gathered 2 x+y+z=1 x-2 y-z=3/2 3 y-5 z=9 gathered

x=1, y=1/2, z=-3/2. Write AX=B; the third equation has no x -term, so its first coefficient is 0. A=[ ], X=[ ], B=[ ] A =2(10+3)-1(-5-0)+1(3-0)=26+5+3=34≠ 0 So A^-1 exists. Cofactors: A 11=13, A 12=5, A 13=3, A 21=8, A 22=-10, A 23=-6, A 31=1, A 32=3, A 33=-5 Transposing the cofactor matrix, adj A=[ ], A^-1=1/34[ ] X=A^-1B=1/34[ ]=1/34[ ] x=1, y=1/2, z=-3/2

gathered x-y+z=4 2 x+y-3 z=0 x+y+z=2 gathered

x=2, y=-1, z=1. Write AX=B and use X=A^-1B. A=[ ], X=[ ], B=[ ] A =1(1+3)-(-1)(2+3)+1(2-1)=4+5+1=10≠ 0 Cofactors: A 11=4, A 12=-5, A 13=1, A 21=2, A 22=0, A 23=-2, A 31=2, A 32=5, A 33=3 adj A=[ ], A^-1=1/10[ ] X=A^-1B=1/10[ ]=1/10[ ] x=2, y=-1, z=1

If A =[ ], find A^-1. Using A^-1 solve the system of equations

Use A^-1=1/( A ) adj A, where adj A is the transpose of the matrix of cofactors. Expanding A along the first row: A =2 (-4+4)-(-3)(-6+4)+5 (3-2)=0-6+5=-1≠ 0 Cofactors of A: A 11=0, A 12=2, A 13=1 A 21=-1, A 22=-9, A 23=-5 A 31=2, A 32=23, A 33=13 Transposing the cofactor matrix, adj A=[ ] A^-1=1/-1[ ]=[ ] The coefficient matrix of the given system is exactly this A, so with B=[ ] the system is AX=B and X=A^-1B: X=[ ][ ]=[ ]=[ ] A^-1=[ ], x=1, y=2, z=3

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹ 70. Find cost of each item per kg by matrix method.

cost of onions per kg =₹ 5 cost of wheat per kg =₹ 8 cost of rice per kg =₹ 8. Let the cost per kg be ₹x for onion, ₹y for wheat and ₹z for rice. Translating the three purchases: In matrix form AX=B: A=[ ], X=[ ], B=[ ] A =4(12-12)-3(6-36)+2(4-24)=0+90-40=50≠ 0 So A^-1 exists. Cofactors: A 11=0, A 12=30, A 13=-20, A 21=-5, A 22=0, A 23=10, A 31=10, A 32=-20, A 33=10 adj A=[ ], A^-1=1/50[ ] X=A^-1B=1/50[ ]=1/50[ ]=[ ] Onion costs ₹5 per kg, wheat ₹8 per kg and rice ₹8 per kg.

EXERCISE 4.5 11–16 — NCERT Solutions for Class 12 Mathematics Chapter 4

EXERCISE 4.5 11–16 (part 7 of 8)

