Exercise-4.5, Class 12th, Maths, Chapter 4, NCERT

Examine the consistency of the system of equations in Exercises 1 to 6.

1.
$\{\begin{array}{l}x+2y=2\\ 2x+3y=3\end{array}$

Solution. From the first, $x=2-2y$. Substitute: $2(2-2y)+3y=3\Rightarrow 4-y=3\Rightarrow y=1$. Then $x=0$.
Unique solution: $(x,y)=(0,1)$

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2.
$\{\begin{array}{l}2x-y=5\\ x+y=4\end{array}$

Solution. From second $y=4-x$. Substitute: $2x-(4-x)=5\Rightarrow 3x=9\Rightarrow x=3$. Then $y=1$
Solution: $(3,1)$

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3.
$\{\begin{array}{l}x+3y=5\\ 2x+6y=8\end{array}$

Solution. The second equation is $2(x+3y)=10$, but RHS is 8, not 10. The two equations are inconsistent (parallel, no solution).
No solution. 

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4.
$\{\begin{array}{l}x+y+z=1\\ 2x+3y+2z=2\\ ax+ay+2az=4\end{array}$

Solution. From (1) & (2): subtract $2$(1) from (2) → $y=0$. Then $x+z=1$.
Third equation is $a(x+y+2z)=4$. With $y=0$ this becomes $a(x+2z)=4$. Using $x=1-z$:

$$a((1-z)+2z)=a(1+z)=4\Rightarrow 1+z=\frac{4}{a}(\text{if }a\mathrm{\ne }0)\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$$

So for $a\mathrm{\ne }0$: $z={\displaystyle \frac{4}{a}}-1,x=2-{\displaystyle \frac{4}{a}},y=0$ (unique solution).
If $a=0$: third eqn becomes $0=4$ impossible → inconsistent.

So: consistent (unique) for $a\mathrm{\ne }0$ with $(x,y,z)=(2-\frac{4}{a},0,\frac{4}{a}-1)$. Inconsistent when $a=0$.  depending on $a$.

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5.
$\{\begin{array}{l}3x-y-2z=2\\ 2y-z=-1\\ 3x-5y=3\end{array}$

Solution. From second: $z=2y+1$. From third: $x=(3+5y)\mathrm{/}3$. Substitute into first:

$$3\cdot \frac{3+5y}{3}-y-2(2y+1)=2\Rightarrow (3+5y)-y-4y-2=2\Rightarrow 1=2,$$

contradiction. So inconsistent (no solution). 

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6.
$\{\begin{array}{l}5x-y+4z=5\\ 2x+3y+5z=2\\ 5x-2y+6z=-1\end{array}$

Solution. Subtract (1) from (3): $-y+2z=-6\Rightarrow y-2z=6$$<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.\; Put\; in\; (2):</span>$

$$2x+3(6+2z)+5z=2\Rightarrow 2x+18+11z=2\Rightarrow x=-8-\frac{11}{2}z\mathrm{}$$

Put $x,y$ into (1): solving gives $z=-2$, then $y=2$, $x=3$
Unique solution: (x,y,z) = (3,2,-2)

Solve system of linear equations, using matrix method, in Exercises 7 to 14.

Question 7 

Solve by matrix method:

$$\{\begin{array}{l}5x+2y=4,\\ 7x+3y=5.\end{array}$$

Matrix form: $A\mathbf{x}=\mathbf{b}$ with

$$A=\left[\begin{array}{cc}5& 2\\ 7& 3\end{array}\right],\mathbf{x}=\left[\begin{array}{c}x\\ y\end{array}\right],\mathbf{b}=\left[\begin{array}{c}4\\ 5\end{array}\right]\mathrm{}$$

$\det A=5\cdot 3-7\cdot 2=15-14=1$
$A^{-1}=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]$ (since $1\mathrm{/}\det A=1$).
Hence $\mathbf{x}=A^{-1}\mathbf{b}=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\left[\begin{array}{c}4\\ 5\end{array}\right]=\left[\begin{array}{c}2\\ -3\end{array}\right]\mathrm{}$

Answer: $x=2,y=-3.$

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Question 8 

$$\{\begin{array}{l}2x-y=-2,\\ 3x+4y=3.\end{array}$$

Matrix form: $A=\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right],\mathbf{b}=\left[\begin{array}{c}-2\\ 3\end{array}\right]\mathrm{}$
$\det A=2\cdot 4-3\cdot (-1)=8+3=11$

