4 Marks

Question 514 Marks

Show that the following system of linear equation is inconsistent:
2x + 5y = 7
6x + 15y = 13

Answer

The given system of equations can be expresed as follows:
$\text{AX = B}$
Here,
$\text{A}=\begin{bmatrix}2&5\\ 6&15\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7\\ 13\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}2&5\\ 6&15\end{vmatrix}$
$={(30-30)}$
$=0$
Let C_ij be the co-factors of the elemennts a_ij in A = [a_ij]. Then,
$\text{C}_{11}=-1^{1+1}{(15)}=15,\\ \text{C}_{12}=-1^{1+2}{(6)}=-6\\ $
$\text{C}_{21}=-1^{2+1}{(5)}=-5,\\ \text{C}_{22}=-1^{2+2}{(6)}=2\\ $
$\text{adj A}=\begin{bmatrix}15&-6\\ -5&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}15&-5\\ -6&2\end{bmatrix}$
$(\text{adj A) B}=\begin{bmatrix}15&-5\\ -6&2\end{bmatrix}\begin{bmatrix}7\\ 13\end{bmatrix}$
$=\begin{bmatrix}105-65\\ -42+26\end{bmatrix}$
$=\begin{bmatrix}40\\ -16\end{bmatrix}\neq0$
Hence, the given system of equations is inconsitent.

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Question 524 Marks

Solve the follwing system of equations by matrix method:
5x + 2y = 3
3x + 2y = 5

Answer

The above system can be written in matrix form as:

$\begin{bmatrix}5&2\\ 3&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}3\\ 2\end{bmatrix}$

Or $\text{AX = B}$

Where,

$\text{A}=\begin{bmatrix}5&2\\ 3&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix},\text{B}=\begin{bmatrix}3\\ 5\end{bmatrix}$

Now, $\text{|A|}=10-6=4\neq0$

So, the above system has a unique solution, given by

$\text{X}=\text{A}^{-1}\text{B}$

Let C_ijbe the co factor of a_ij in A, then

$\text{C}_{11} = 2,\text{C}_{12} = -3$

$\text{C}_{21} = -2,\text{C}_{22} = 5$

Also,

$\text{Adj A}=\begin{bmatrix}2&-3\\ -2&5\end{bmatrix}^\text{T}=\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}$

$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj A}=\frac{1}{4}\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}$

Now, X = A^-1B

$=\frac{1}{4}\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}\begin{bmatrix}3\\ 5\end{bmatrix}$

$=\frac{1}{4}\begin{bmatrix}-4\\ 16\end{bmatrix}$

$\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-1\\ 4\end{bmatrix}$

Hence, x = -1

y = 4

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Question 534 Marks

Show that the following system of linear equations is consistent and also find solution:
6x + 4y = 2
9x + 6y =3

Answer

Here,
$6\text{x}+4\text{y}=2\ \dots(1)$
$9\text{x}+6\text{y}\ \dots(2)$
AX = B
Here,
$\text{A}=\begin{bmatrix}6&4\\ 9&6\end{bmatrix},\text{X}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\ 3\end{bmatrix}$
$\begin{bmatrix}6&4\\ 9&6\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}2\\ 3\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}6&4\\ 9&6\end{vmatrix}$
$= 36 -36$
$= 0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because (adj A)B ≠ 0 or (adj A) = 0.
Let C_ij be the co-factors of the elements a_ij in A [a_ij]. Then,
$\text{C}_{11}=6,\\ \text{C}_{12}=-9,\\ \text{C}_{21}=-4,\\ \text{C}_{22}=6$
$\text{adj A}=\begin{bmatrix}6&-9\\ -4&6\end{bmatrix}^\text{T}$
$=\begin{bmatrix}6&-4\\ -9&6\end{bmatrix}$
$\text{(adj A) B}=\begin{bmatrix}6&-4\\ -9&6\end{bmatrix}\begin{bmatrix}2\\ 3\end{bmatrix}$
$=\begin{bmatrix}12-12\\ -18+18\end{bmatrix}$
$=\begin{bmatrix}0\\ 0\end{bmatrix}$
If |A| = 0 and (adj A) B = 0, then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions.
Substituting y = k in the eq. (1), we get
$6\text{x}+4\text{k}=2$
$\Rightarrow6\text{x}=2-4\text{k}$
$\Rightarrow\text{x}=\frac{2-\text{4k}}{6}$
$\Rightarrow\text{x}=\frac{1-\text{2k}}{3}$
$\therefore \text{x}=\frac{1-2\text{k}}{3}\text{ and }\text{y}=\text{k}$
These values of x and y satisfy the third equation.
Thus, $\text{x}=\frac{1-2\text{k}}{3}$ and y = k ( where k is a real number ) satisfy the given system of equations.

