13}=\left|\[\begin{array}{cc}-2 & 1 \\-1 & -1\end{array}\]\right|
\begin{aligned} & A_{21}=-\left|\begin{array}{ll}-4 & 2 \\ -1 & 1\end{aligned}\right| A_{21}=-\left|\[\begin{array}{ll}-4 & 2 \\ -1 & 1\end{array}\]\right| \\ & =-(-4+2) \\ & =2 \\ & A_{22}=\left|\[\begin{array}{cc}3 & 2 \\ -1 & 1\end{array}\]\right| \\ & =3+2 \\ & =5 \\ & A_{z 3}=-\left|\[\begin{array}{cc}3 & -4 \\ -1 & -1\end{array}\]\right| \\ & =-(-3-4) \\ & 7 \\ & A_{31}=\left|\[\begin{array}{cc}-4 & 2 \\ 1 & 0\end{array}\]\right| \\ & =0-2 \\ & =-2 \\ & A_{32}=\left|\[\begin{array}{cc}3 & 2 \\ -2 & 0\end{array}\]\right| \\ & =-(0+4) \\ & =-4 \\ & A_{33}=\left|\[\begin{array}{cc}3 & -4 \\ & 16 \\ -2 & 1\end{array}\]\right| \\ & =3-8 \\ & =-5 \\ & \therefore \text{ Adj } A=\left[\[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\]\right]^{\top} \\ & =\left[\[\begin{array}{lll}1 & 2 & -2 \\ 2 & 5 & -4 \\ 3 & 7 & -5\end{array}\]\right] \\ & \therefore \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \text{ Adj } \mathrm{A} \\ & =\frac{1}{1}\left[\[\begin{array}{lll}1 & 2 & -2 \\ 2 & 5 & -4 \\ 3 & 7 & -5\end{array}\]\right] \\ & =\left[\[\begin{array}{lll}1 & 2 & -2 \\ 2 & 5 & -4 \\ 3 & 7 & -5\end{array}\]\right] \\ & \end{aligned}
\end{array}\end{array}
উত্তর
\begin{aligned} & \therefore\left[\begin{array}{ccc}3 & -4 & 2 \\ -2 & 1 & 0 \\ -1 & -1 & 1\end{aligned}\right]\left[\[\begin{array}{lll}x & 2 & -2 \y & 5 & -4 \z & 7 & -5\end{array}\]\right]=\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right] \\ & =\left[\[\begin{array}{ccc}3 x-4 y+2 z & 2 \cdot 3+(-4) \cdot 5+2 \cdot 7 & 3(-2)+(-4) \cdot(-4)+2(-5) \\ -2 x+y+0 \cdot z & -2 \cdot 2+1 \cdot 5+0 \cdot 7 & -2(-2)+(-4) \cdot 1+0(-5) \\ -x-y+z & -1 \cdot 2+(-1) \cdot 5+1 \cdot 7 & -1(-2)+(-1) \cdot(-4)+(1)(-5)\end{array}\]\right] \\ & =\left[\[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right] \\ & =\left[\[\begin{array}{lll}3 x-4 y+z z & 0 & 0 \\ -2 x+y+0 . z & 1 & 0 \\ -x-y+z & 0 & 1\end{array}\]\right]=\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right]\end{aligned}
\end{array}
ম্যাট্রিক্সর সমতা অনুসারে,
x, y, z এর সহগ নিয়ে গঠিত নির্ণায়ক
\begin{aligned} & D=\left|\begin{array}{ccc}3 & -4 & 2 \\ -2 & 1 & 0 \\ -1 & -1 & 1\end{aligned}\right| \\ & =3(1+0)+4(-2+0)+2(2+1) \\ & =3-8+6 \\ & =1 \\ & D_x=\left|\[\begin{array}{ccc}1 & -4 & 2 \\ 0 & 1 & 0 \\ 0 & -1 & 1\end{array}\]\right|=1\left|\[\begin{array}{cc}1 & 0 \\ -1 & 1\end{array}\]\right| \\ & =1(1+0) \\ & =1 \\ & \text{ Dy }=\left|\[\begin{array}{ccc}3 & 1 & 2 \\ -2 & 0 & 0 \\ -1 & 0 & 1\end{array}\]\right| \\ & =3(0-0)-1(-2-0)+2(0-0) \\ & =0+2+0=2 \\ & \mathrm{D} z=\left|\[\begin{array}{ccc}3 & -4 & 1 \\ -2 & 1 & 0 \\ -1 & -1 & 0\end{array}\]\right| \\ & =3(0-0)+4(0-0)+1(2+1) \\ & =3 \\ & \therefore \mathrm{x}=\frac{\mathrm{Dx}}{\mathrm{x}}=\frac{1}{1}=1 \\ & \therefore y=\frac{D y}{y}=\frac{2}{7}=2 \\ & \therefore \mathrm{z}=\frac{\mathrm{Dz}}{\mathrm{z}}=\frac{3}{1}=3 \\ & \therefore(x, y, z)=(1,2,3) \\ & \end{aligned}
\end{array}\end{array}