$A^{-1}={\displaystyle \frac{1}{11}}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\mathrm{}$
$\mathbf{x}=A^{-1}\mathbf{b}={\displaystyle \frac{1}{11}}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\left[\begin{array}{c}-2\\ 3\end{array}\right]=\left[\begin{array}{c}-\frac{5}{11}\\ \frac{12}{11}\end{array}\right]\mathrm{}$

Answer: $x=-{\displaystyle \frac{5}{11}},y={\displaystyle \frac{12}{11}}\mathrm{.}$

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Question 9 

$$\{\begin{array}{l}4x-3y=3,\\ 3x-5y=7.\end{array}$$

Matrix form: $A=\left[\begin{array}{cc}4& -3\\ 3& -5\end{array}\right],\mathbf{b}=\left[\begin{array}{c}3\\ 7\end{array}\right]\mathrm{}$
$\det A=4(-5)-3(-3)=-20+9=-11$

$A^{-1}={\displaystyle \frac{1}{-11}}\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]={\displaystyle \frac{1}{11}}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]\mathrm{}$$\mathbf{x}=A^{-1}\mathbf{b}={\displaystyle \frac{1}{11}}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]\left[\begin{array}{c}3\\ 7\end{array}\right]=\left[\begin{array}{c}-\frac{6}{11}\\ -\frac{19}{11}\end{array}\right]\mathrm{}$

Answer: $x=-{\displaystyle \frac{6}{11}},y=-{\displaystyle \frac{19}{11}}\mathrm{.}$

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Question 10 

$$\{\begin{array}{l}5x+2y=3,\\ 3x+2y=5.\end{array}$$

Matrix form: $A=\left[\begin{array}{cc}5& 2\\ 3& 2\end{array}\right],\mathbf{b}=\left[\begin{array}{c}3\\ 5\end{array}\right]\mathrm{}$
Subtract second from first: $2x=-2\Rightarrow x=-1$Then $5(-1)+2y=3\Rightarrow -5+2y=3\Rightarrow 2y=8\Rightarrow y=4.$

(Or use inverse: $\det A=5\cdot 2-3\cdot 2=10-6=4$, $A^{-1}=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]$ giving same result.)

Answer: $x=-1,y=4.$

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Question 11 

Solve the system

$$\{\begin{array}{l}2x+y+z=1,\\ x-2y-z={\displaystyle \frac{3}{2}},\\ 2x+y-3z=0.\end{array}$$

Solution (matrix / elimination).

Augmented matrix:

$$\left[\begin{array}{cccc}2& 1& 1& 1\\ 1& -2& -1& \frac{3}{2}\\ 2& 1& -3& 0\end{array}\right]$$

To avoid fractions, multiply row2 by 2 first:

$$R_2\leftarrow 2R_2:\left[\begin{array}{cccc}2& 1& 1& 1\\ 2& -4& -2& 3\\ 2& 1& -3& 0\end{array}\right]$$

Now eliminate using $R_1$:

• $R_2\leftarrow R_2-R_1$
  $[0,-5,-3\mid 2]$
  

• $R_3\leftarrow R_3-R_1$
  $[0,0,-4\mid -1]$

Matrix becomes

$$\left[\begin{array}{cccc}2& 1& 1& 1\\ 0& -5& -3& 2\\ 0& 0& -4& -1\end{array}\right]$$

Back-substitution:

From row3: $-4z=-1\Rightarrow z={\displaystyle \frac{1}{4}}\mathrm{}$

Row2: $-5y-3z=2\Rightarrow -5y-3\cdot \frac{1}{4}=2\Rightarrow -5y={\displaystyle \frac{11}{4}}\Rightarrow y=-{\displaystyle \frac{11}{20}}\mathrm{}$

Row1: $2x+y+z=1\Rightarrow 2x+(-{\displaystyle \frac{11}{20}})+{\displaystyle \frac{1}{4}}=1$
Compute $y+z=-{\displaystyle \frac{11}{20}}+{\displaystyle \frac{5}{20}}=-{\displaystyle \frac{6}{20}}=-{\displaystyle \frac{3}{10}}\mathrm{}$
So $2x-{\displaystyle \frac{3}{10}}=1\Rightarrow 2x={\displaystyle \frac{13}{10}}\Rightarrow x={\displaystyle \frac{13}{20}}\mathrm{}$

Answer: ${\displaystyle \boxed{{\displaystyle x=\frac{13}{20},y=-\frac{11}{20},z=\frac{1}{4}}}}$

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Question 12 

Solve the system

$$\{\begin{array}{l}x-y+z=4,\\ 2x+y-3z=0,\\ 3y-5z=9.\end{array}$$

Solution (matrix / substitution).