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Question 544 Marks

Show that the following system of linear equations is consistent and also find solution:
2x + 3y = 5
6x + 9y = 15

Answer

$2\text{x}+3\text{y}=5\dots(1)$
$6\text{x}+9\text{y}=15\dots(2)$
Or , AX = B
Where,
$\text{A}=\begin{bmatrix}2&3\\ 6&9\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\begin{bmatrix}2&3\\ 6&9\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\therefore\ |\text{A}|=\begin{vmatrix}2&3\\6&9\end{vmatrix}$
$=18-18$
$=0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\text{adj A})\text{B}\neq0\text{ or }(\text{adj A})=0$.
C₁₁ = 9, C₁₂ = -6, C₂₁ = -3 and C₂₂ = 2
$\therefore\ \text{adj A}=\begin{bmatrix}9&-6\\-3&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}2&-3\\-6&9\end{bmatrix}$
$\Rightarrow(\text{adj A})\text{B}=\begin{bmatrix}9&-3\\-6&2\end{bmatrix}\begin{bmatrix}5\\15\end{bmatrix}$
$=\begin{bmatrix}45-45\\-30+30\end{bmatrix}$
$=\begin{bmatrix}0\\0\end{bmatrix}$
If |A| = 0 and (adj A) B = 0, then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions.
Substituting y = k in eq. (1), we get
$2\text{x} + 3\text{k}=5$
$\Rightarrow2\text{x}=5-3\text{k}$
$\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$
These values of x and y satisfy the third equation.
Thus, $\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$ (where k is a real number) satisfy the given system of equations.

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Question 554 Marks

Show that the following system of linear equation is inconsistent:

4x − 2y = 3

6x − 3y = 5

Answer

This system can be written as:
$\begin{bmatrix}4&-2\\ 6&-3\end{bmatrix}\begin{bmatrix} \text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}3\\ 5\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=-12+12=0$
So, A is singular, Now system will be inconsisent, if
$(\text{adj A})\times\text{B}\neq0$
$\text{C}_{11}=-3$
$\text{C}_{12}=-6$
$\text{C}_{21}=2$
$\text{C}_{22}=4$
$\text{adj A}=\begin{bmatrix}-3&-6\\ 2&4\end{bmatrix}^\text{T}\begin{bmatrix}-3&2\\ -6&4\end{bmatrix}$
$(\text{adj A})\times(\text{B})=\begin{bmatrix}-3&2\\ -6&4\end{bmatrix}\begin{bmatrix}3\\ 5\end{bmatrix}$
$=\begin{bmatrix}-9+10\\ -18+20\end{bmatrix}$
$=\begin{bmatrix}1\\ 2\end{bmatrix}$
$\neq0$
Hence, the above system is inconsisent.

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Question 564 Marks

Solve the following systems of homogeneous linear equations by matrix method:

2x - y + z = 0

3x + 2y - z = 0

x + 4y + 3z = 0

Answer

The given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}2&-1&1\\3&2&-1\\1&4&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
or, $\text{AX}=\text{O}$
where, $\text{A}=\begin{bmatrix}2&-1&1\\3&2&-1\\1&4&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
$|\text{A}|=\begin{vmatrix}2&-1&1\\3&2&-1\\1&4&3\end{vmatrix}$
$=2(6+4)+1(9+1)+1(12-2)$
$=40$
$\therefore\ |\text{A}|\neq0$
So, the given system has only trivial solution, which is given below:
x = y = z = 0

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Question 574 Marks

Show that the following system of linear equation is inconsistent:
2x + 3y = 5
6x + 9y = 10

