Applications of Matrices and Determinants

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Questions:

Generic 5

6 x 5 = 30

Solve the following equation by using Cramer’s rule
x + 4y + 3z = 2, 2x−6y + 6z = −3, 5x− 2y + 3z = −5

Solve the following equation by using Cramer’s rule
x + y + z = 6, 2x + 3y− z =5, 6x−2y− 3z = −7

Solve the following equation by using Cramer’s rule
2x + y −z = 3, x + y + z =1, x− 2y− 3z = 4

80% of students who do maths work during one study period, will do the maths work at the next study period. 30% of students who do english work during one study period, will do the english work at the next study period. Initially there were 60 students do maths work and 40 students do english work.
Calculate,
(i) The transition probability matrix
(ii) The number of students who do maths work, english work for the next subsequent 2 study periods.

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.

Solve by Cramer’s rule x + y + z = 4, 2x − y + 3z = 1, 3x + 2y − z = 1

Generic 3

2 x 3 = 6

Akash bats according to the following traits. If he makes a hit (S), there is a 25% chance that he will make a hit his next time at bat. If he fails to hit (F), there is a 35% chance that he will make a hit his next time at bat. Find the transition probability matrix for the data and determine Akash’s long- range batting average.

Parithi is either sad (S) or happy (H) each day. If he is happy in one day, he is sad on the next day by four times out of five. If he is sad on one day, he is happy on the next day by two times out of three. Over a long run, what are the chances that Parithi is happy on any given day?

Multiple Choice Question

5 x 1 = 5

Rank of a null matrix is _______.

(a)

(b)

-1

(c)

\(\infty \)

(d)

10.

\(\left| { A }_{ n\times n } \right| \) = 3 \(\left| adjA \right| \) = 243 then the value n is _______.

(a)

(b)

(c)

(d)

11.

If \(\frac { { a }_{ 1 } }{ x } +\frac { { b }_{ 1 } }{ y } ={ c }_{ 1 },\frac { { a }_{ 2 } }{ x } +\frac { { b }_{ 2 } }{ y } ={ c }_{ 2 },\) \({ \triangle }_{ 1= }\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } \end{vmatrix}, \ { \triangle }_{ 2 }=\begin{vmatrix} { b }_{ 1 } & { c }_{ 1 } \\ { b }_{ 2 } & { c }_{ 2 } \end{vmatrix}{ \triangle }_{ 3 }=\begin{vmatrix} { c }_{ 1 } & { a }_{ 1 } \\ { c }_{ 2 } & a_{ 2 } \end{vmatrix}\) then (x, y) is _______.

(a)

\(\left( \frac { { \triangle }_{ 2 } }{ { \triangle }_{ 1 } } ,\frac { { \triangle }_{ 3 } }{ { \triangle }_{ 1 } } \right) \)

(b)

\(\left( \frac { { \triangle }_{ 3 } }{ { \triangle }_{ 1 } }, \frac { { \triangle }_{ 2 } }{ { \triangle }_{ 1 } } \right) \)

(c)

\(\left( \frac { { \triangle }_{ 1 } }{ { \triangle }_{ 2 } } ,\frac { { \triangle }_{ 1 } }{ { \triangle }_{ 3 } } \right) \)

(d)

\(\left( \frac { { -\triangle }_{ 1 } }{ { \triangle }_{ 2 } }, \frac { {- \triangle }_{ 1 } }{ { \triangle }_{ 3 } } \right) \)

12.

Cramer’s rule is applicable only to get an unique solution when _______.

(a)

\({ \triangle }_{ z }\neq 0\)

(b)

\({ \triangle }_{ x }\neq 0\)

(c)

\({ \triangle } \neq 0\)

(d)

\({ \triangle }_{ y }\neq 0\)

13.

The system of linear equations x + y + z = 2, 2x + y − z = 3, 3x + 2y + k = 4 has unique solution, if k is not equal to _______.

(a)

(b)

(c)

-4

(d)

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Answers Key:

Generic 5

6 x 5 = 30

\(\Delta =\left| \begin{matrix} 1 & 4 & 3 \\ 2 & -6 & 6 \\ 5 & -2 & 3 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -6 & 6 \\ -2 & 3 \end{matrix} \right| -4\left| \begin{matrix} 2 & 6 \\ 5 & 3 \end{matrix} \right| +3\left| \begin{matrix} 2 & -6 \\ 5 & -2 \end{matrix} \right| \)
= 1(-18 + 12) - 4(6 - 30) +3 (- 4 +30)
= 1(- 6) - 4(- 24) + 3(26)
= - 6 + 96 + 78 = 168 \(\neq \) 0
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied
\(\Delta x=\left| \begin{matrix} 2 & 4 & 3 \\ -3 & -6 & 6 \\ -5 & -2 & 3 \end{matrix} \right| \)
= \(2\left| \begin{matrix} -6 & 6 \\ -2 & 3 \end{matrix} \right| -4\left| \begin{matrix} -3 & 6 \\ -5 & 3 \end{matrix} \right| +3\left| \begin{matrix} -3 & -6 \\ -5 & -2 \end{matrix} \right| \)
= 2 (- 18 + 12) - 4(- 9 +30) + 3(6 -30)
= 2(- 6) - 4(21) + 3(- 24)
= -12-84-72 =-168
\(\Delta y=\left| \begin{matrix} 1 & 2 & 3 \\ 2 & - & 6 \\ 5 & -5 & 3 \end{matrix} \right| =1\left| \begin{matrix} -3 & 6 \\ -5 & 3 \end{matrix} \right| -2\left| \begin{matrix} 2 & 6 \\ 5 & 3 \end{matrix} \right| +3\left| \begin{matrix} 2 & -3 \\ 5 & -5 \end{matrix} \right| \)
= 1 (-9+30)-2(6-30)+3(- 10+ 15)
= 1(21) - 2(- 24) + 3(5)
= 21 + 48 + 15 = 84
\(\Delta z=\left| \begin{matrix} 1 & 4 & 2 \\ 2 & -6 & -3 \\ 5 & -2 & -5 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -6 & -3 \\ -2 & -5 \end{matrix} \right| -4\left| \begin{matrix} 2 & -3 \\ 5 & -5 \end{matrix} \right| +2\left| \begin{matrix} 2 & -6 \\ 5 & -2 \end{matrix} \right| \)
= 1(30-6)-4(-10+ 15)+2(-4+30)
= 24 - 4(5) + 2(26)
= 24 - 20 + 52 = 56