From the third equation:

$$3y-5z=9\Rightarrow y=\frac{9+5z}{3}\mathrm{}$$

Substitute $y$ into the first equation:

$$x-\frac{9+5z}{3}+z=4$$

Multiply by 3:

$$3x-(9+5z)+3z=12\Rightarrow 3x-9-2z=12\Rightarrow 3x-2z=21.$$

So$$\begin{array}{cccc}& x=7+\frac{2}{3}z\mathrm{.}& & \text{(*)}\end{array}$$

Now substitute $x$ and $y$ into the second equation $2x+y-3z=0$

$$2(7+\frac{2}{3}z)+\frac{9+5z}{3}-3z=0.$$

Compute left side:

$$14+\frac{4}{3}z+\frac{9+5z}{3}-3z=14+\frac{4z+9+5z}{3}-3z$$

$$=14+\frac{9+9z}{3}-3z=14+3+3z-3z=17.$$

That gives $17=0$, a contradiction.

Conclusion: The system is inconsistent; no solution.

Answer: $\boxed{{\displaystyle \text{No solution (inconsistent system).}}}$

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Question 13 

Solve the system

$$\{\begin{array}{l}2x+3y+3z=5,\\ x-2y+z=-4,\\ 3x-y-2z=3.\end{array}$$

Solution (matrix / elimination).

Augmented matrix:

$$\left[\begin{array}{cccc}2& 3& 3& 5\\ 1& -2& 1& -4\\ 3& -1& -2& 3\end{array}\right]$$

Use row2 as pivot to eliminate others:

• $R_1\leftarrow R_1-2R_2$
  $[0,7,1\mid 13]$

• $R_3\leftarrow R_3-3R_2$
  $[0,5,-5\mid 15]$

So we have

$$\left[\begin{array}{cccc}1& -2& 1& -4\\ 0& 7& 1& 13\\ 0& 5& -5& 15\end{array}\right]$$

(reordered so row2 is the original $R_2$).

Eliminate $y$ from row3 using row2:

• $R_3\leftarrow 7R_3-5R_2$
  

$7R_3=[0,35,-35\mid 105],5R_2=[0,35,5\mid 65]$

So $R_3\leftarrow [0,0,-40\mid 40]$ ⇒ $-40z=40\Rightarrow z=-1$

Back-substitute:

Row2: $7y+z=13\Rightarrow 7y-1=13\Rightarrow 7y=14\Rightarrow y=2$

Row1 (original row2): $x-2y+z=-4\Rightarrow x-4-1=-4\Rightarrow x=1$

Answer: ${\displaystyle \boxed{{\displaystyle x=1,y=2,z=-1}}}$

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Question 14 (Plain text)

Solve the system

$$\{\begin{array}{l}x-y+2z=7,\\ 3x+4y-5z=-5,\\ 2x-y+3z=12.\end{array}$$

Solution (matrix / elimination).

Augmented matrix:

$$\left[\begin{array}{cccc}1& -1& 2& 7\\ 3& 4& -5& -5\\ 2& -1& 3& 12\end{array}\right]$$

Eliminate below first pivot $R_1$:

• $R_2\leftarrow R_2-3R_1:[0,7,-11\mid -26]$
  

• $R_3\leftarrow R_3-2R_1:[0,1,-1\mid -2]$
  

Swap $R_2$ and $R_3$ to use the simpler pivot:

$$\left[\begin{array}{cccc}1& -1& 2& 7\\ 0& 1& -1& -2\\ 0& 7& -11& -26\end{array}\right]$$

Eliminate $y$ from row3:

• $R_3\leftarrow R_3-7R_2:[0,0,-4\mid -12]$

So $-4z=-12\Rightarrow z=3$

Back-substitute:

Row2: $y-z=-2\Rightarrow y-3=-2\Rightarrow y=1$

Row1: $x-y+2z=7\Rightarrow x-1+6=7\Rightarrow x=2$

Answer: ${\displaystyle \boxed{{\displaystyle x=2,y=1,z=3}}\mathrm{}}$

 

$\mathrm{}$