Answer

The given system of equations can be expresesed as follows:
$\text{AX = B}$
Here,
$\text{A}=\begin{bmatrix}2&3\\ 6&9\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}5\\ 10\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}2&3\\ 6&9\end{vmatrix}$
$={(18-18)}$
$=0$
Let C_ijbe the co-factors of the elements a_ij in A = [a_ij]. Then,
$\text{C}_{11}={(-1)}^{1+1}{(9)}=9,\\ \text{C}_{12}={(-1)}^{1+2}{(6)}=-6$
$\text{C}_{21}={(-1)}^{2+1}{(3)}=-3,\\ \text{C}_{22}={(-1)}^{2+2}{(6)}=2$
$\text{adj A}=\begin{bmatrix}9&-6\\ -3&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}9&-3\\ -6&2\end{bmatrix}$
$\text{(adj A) = B}=\begin{bmatrix}9&-3\\ -6&2\end{bmatrix}\begin{bmatrix}5\\ 10\end{bmatrix}$
$=\begin{bmatrix}45-30\\ -30+30\end{bmatrix}$
$=\begin{bmatrix}15\\ -10\end{bmatrix}\neq0$
Hence, the given system of equations is inconsistent.

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Question 584 Marks

Solve the following system of equations by matrix method:
5x + 7y + 2 = 0

4x + 6y + 3 = 0

Answer

The given system of equations can be written in matrix from as follws:

$\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-2\\ -3\end{bmatrix}$

$\text{AX = B}$

Here,

$\text{A}=\begin{bmatrix}5&7\\ 4&6\end{bmatrix},\text{X=}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}$ And$\text{ B}=\begin{bmatrix}-2\\ -3\end{bmatrix}$

Now,

$|\text{A}|=\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\\ $

$=30-28$

$=2\neq0$

The given system has a unique solution given by $\text{X}=\text{A}^{-1 }\text{B.}$

Let C_ijbe the cofactors of the elements a_ij in A = [a_ij]. Then,

$\text{C}_{11}=(-1)^{1+1}(6)=6,\text{C}_{12}=(-1)^{1+2}(4)=-4$

$\text{C}_{21}=(-1)^{2+1}(7)=-7,\text{C}_{22}=(-1)^{2+2}(5)$

$\text{adj}\ \text{A}=\begin{bmatrix}6&-4\\ -7&5\end{bmatrix}=\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$

$\text{A}^{-1}=\frac{1}{\text{|A|}}\text{ adj}\text{ A}$

$\Rightarrow \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$

$\text{X}=\text{A}^{-1}\text{B}$

$=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}\begin{bmatrix}-2\\-3\end{bmatrix} $

$=\frac{1}{2}\begin{bmatrix}-12+21\\ 8-15\end{bmatrix} $

$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{2}\\ \frac{-7}{2}\end{bmatrix}$

$\therefore\text{X}=\frac{9}{2}$ And $\text{ y }=\frac{-7}{2}$

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Question 594 Marks

Solve the following system of equations by matrix method:
3x + y = 19
3x - y = 23

Answer

The given system of equations can be written in matrix form as follows:
$\begin{bmatrix}3&1\\ 3&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}19\\ 23\end{bmatrix}$
AX = B
Here,
$\text{A}=\begin{bmatrix}3&1\\ 3&-1\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}19\\ 23\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&1\\ 3&-1\end{vmatrix}$
$= - 3 - 3$
$=-6\neq0$
So, the given system has a unique solution given by X = A^-1B.
Let C_ijbe the co-factors of the elements a_ij in A = [a_ij]. Then,
$\text{C}_{11}=(-1)^{1+1}(-1)=-1,\text{C}_{12}=(-1)^{1+2}(3)=-3$
$\text{C}_{21}=(-1)^{2+1}(1)=-1,\text{C}_{22}=(-1)^{2+2}(3)=3$
$\text{adj A}=\begin{bmatrix}-1&-3\\ -1&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}$

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Question 604 Marks

Show that the following system of linear equations is consistent and also find solution:
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3

Answer

This system can be written as:

$\begin{bmatrix}2&2&-2\\ 4&4&-1\\ 6&6&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$

or $\text{AX = B}$

$\text{|A|}=2{(14)}-2(14)-2{(0)}=0$

So, A is singular and the system has either no solution or infinite solutions according as