Solve system of linear equations, using matrix method, in Exercises $\displaystyle 7$ to 14.
Exercise 11
$\begin{gathered} 2 x+y+z=1 \\ x-2 y-z=\frac{3}{2} \\ 3 y-5 z=9 \end{gathered}$
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
$\displaystyle x=1, y=\frac{1}{2}, z=\frac{-3}{2}$
Write $\displaystyle AX=B$; the third equation has no $\displaystyle x$-term, so its first coefficient is $\displaystyle 0$.\[A=\left[\begin{array}{rrr}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y \\ z\end{array}\right],\qquad B=\left[\begin{array}{r}1 \\ \tfrac{3}{2} \\ 9\end{array}\right] \]\[|A|=2(10+3)-1(-5-0)+1(3-0)=26+5+3=34\neq 0 \]So $\displaystyle A^{-1}$ exists. Cofactors:\[A_{11}=13,\ A_{12}=5,\ A_{13}=3,\quad A_{21}=8,\ A_{22}=-10,\ A_{23}=-6,\quad A_{31}=1,\ A_{32}=3,\ A_{33}=-5 \]Transposing the cofactor matrix,\[\mathrm{adj}\,A=\left[\begin{array}{rrr}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{array}\right],\qquad A^{-1}=\frac{1}{34}\left[\begin{array}{rrr}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{array}\right] \]\[X=A^{-1}B=\frac{1}{34}\left[\begin{array}{r}13(1)+8\left(\tfrac{3}{2}\right)+1(9) \\ 5(1)-10\left(\tfrac{3}{2}\right)+3(9) \\ 3(1)-6\left(\tfrac{3}{2}\right)-5(9)\end{array}\right]=\frac{1}{34}\left[\begin{array}{r}34 \\ 17 \\ -51\end{array}\right] \]\[x=1,\qquad y=\frac{1}{2},\qquad z=-\frac{3}{2} \]
Exercise 12
$\begin{gathered} x-y+z=4 \\ 2 x+y-3 z=0 \\ x+y+z=2 \end{gathered}$
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This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
$\displaystyle x=2, y=-1, z=1$
Write $\displaystyle AX=B$ and use $\displaystyle X=A^{-1}B$.\[A=\left[\begin{array}{rrr}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y \\ z\end{array}\right],\qquad B=\left[\begin{array}{r}4 \\ 0 \\ 2\end{array}\right] \]\[|A|=1(1+3)-(-1)(2+3)+1(2-1)=4+5+1=10\neq 0 \]Cofactors:\[A_{11}=4,\ A_{12}=-5,\ A_{13}=1,\quad A_{21}=2,\ A_{22}=0,\ A_{23}=-2,\quad A_{31}=2,\ A_{32}=5,\ A_{33}=3 \]\[\mathrm{adj}\,A=\left[\begin{array}{rrr}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right],\qquad A^{-1}=\frac{1}{10}\left[\begin{array}{rrr}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right] \]\[X=A^{-1}B=\frac{1}{10}\left[\begin{array}{r}16+0+4 \\ -20+0+10 \\ 4-0+6\end{array}\right]=\frac{1}{10}\left[\begin{array}{r}20 \\ -10 \\ 10\end{array}\right] \]\[x=2,\qquad y=-1,\qquad z=1 \]
Exercise 13
$\begin{aligned} & 2 x+3 y+3 z=5 \\ & x-2 y+z=-4 \\ & 3 x-y-2 z=3 \end{aligned}$
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This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
$\displaystyle x=1, y=2, z=-1$
Write $\displaystyle AX=B$ and use $\displaystyle X=A^{-1}B$.\[A=\left[\begin{array}{rrr}2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y \\ z\end{array}\right],\qquad B=\left[\begin{array}{r}5 \\ -4 \\ 3\end{array}\right] \]\[|A|=2(4+1)-3(-2-3)+3(-1+6)=10+15+15=40\neq 0 \]Cofactors:\[A_{11}=5,\ A_{12}=5,\ A_{13}=5,\quad A_{21}=3,\ A_{22}=-13,\ A_{23}=11,\quad A_{31}=9,\ A_{32}=1,\ A_{33}=-7 \]\[\mathrm{adj}\,A=\left[\begin{array}{rrr}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right],\qquad A^{-1}=\frac{1}{40}\left[\begin{array}{rrr}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right] \]\[X=A^{-1}B=\frac{1}{40}\left[\begin{array}{r}25-12+27 \\ 25+52+3 \\ 25-44-21\end{array}\right]=\frac{1}{40}\left[\begin{array}{r}40 \\ 80 \\ -40\end{array}\right] \]\[x=1,\qquad y=2,\qquad z=-1 \]
Exercise 14
$\begin{aligned} & x-y+2 z=7 \\ & 3 x+4 y-5 z=-5 \\ & 2 x-y+3 z=12 \end{aligned}$
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This solution has not been cross-checked against the answer printed in NCERT.
Write $\displaystyle AX=B$ and use $\displaystyle X=A^{-1}B$.\[A=\left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y \\ z\end{array}\right],\qquad B=\left[\begin{array}{r}7 \\ -5 \\ 12\end{array}\right] \]\[|A|=1(12-5)-(-1)(9+10)+2(-3-8)=7+19-22=4\neq 0 \]Cofactors:\[A_{11}=7,\ A_{12}=-19,\ A_{13}=-11,\quad A_{21}=1,\ A_{22}=-1,\ A_{23}=-1,\quad A_{31}=-3,\ A_{32}=11,\ A_{33}=7 \]Transposing gives\[\mathrm{adj}\,A=\left[\begin{array}{rrr}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right],\qquad A^{-1}=\frac{1}{4}\left[\begin{array}{rrr}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right] \]\[X=A^{-1}B=\frac{1}{4}\left[\begin{array}{r}49-5-36 \\ -133+5+132 \\ -77+5+84\end{array}\right]=\frac{1}{4}\left[\begin{array}{r}8 \\ 4 \\ 12\end{array}\right] \]\[x=2,\qquad y=1,\qquad z=3 \]
