Exercise 8.1
Question 1
आव्यूह का प्रयोग कर निम्नलिखित समीकरण निकाय को हल करे.
[Using matrices solve the following system of equations:]
(i)
2 x+5 y=1
3 x+2 y=7
Sol :
The system of equation is non-homogeneous
Let $A=\left[\begin{array}{ll}2 & 5 \\ 3 & 2\end{array}\right]$ ,$x=\left[\begin{array}{l}x \\ y\end{array}\right]$ , $B=\left[\begin{array}{l}1 \\ 7\end{array}\right]$
|A|=4-15
=-11≠0
समीकरण निकाय के अद्विती हल है
$\therefore \quad x=A^{-1} B$
$a d j A=\left[\begin{array}{rr}2 & -3 \\ -5 & 2\end{array}\right]^{\prime}=\left[\begin{array}{rr}2 & -5 \\ -3 & 2\end{array}\right]$
$A^{-1}=\frac{1}{|A|} \cdot(a d j A)=\frac{1}{-11}\left[\begin{array}{cc}2 & -5 \\ -3 & 2\end{array}\right]$
x=A-1B
$a d j A=\left[\begin{array}{cc}2 & -3 \\ -5 & 2\end{array}\right]^{\prime}$
$A^{-1}=\dfrac{1}{|A|} \cdot(a d j A)$
$=\frac{1}{-11}\left[\begin{array}{cc}2 & -5 \\ -3 & 2\end{array}\right]$
x=A-1B
$\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{-11}\left[\begin{array}{cc}2 & -5 \\ -3 & 2\end{array}\right]\left[\begin{array}{l}1 \\ 7\end{array}\right]$
$=\frac{-1}{11}\left[\begin{array}{cc}2 & -35 \\ -3 & +14\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=-\frac{1}{11}\left[\begin{array}{c}-33 \\ 11\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}3 \\ -1\end{array}\right]$
x=3 , y=-1
(ii)
5x+2y=3
3x+2y=5
Sol :
<to be added>
Let $A=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right]$ , $x=\left[\begin{array}{l}x \\ y\end{array}\right]$, $B=\left[\begin{array}{l}3 \\ 5\end{array}\right]$
|A|=10-6=4
समीकरण निकाय के अद्विती हल होगा
$a d j A=\left[\begin{array}{cc}2 & -3 \\ -2 & 5\end{array}\right]^{\prime}=\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$
$A^{-1}=\frac{1}{|A|} \cdot(a d j A)=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$
$\therefore x=A^{-1}B$
|A|=10-6=4
$a d j A=\left[\begin{array}{cc}2 & -3 \\ -2 & 5\end{array}\right]$
$=\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$
$A^{-1}=\dfrac{1}{|A |} \cdot(a d j A)$
$=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$
∴x=A-1B
$\left[\begin{array}{c}x \\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]\left[\begin{array}{l}3 \\ 5\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}6-10 \\ -9+25\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{l}-4 \\ 160\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-1 \\ 4\end{array}\right]$
x=-1 , y=4
(iii) 5x+2y=4
आव्यूह का प्रयोग कर निम्नलिखित समीकरण निकाय को हल करे.