$\text{(Adj A)}\times\text{(B)}\neq0$ or $\text{(Adj A)}\times\text{(B)}=0$

Let C_ijbe the co-factors of a_ijin A

$\text{C}_{11}=14\\ \text{C}_{21}=-16\\ \text{C}_{31}=6$

$\text{C}_{12}=-14\\ \text{C}_{22}=16\\ \text{C}_{32}=-6$

$\text{C}_{1}=0\\ \text{C}_{23}=0\\ \text{C}_{33}=0$

$\text{adj A}=\begin{bmatrix}14&-14&0\\ -16&16&0\\ 0&0&0\end{bmatrix}^\text{T}=\begin{bmatrix}14&-16&6\\ -14&16&-6\\ 0&0&0\end{bmatrix}$

$(\text{adj A})\times\text{B}=\begin{bmatrix}14&-16&0\\ -14&16&-6\\ 0&0&0\end{bmatrix}\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}=\begin{bmatrix}14-32+18\\ -14+32-18\\ 0+0+0\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

So, $\text{AX}=\text{B}$ has infinite solutions.

Now, let z = k

So, 2x + 2y = 1 + 2k

4x + 4y = 2 + k

which can be written as:

$\begin{bmatrix}2&2\\ 4&4\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}1+2\text{k}\\ 2+\text{k}\end{bmatrix}$

or $\text{AX = B}$

|A| = 0, z = 0

Again,

$2\text{x}+2\text{y}=1$

$4\text{x}+4\text{y}=2$

Let $\text{y = k}$

$2\text{x}=1-2\text{k}$

$\text{x}=\frac{1}{2}-\text{k}$

Hence, $\text{x}=\frac{1}{2}-\text{k}$

$\text{y}=\text{k}$

$\text{z}=0$

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Question 614 Marks

Solve the following systems of homogeneous linear equations by matrix method:

3x + y - 2z = 0

x + y + z = 0

x - 2y + z = 0

Answer

The given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}3&1&-2\\1&1&1\\1&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
or, $\text{AX}=\text{O}$
where, $\text{A}=\begin{bmatrix}3&1&-2\\1&1&1\\1&-2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}3&1&-2\\1&1&1\\1&-2&1\end{vmatrix}$
$=3(1+2)-1(1-1)-2(-2-1)$
$=9-0+6$
$=15\neq0$
So, the given system has only trivial solution, which is given below:
x = y = z = 0

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Question 624 Marks

Solve the following systems of homogeneous linear equations by matrix method:

3x - y + 2z = 0

4x + 3y + 3z = 0

5x + 7y + 4z =0

Answer

Here,

3x - y + 2z = 0 .....(1)

4x + 3y + 3z = 0 .....(2)

5x + 7y + 4z =0 .....(3)

The given system of homogeneous equaions can be written in matrix form as follows:

$\begin{bmatrix}3&-1&2\\4&3&3\\5&7&4\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} $

$\text{AX}=\text{O}$

Here,

$\text{A}=\begin{bmatrix}3&-1&2\\4&3&3\\5&7&4\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix} $

Now,

$|\text{A}|=\begin{vmatrix}3&-1&2\\4&3&3\\5&7&4\end{vmatrix}$

$=3(12-21)+1(16-15)+2(28-15)$

$=0$

$\therefore\ |\text{A}|\neq0$

So, the given system of homogeneous equations has non-trivial solutions.

Substituting z = k in eq. (1) & eq. (2), we get

3x - y = -2k and 4x + 3y = -3k

$\text{AX}=\text{B}$

Here,

$\text{A}=\begin{bmatrix}3&-1\\4&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}-2\text{k}\\-3\text{k}\end{bmatrix} $

$\Rightarrow\begin{bmatrix}3&-1\\4&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}-2\text{k}\\-3\text{k}\end{bmatrix} $

$|\text{A}|=\begin{vmatrix}3&-1\\4&3\end{vmatrix}$

$=(3\times3+4\times1)$

$=13$

So, A^-1 exists.