Exercise 15
If $\displaystyle \mathrm{A}=\left[\begin{array}{rrr}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$ , find $\displaystyle \mathrm{A}^{-1}$ . Using $\displaystyle \mathrm{A}^{-1}$ solve the system of equations $\begin{aligned} 2 x-3 y+5 z & =11 \\ 3 x+2 y-4 z & =-5 \\ x+y-2 z & =-3 \end{aligned}$
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This solution has not been cross-checked against the answer printed in NCERT.
Use $\displaystyle A^{-1}=\dfrac{1}{|A|}\,\mathrm{adj}\,A$, where $\displaystyle \mathrm{adj}\,A$ is the transpose of the matrix of cofactors.Expanding $\displaystyle |A|$ along the first row:\[|A|=2\,(-4+4)-(-3)(-6+4)+5\,(3-2)=0-6+5=-1\neq 0 \]Cofactors of $\displaystyle A$:\[A_{11}=0,\quad A_{12}=2,\quad A_{13}=1 \] \[A_{21}=-1,\quad A_{22}=-9,\quad A_{23}=-5 \] \[A_{31}=2,\quad A_{32}=23,\quad A_{33}=13 \]Transposing the cofactor matrix,\[\mathrm{adj}\,A=\left[\begin{array}{rrr}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right] \]\[A^{-1}=\frac{1}{-1}\left[\begin{array}{rrr}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right]=\left[\begin{array}{rrr}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right] \]The coefficient matrix of the given system is exactly this $\displaystyle A$, so with $\displaystyle B=\left[\begin{array}{r}11 \\ -5 \\ -3\end{array}\right]$ the system is $\displaystyle AX=B$ and $\displaystyle X=A^{-1}B$:\[X=\left[\begin{array}{rrr}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]\left[\begin{array}{r}11 \\ -5 \\ -3\end{array}\right]=\left[\begin{array}{r}0-5+6 \\ -22-45+69 \\ -11-25+39\end{array}\right]=\left[\begin{array}{r}1 \\ 2 \\ 3\end{array}\right] \]\[A^{-1}=\left[\begin{array}{rrr}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right],\qquad x=1,\ y=2,\ z=3 \]
Exercise 16
The cost of $\displaystyle 4$ kg onion, $\displaystyle 3$ kg wheat and $\displaystyle 2$ kg rice is ₹ $\displaystyle 60$ . The cost of $\displaystyle 2$ kg onion, $\displaystyle 4$ kg wheat and $\displaystyle 6$ kg rice is ₹ $\displaystyle 90$ . The cost of $\displaystyle 6$ kg onion $\displaystyle 2$ kg wheat and $\displaystyle 3$ kg rice is ₹ 70. Find cost of each item per kg by matrix method.
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
cost of onions per $\displaystyle \mathrm{kg}=₹ 5$ cost of wheat per $\displaystyle \mathrm{kg}=₹ 8$ cost of rice per kg =₹ $\displaystyle 8$
Let the cost per kg be $\displaystyle ₹x$ for onion, $\displaystyle ₹y$ for wheat and $\displaystyle ₹z$ for rice. Translating the three purchases:\[\begin{aligned} 4x+3y+2z &= 60 \\ 2x+4y+6z &= 90 \\ 6x+2y+3z &= 70 \end{aligned} \]In matrix form $\displaystyle AX=B$:\[A=\left[\begin{array}{rrr}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y \\ z\end{array}\right],\qquad B=\left[\begin{array}{r}60 \\ 90 \\ 70\end{array}\right] \]\[|A|=4(12-12)-3(6-36)+2(4-24)=0+90-40=50\neq 0 \]So $\displaystyle A^{-1}$ exists. Cofactors:\[A_{11}=0,\ A_{12}=30,\ A_{13}=-20,\quad A_{21}=-5,\ A_{22}=0,\ A_{23}=10,\quad A_{31}=10,\ A_{32}=-20,\ A_{33}=10 \]\[\mathrm{adj}\,A=\left[\begin{array}{rrr}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right],\qquad A^{-1}=\frac{1}{50}\left[\begin{array}{rrr}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right] \]\[X=A^{-1}B=\frac{1}{50}\left[\begin{array}{r}0-450+700 \\ 1800+0-1400 \\ -1200+900+700\end{array}\right]=\frac{1}{50}\left[\begin{array}{r}250 \\ 400 \\ 400\end{array}\right]=\left[\begin{array}{r}5 \\ 8 \\ 8\end{array}\right] \]Onion costs $\displaystyle ₹5$ per kg, wheat $\displaystyle ₹8$ per kg and rice $\displaystyle ₹8$ per kg.

EXERCISE 4.5 11–16 (part 7 of 8)

Exercise 11

Exercise 12

Exercise 13

Exercise 14

Exercise 15

Exercise 16