[Using matrices solve the following system of equations:]
(i)
2 x+5 y=1
3 x+2 y=7
Sol :
The system of equation is non-homogeneous
Let $A=\left[\begin{array}{ll}2 & 5 \\ 3 & 2\end{array}\right]$ ,$x=\left[\begin{array}{l}x \\ y\end{array}\right]$ , $B=\left[\begin{array}{l}1 \\ 7\end{array}\right]$
|A|=4-15
=-11≠0
समीकरण निकाय के अद्विती हल है
$\therefore \quad x=A^{-1} B$
$a d j A=\left[\begin{array}{rr}2 & -3 \\ -5 & 2\end{array}\right]^{\prime}=\left[\begin{array}{rr}2 & -5 \\ -3 & 2\end{array}\right]$
$A^{-1}=\frac{1}{|A|} \cdot(a d j A)=\frac{1}{-11}\left[\begin{array}{cc}2 & -5 \\ -3 & 2\end{array}\right]$
x=A-1B
$a d j A=\left[\begin{array}{cc}2 & -3 \\ -5 & 2\end{array}\right]^{\prime}$
$A^{-1}=\dfrac{1}{|A|} \cdot(a d j A)$
$=\frac{1}{-11}\left[\begin{array}{cc}2 & -5 \\ -3 & 2\end{array}\right]$
x=A-1B
$\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{-11}\left[\begin{array}{cc}2 & -5 \\ -3 & 2\end{array}\right]\left[\begin{array}{l}1 \\ 7\end{array}\right]$
$=\frac{-1}{11}\left[\begin{array}{cc}2 & -35 \\ -3 & +14\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=-\frac{1}{11}\left[\begin{array}{c}-33 \\ 11\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}3 \\ -1\end{array}\right]$
x=3 , y=-1
(ii)
5x+2y=3
3x+2y=5
Sol :
<to be added>
Let $A=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right]$ , $x=\left[\begin{array}{l}x \\ y\end{array}\right]$, $B=\left[\begin{array}{l}3 \\ 5\end{array}\right]$
|A|=10-6=4
समीकरण निकाय के अद्विती हल होगा
$a d j A=\left[\begin{array}{cc}2 & -3 \\ -2 & 5\end{array}\right]^{\prime}=\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$
$A^{-1}=\frac{1}{|A|} \cdot(a d j A)=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$
$\therefore x=A^{-1}B$
|A|=10-6=4
$a d j A=\left[\begin{array}{cc}2 & -3 \\ -2 & 5\end{array}\right]$
$=\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$
$A^{-1}=\dfrac{1}{|A |} \cdot(a d j A)$
$=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$
∴x=A-1B
$\left[\begin{array}{c}x \\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]\left[\begin{array}{l}3 \\ 5\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}6-10 \\ -9+25\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{l}-4 \\ 160\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-1 \\ 4\end{array}\right]$
x=-1 , y=4
(iii) 5x+2y=4
7x+3y=5
Sol :
(iv) 4x-3y=3
3x-5y=7
Sol :
(v) 2x-y=-2
3x+4y=3
Sol :
Question 2
Solve the following system of equations by matrix method:
(i)
x-y+z=4
x-2 y-2 z=9
2 x+y+3 z=1
Sol :
The system of equation is non-homogeneous
Let $A=\left[\begin{array}{rrr}1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3\end{array}\right]$ ,$x=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ ,$B=\left[\begin{array}{l}4 \\ 9 \\ 1\end{array}\right]$
$|A|=1\left.\left|\begin{array}{cc}-2 & -2 \\ 1 & 3\end{array}\bigg|+1\bigg| \begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\bigg|+1\bigg| \begin{array}{c}1&-2 \\ 2&1\end{array} \bigg|\right.\right.$
