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Questions:

Generic 5

5 x 5 = 25

An automobile company uses three types of Steel S₁, S₂ and S₃ for providing three different types of Cars C₁, C₂and C₃. Steel requirement R (in tonnes) for each type of car and total available steel of all the three types are summarized in the following table.

Types of Steel	Types of Car			Total Steel available
Types of Steel	C₁	C₂	C₃	Total Steel available
S₁	3	2	4	28
S₂	1	1	2	13
S₃	2	2	2	14

Determine the number of Cars of each type which can be produced by Cramer’s rule.

An amount of Rs. 5,000/- is to be deposited in three different bonds bearing 6%, 7% and 8% per year respectively. Total annual income is Rs. 358/-. If the income from first two investments is Rs. 70/- more than the income from the third, then find the amount of investment in each bond by rank method.

For what values of the parameter λ, will the following equations fail to have unique solution: 3x − y+λz = 1, 2x + y + z = 2, x + 2y − λz = −1 by rank method.

Investigate for what values of ‘a’ and ‘b’ the following system of equations x + y + z = 6,x + 2y + 3z = 10, x + 2y + az = b have
(i) no solution
(ii) a unique solution
(iii) an infinite number of solutions.

Find k, if the equations x + y + z = 7, x + 2y + 3z = 18, y + kz = 6 are inconsistent

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Answers Key:

Generic 5

5 x 5 = 25

Let ‘x’ be the number of cars of type C₁
Let ‘y’ be the number of cars of type C₂
Let ‘z’ be the number of cars of type C₃
3x + 2y + 4z = 28
x + y + 2z =13
2x + 2y + z =14
Here \({ \triangle }=\left| \begin{matrix} 3 & 2 & 4 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right| =-3\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 28 & 2 & 4 \\ 13 & 1 & 2 \\ 14 & 2 & 1 \end{matrix} \right| =-6\)
\({ \triangle }_{ y }=\left| \begin{matrix} 3 & 28 & 4 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right| =-9\)
\({ \triangle }_{ z }=\left| \begin{matrix} 3 & 2 & 28 \\ 1 & 1 & 13 \\ 2 & 2 & 14 \end{matrix} \right| =-12\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -6 }{ -3 } =2\)
\(y=\frac { { \triangle }y }{ { \triangle } } =\frac { -9 }{ -3 } =3\)
\(z=\frac { { \triangle }z }{ { \triangle } } =\frac { -12 }{ -3 } =4\)
\(\therefore \) The number of cars of each type which can be produced are 2, 3 and 4.

Let the amount of investment in each bond be
Rs. x, Rs. y, Rs. z respectively.
Given x + y + z = 5000 ..(1)
Also \(\cfrac { 6x }{ 100 } +\cfrac { 7y }{ 100 } +\cfrac { 8z }{ 100 } =358\)
∴ Interes \(= \cfrac { PNR }{ 100 } =\cfrac { x\times 1\times 6 }{ 100 } =\cfrac { 6x }{ 100 } \)
\(\Rightarrow \cfrac { 6x+7y+8z }{ 100 } =358\)
\(\Rightarrow 6x+7y++8z=35800\)
Given that \(\cfrac { 6x }{ 100 } +\cfrac { 7y }{ 100 } =70+\cfrac { 8z }{ 100 } \)
\(\Rightarrow 6x+7y=7008z\)
\(\Rightarrow 6x+7y-8z=7000\)
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right) \)

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix}\begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & -14 \end{matrix}\begin{matrix} 5000 \\ 5800 \\ -2300 \end{matrix} \right) \)	\(\ { R }_{ 2 }\rightarrow { R }_{ 2 }-6{ R }_{ 1 }\) \({ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ 6R }_{ 1 } }\)
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & -16 \end{matrix}\begin{matrix} 5000 \\ 5800 \\ -28800 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

The last equivalent matrix is in echelon form and ρ(A) = ρ([A,B]) = 3 = Number of unknowns
Thus, the given system is consistent with unique solution. To find the solution, let us rewrite the above echelon form into the matrix form
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & -16 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 5000 \\ 5800 \\ -28800 \end{matrix} \right) \)
\(\Rightarrow x+y+z=5000\) ...(1)
\(\Rightarrow y+2z=5800\) ..(2)
\(\Rightarrow -16z=-28800\) ...(3)
\((3)\Rightarrow -16z=-28800\)
\(\Rightarrow z=\cfrac { 28800 }{ -16 } =1800\)
Substituting z = 1800 in (2) we get,
y + 2(1800) = 5800
\(\Rightarrow\) y + 3600 = 5800
\(\Rightarrow\) y=5800-3600
\(\Rightarrow\) y = 2200
Substituting y = 2200 and z = 1800 in (1) we get
\(\Rightarrow\) x + 2200 + 1800 = 5000
\(\Rightarrow\)x + 4000 = 5000
\(\Rightarrow\)x = 5000 - 4000
\(\Rightarrow\)x = 1000
Hence, the amount of investment in each bond is Rs. 1000, Rs. 2200 and Rs. 1800 respectively