We have

$\text{adj }\text{A}=\begin{bmatrix}3&1\\-4&3\end{bmatrix}$

$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$

$\Rightarrow\text{A}^{-1}=\frac{1}{13}\begin{bmatrix}3&1\\-4&3\end{bmatrix}$

$\text{X}=\text{A}^{-1}\text{B}$

$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{13}\begin{bmatrix}3&1\\-4&3\end{bmatrix}\begin{bmatrix}-2\text{k}\\-3\text{k}\end{bmatrix}$

$=\frac{1}{13}\begin{bmatrix}-6\text{k}-3\text{k}\\8\text{k}-9\text{k}\end{bmatrix} $

Thus, $\text{x}=\frac{-9\text{k}}{13}, \text{y}=\frac{-\text{k}}{13}\text{ and }\text{z}=\text{k}$ (where k is any real number) satisfy the given system of equations.

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Question 634 Marks

Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹ 1,600. School B wants to spend ₹ 2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ₹ 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.

Answer

Let the award money given for sincerity, truthfulness and helpfulness be ₹ x, ₹ y and ₹ z respectively.

Since, the total cash award is ₹ 900.

$\therefore$ x + y + z = 900 .....(1)

Award money given by school A is ₹ 1,600.

$\therefore$ 3x + 2y + z = 1600 .....(2)

Award money given by school B is ₹ 2,300.

$\therefore$ 4x + y + 3z = 2300 .....(3)

The above system of equations can be written in matrix form CX = D as:

$\begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}900\\1600\\2300\end{bmatrix}$

Where, $\text{C}=\begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{D}=\begin{bmatrix}900\\1600\\2300\end{bmatrix}$

Now,

$|\text{C}|=\begin{vmatrix}1&1&1\\3&2&1\\4&1&3\end{vmatrix}$

$=1(6-1)-1(9-4)+1(3-8)$

$=5-5-5$

$=-5$

Let C_ij be the co-factors of elements c_ij in C = [c_ij]. Then,

$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}2&1\\1&3\end{vmatrix}=5,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}3&1\\4&3\end{vmatrix}=-5,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}3&2\\4&1\end{vmatrix}=-5$

$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&1\\1&3\end{vmatrix}=-2,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&1\\4&3\end{vmatrix}=-1,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&1\\1&1\end{vmatrix}=3$

$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&1\\2&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&1\\3&1\end{vmatrix}=2,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&1\\3&2\end{vmatrix}=-1$

$\text{adj }\text{C}=\begin{bmatrix}5&-5&-5\\-2&-1&3\\-1&2&-1\end{bmatrix}^\text{T}$

$=\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}$

$\Rightarrow\text{C}^{-1}=\frac{1}{|\text{C}|}\text{adj }\text{C}$

$=\frac{1}{-5}\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}$

$\text{X}=\text{C}^{-1}\text{D}$

$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}4500-3200-2300\\-4500-4600+4600\\-4500+4800-2300\end{bmatrix}$

$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}-1000\\-1500\\-2000\end{bmatrix}$

$\Rightarrow\text{x}=\frac{-1000}{-5},\text{y}=\frac{-1500}{-5}\ \text{and }\text{z}=\frac{-2000}{-5}$

$\therefore$ x = 200, y = 300 and z = 400.

Hence, the award money for each value of sincerity, truthfulness and helpfulness is ₹ 200, ₹ 300 and ₹ 400.

One more value which should be considered for award hardwork.

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Question 644 Marks

Solve the following system of equations by matrix method:
3x + 7y = 4
x + 2y = -1

Answer

The above system can be written in matrix form as:

$\begin{bmatrix}3&7\\ 1&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}4\\ -1\end{bmatrix}$

Or AX = B

Where $\text{A}=\begin{bmatrix}3&7\\ 1&2\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}4\\ -1\end{bmatrix}$

Now,

$\text{|A|}=-1\neq0$

So, the above system has a unique solution, given by

X = A^-1 B

Now, let C_ij be the co-factors of a_ijin A

$\text{C}_{11} = 2,\text{C}_{12} = -1$

$\text{C}_{21} = -7,\text{C}_{22} = 3$

$\text{Adj A}=\begin{bmatrix}2&-1\\ -7&3\end{bmatrix}^\text{T}=\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}$

$\therefore\text{A}^{-1}=\frac{1}{\text{|A|}}.\text{adj A}=\frac{1}{(-1)}\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}$

Now, X = A^-1 B

$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}\begin{bmatrix}4\\ -1\end{bmatrix}$

$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}15\\ -7\end{bmatrix}=\begin{bmatrix}-15\\ 7\end{bmatrix}$

Hence, x = -15

y = 7

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Maths 4 Marks