=1(-6+2)+1(3+4)+1(1+4)
=-4+7+5
=8≠0
समीकरण निकाय के अद्विती हल होगा
$a d j A=\left[\begin{array}{rrr}-4 & -7 & 5 \\ 4 & 1 & -3 \\ 4 & 3 & -1\end{array}\right]^,=\left[\begin{array}{ccc}-4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1\end{array}\right]$
$A^{-1}=\frac{1}{|A|} \cdot a d j A$
$=\frac{1}{8}\left[\begin{array}{ccc}-4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1\end{array}\right]$
∴X=A-1B
$\left[\begin{array}{c}x \\ y \\ z\end{array}\right]=\frac{1}{8}\left[\begin{array}{ccc}-4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1\end{array}\right]\left[\begin{array}{l}4 \\ 9 \\ 1\end{array}\right]$
$=\frac{1}{8}\left[\begin{array}{c}-16+36+4 \\ -28+9+3 \\ 20-27-1\end{array}\right]$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{8}\left[\begin{array}{c}24 \\ -16 \\ -8\end{array}\right]$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}3 \\ -2 \\ -1\end{array}\right]$
∴x=3 , y=-2 , z=-1
(ii)
x-2 y-2 z=9
2 x+y+3 z=1
Sol :
The system of equation is non-homogeneous
Let $A=\left[\begin{array}{rrr}1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3\end{array}\right]$ ,$x=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ ,$B=\left[\begin{array}{l}4 \\ 9 \\ 1\end{array}\right]$
$|A|=1\left.\left|\begin{array}{cc}-2 & -2 \\ 1 & 3\end{array}\bigg|+1\bigg| \begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\bigg|+1\bigg| \begin{array}{c}1&-2 \\ 2&1\end{array} \bigg|\right.\right.$
=1(-6+2)+1(3+4)+1(1+4)
=-4+7+5
=8≠0
समीकरण निकाय के अद्विती हल होगा
$a d j A=\left[\begin{array}{rrr}-4 & -7 & 5 \\ 4 & 1 & -3 \\ 4 & 3 & -1\end{array}\right]^,=\left[\begin{array}{ccc}-4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1\end{array}\right]$
$A^{-1}=\frac{1}{|A|} \cdot a d j A$
$=\frac{1}{8}\left[\begin{array}{ccc}-4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1\end{array}\right]$
∴X=A-1B
$\left[\begin{array}{c}x \\ y \\ z\end{array}\right]=\frac{1}{8}\left[\begin{array}{ccc}-4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1\end{array}\right]\left[\begin{array}{l}4 \\ 9 \\ 1\end{array}\right]$
$=\frac{1}{8}\left[\begin{array}{c}-16+36+4 \\ -28+9+3 \\ 20-27-1\end{array}\right]$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{8}\left[\begin{array}{c}24 \\ -16 \\ -8\end{array}\right]$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}3 \\ -2 \\ -1\end{array}\right]$
∴x=3 , y=-2 , z=-1
(ii)
x+2 y-3 z=-4
2 x+3 y+2 z=2
3 x-3 y-4 z=11
Sol :
(iii)
2 x+3 y+2 z=2
3 x-3 y-4 z=11
Sol :
(iii)
2x-y-z=1
x+y+2y=1
3x-2y-2z=1
Sol :
(iv)
x+y+z=3
2x-y-z=2
x-2y+3z=2
Sol :
(v)
2x-3y+5z=11
3x+2y-4z=-5
x-y-2z=-3
Sol :
(vi)
x-y+z=1
x-2y+3z=2
x-3y+5z=3
Sol :
(vii)
2x-3y+z=-1
x-2y+3z=6
-3y+2z=0
Sol :
(viii)
2x+y-z=1
x-y+z=2
3x+y-2z=-1
Sol :
(ix)
2 x-3 y=1
x+3 z=11
x+2 y+z=7
Sol :
The system of equation is non-homogeneous
$A=\left[\begin{array}{ccc}2 & -3 & 0 \\ 1 & 0 & 3 \\ 1 & 2 & 1\end{array}\right]$,$x=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$,$B=\left[\begin{array}{l}1 \\ 11 \\ 7\end{array}\right]$