Given non-homogeneous equations are
\(3x-y+\lambda z=1\)
\(2x+y+z=2\)
\(x+2y-\lambda z=-1\)
The matrix equation corresponding to the given system is

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 3 & -1 & \lambda \\ 2 & 1 & 1 \\ 1 & 2 & - \end{matrix}\begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 2 & -\lambda \\ 2 & 1 & 1 \\ 3 & -1 & \lambda \end{matrix}\begin{matrix} -1 \\ 2 \\ 1 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -3 & 1+2\lambda \\ 3 & -1 & \lambda \end{matrix}\begin{matrix} -1 \\ 4 \\ 1 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\)
\(\left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -3 & 1+2\lambda \\ 0 & -7 & 4\lambda \end{matrix}\begin{matrix} -1 \\ 4 \\ 4 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -1 & \frac { 1+2\lambda }{ 3 } \\ 0 & - & \frac { 4\lambda }{ 7 } \end{matrix}\begin{matrix} -1 \\ \frac { 4 }{ 3 } \\ \frac { 4 }{ 7 } \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }\div 3\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }\div 7\)
\(\left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -1 & \frac { 1+2 }{ 3 } \\ 0 & 0 & \frac { -7-2\lambda }{ 21 } \end{matrix}\begin{matrix} -1 \\ \frac { 4 }{ 3 } \\ \frac { -16 }{ 21 } \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

Since
\(\cfrac { 4\lambda }{ 7 } -\cfrac { 1+2\lambda }{ 3 } \)
= \(\cfrac { 12\lambda -7-14\lambda }{ 21 } =\cfrac { -7-2\lambda }{ 21 } \)
and \(\cfrac { 4 }{ 7 } -\cfrac { 4 }{ 3 } =\cfrac { 12-28 }{ 21 } \)
= \(\cfrac { -16 }{ 21 } \)
\(\therefore\) Since the system is fail to have unique solution either it can have infinitely many solution or it may be inconsistent.
This can happen only when \(\cfrac { -7-2\lambda }{ 21 } =0\)
\(\Rightarrow -7-2\lambda =0\)
\(\Rightarrow -7=2\lambda \)
\(\Rightarrow \lambda =\cfrac { -7 }{ 2 } \)

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 6 \\ 10 \\ b \end{matrix} \right) \)
AX = B

Augmented matrix [A,B]

Elementary Transformation

\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{matrix}\begin{matrix} 6 \\ 10 \\ b \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & a-1 \end{matrix}\begin{matrix} 6 \\ 4 \\ b-6 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & a-3 \end{matrix}\begin{matrix} 6 \\ 4 \\ b-10 \end{matrix} \right) \)

\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\)

\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\)

Case (i) For no solution:
The system possesses no solution only when \(\rho (A)\neq ([A,B])\) which is possible only when a−3 = 0 and b −10 \(\neq \) 0.
Hence for a = 3, b \(\neq \) 10, the system possesses no solution.
Case (ii) For a unique solution:
The system possesses a unique solution only when \(\rho (A)= ([A,B])\)=number of unknowns.
i.e when \(\rho (A)=\rho ([A,B])\) = 3
Which is possible only when a−3 \(\neq \) 0 and b may be any real number as we can observe .
Hence for a \(\neq \) and b \(\in \) R, the system possesses a unique solution.
Case (iii) For an infinite number of solutions:
The system possesses an infinite number of solutions only when
\(\rho (A)=\rho ([A,B])\) i,e when \(\rho (A)=\rho ([A,B])\) = 2 < 3 ( number of unknowns) which is possible only when a − 3 = 0, b−10 = 0
Hence for a = 3, b = 10, the system possesses infinite number of solutions.

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & k \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 7 \\ 18 \\ 6 \end{matrix} \right) \)
AX = B

Augmented matrix [A,B]

Elementary Transformation

\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & k \end{matrix}\begin{matrix} 7 \\ 18 \\ 6 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & k \end{matrix}\begin{matrix} 7 \\ 11 \\ 6 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & k-2 \end{matrix}\begin{matrix} 7 \\ 11 \\ -5 \end{matrix} \right) \)

ρ(A) = 2 or 3, ρ([A]) = 3

\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\)

\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

For the equations to be inconsistent
\(\rho ([A,B])\neq \rho (A)\)
It is possible if k − 2 = 0.
\(\therefore \) k = 2