$|A|=2\left|\begin{array}{cc}0 & 3 \\ 2 & 1\end{array}\right|+3\left|\begin{array}{cc}1 & 3 \\ 1 & 1\end{array}\right|$
=2(0-6)+3(1-3)
=2(-6)+3(-2)
=-12-6
=-18≠0
समीकरण निकाय के अद्विती हल होगा
$\operatorname{adj} A=\left[\begin{array}{crr}-6 & 2 & 2 \\ 3 & 2 & -7 \\ -9 & -6 & 3\end{array}\right]^{\prime}=\left[\begin{array}{rrr}-6 & 3 & -9 \\ 2 & 2 & -6 \\ 2 & -7 & 3\end{array}\right]$
$A^{-1}=\frac{1}{|A|} \cdot a d j A$
$=\frac{1}{-18}\left[\begin{array}{ccc}-6 & 3 & -9 \\ 2 & 2 & -6 \\ 2 & -7 & 3\end{array}\right]$
∴X=A-1B
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-18}\left[\begin{array}{ccc}-6 & 3 & -9 \\ 2 & 2 & -6 \\ 2 & -7 & 3\end{array}\right]\left[\begin{array}{c}1 \\ 11 \\ 7\end{array}\right]$
$=\frac{-1}{18}\left[\begin{array}{ccc}-6 +33 -63 \\ 2 +22 -42 \\ 2 -77 +21\end{array}\right]$
$=-\frac{1}{10}\left[\begin{array}{l}-36 \\ -18 \\ -54\end{array}\right]$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}2 \\ 1 \\ 3\end{array}\right]$
∴x=2,y=1,z=3
x+3 z=11
x+2 y+z=7
Sol :
The system of equation is non-homogeneous
$A=\left[\begin{array}{ccc}2 & -3 & 0 \\ 1 & 0 & 3 \\ 1 & 2 & 1\end{array}\right]$,$x=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$,$B=\left[\begin{array}{l}1 \\ 11 \\ 7\end{array}\right]$
$|A|=2\left|\begin{array}{cc}0 & 3 \\ 2 & 1\end{array}\right|+3\left|\begin{array}{cc}1 & 3 \\ 1 & 1\end{array}\right|$
=2(0-6)+3(1-3)
=2(-6)+3(-2)
=-12-6
=-18≠0
समीकरण निकाय के अद्विती हल होगा
$\operatorname{adj} A=\left[\begin{array}{crr}-6 & 2 & 2 \\ 3 & 2 & -7 \\ -9 & -6 & 3\end{array}\right]^{\prime}=\left[\begin{array}{rrr}-6 & 3 & -9 \\ 2 & 2 & -6 \\ 2 & -7 & 3\end{array}\right]$
$A^{-1}=\frac{1}{|A|} \cdot a d j A$
$=\frac{1}{-18}\left[\begin{array}{ccc}-6 & 3 & -9 \\ 2 & 2 & -6 \\ 2 & -7 & 3\end{array}\right]$
∴X=A-1B
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-18}\left[\begin{array}{ccc}-6 & 3 & -9 \\ 2 & 2 & -6 \\ 2 & -7 & 3\end{array}\right]\left[\begin{array}{c}1 \\ 11 \\ 7\end{array}\right]$
$=\frac{-1}{18}\left[\begin{array}{ccc}-6 +33 -63 \\ 2 +22 -42 \\ 2 -77 +21\end{array}\right]$
$=-\frac{1}{10}\left[\begin{array}{l}-36 \\ -18 \\ -54\end{array}\right]$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}2 \\ 1 \\ 3\end{array}\right]$
∴x=2,y=1,z=3
(x)
2x+6y=2
3x-z=-8
2x-y+z=-3
Sol :
Question 3
Using matrices solve the following system of equations:
(i)
3 x-y+z=5
2 x-2 y+3 z=7
x+y-z=-1
Sol :
(ii)
2 x-2 y+3 z=7
x+y-z=-1
Sol :
(ii)
2 x-y+z=0
x+y-z=6
3 x-y-4 z=7
Sol :
x+y-z=6
3 x-y-4 z=7
Sol :
(iii)
2x+y-3z=13
x+y+z=6
2x-y+4z=-12
Sol :
(iv)
x+y+z=4
2 x-y+z=-1
2 x+y-3 z=-9
Sol :
2 x-y+z=-1
2 x+y-3 z=-9
Sol :
(v)
x-y+z=1
x-y-z=-1
3x+y-2z=3
Sol :
(vi)
x+2 y+z=7
x+3 z=11
2 x-3 y=1
x+3 z=11
2 x-3 y=1
Sol :
(vii)
x-y-z=1
(vii)
x-y-z=1
3x+y-2z=3
x-y-z=1
Sol :
(viii)
(viii)
x+y+z=6
2 x-y+z=3
x-2 y+3 z=6
(ix)
x+y-z=1
2 x-y+z=3
x-2 y+3 z=6
Sol :
(ix)
x+y-z=1
3x+y-2z=3
x-y-z=-1
Sol :
Nice Solution
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