Solution set is \(\left\{ -1,\frac { 1 }{ 2 } ,\frac { 1, }{ 3 } \right\} \)

\(\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & -1 \\ 6 & -2 & -3 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 3 & -1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 2 & -1 \\ 6 & 3 \end{matrix} \right| +1\left| \begin{matrix} 2 & 3 \\ 6 & -2 \end{matrix} \right| \)
= 1(-9 -2) -1(-6 +6) + 1(-4 -18)
= 1(-11) -1(0) +1(-22)
= -11 -22 = -33 \(\neq \)0
Since \(\Delta \neq 0\)
Cramer's rule can be applied and the system is consistent with unique solution
\(\Delta x=\left| \begin{matrix} 6 & 1 & 1 \\ 5 & 3 & -1 \\ -7 & -2 & -3 \end{matrix} \right| \)
= \(6\left| \begin{matrix} 3 & -1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 5 & -1 \\ -7 & -3 \end{matrix} \right| +1\left| \begin{matrix} 5 & 3 \\ -7 & -2 \end{matrix} \right| \)
= 6 (-9 -2) -1(-15 -7) + 1(-10 +21)
= 6 (-11) -1 (-22) + 1 (11)
= -66 + 22 + 11= - 33
\(\Delta y=\left| \begin{matrix} 1 & 6 & 1 \\ 2 & 5 & -1 \\ 6 & -7 & -3 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 5 & -1 \\ -7 & -3 \end{matrix} \right| -6\left| \begin{matrix} 2 & -1 \\ 6 & -3 \end{matrix} \right| +1\left| \begin{matrix} 2 & 5 \\ 6 & -7 \end{matrix} \right| \)
= 1(-15 -7) -6(-6 +6) + 1(-14 -30)
= 1(-22) -6(0) + 1 (-44)
= -22 - 44 = - 66
\(\Delta z=\left| \begin{matrix} 1 & 1 & 6 \\ 2 & 3 & 5 \\ 6 & -2 & -7 \end{matrix} \right| \)
= \(=1\left| \begin{matrix} 3 & 5 \\ -2 & -7 \end{matrix} \right| -1\left| \begin{matrix} 2 & 5 \\ 6 & -7 \end{matrix} \right| +6\left| \begin{matrix} 2 & 3 \\ 6 & -2 \end{matrix} \right| \)

= 1(-21 +10) -1(-14 -30) +6 (-4 -18)
=1(-11) -1(-44) +6(-22)
= -11 + 44 - 132 = - 99

\(\therefore\) Solution set is {1, 2, 3}

\(\Delta =\left| \begin{matrix} 2 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{matrix} \right| =2\)
\(\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 2 (-3+2) - 1 (-3 -1) -1 (-2-1)
= 2(-1) -1 (-4) -1 (-3)
= -2 + 4 + 3 = 5.
Since\(\Delta \neq 0\),
we can apply Cramer's rule and the system is consistent with unique solution.
\(x=\left| \begin{matrix} 3 & 1 & -1 \\ 1 & 1 & 1 \\ 4 & -2 & 3 \end{matrix} \right| =3\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 4 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 4 & -2 \end{matrix} \right| \)
= 3(-3 + 2) -1(-3 -4) -1(-2 -4)
= 3 (-1) -1 (-7) -1 (-6)
= -3 + 7 + 6 = 10.
\(\Delta y=\left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & 4 & -3 \end{matrix} \right| =2\left| \begin{matrix} 1 & 1 \\ 4 & -3 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right| \)
= 2(-3 -4) -3 (-3 -1) -1 (4-1)
= 2 (-7) -3 (-4) -1(3)
= 14 + 12 - 3 = -5
\(\Delta z=\left| \begin{matrix} 2 & 1 & 3 \\ 1 & 1 & 1 \\ 1 & -2 & 4 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 1 & 1 \\ -2 & 4 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right| +3\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 2(4 + 2) -1(4 -1) + 3(-2 -1)
= 2(6) -1(3) + 3(-3)
= 12 - 3 - 9
= 0

\(z=\cfrac { \Delta z }{ \Delta } =\cfrac { 0 }{ 5 } =0\)
\(\therefore\)Solution set is (2, -1, 0)

(i) Transition probability matrix T = \(_{ E }^{ M }\left( \begin{matrix} \overset { M }{ 0.8 } & \overset { E }{ 0.2 } \\ 0.7 & 0.3 \end{matrix} \right) \)
After one study period, \(\left( \overset { M }{ 60\quad } \overset { E }{ 40 } \right) \) \(_{ E }^{ M }\left( \begin{matrix} \overset { M }{ 0.8 } & \overset { E }{ 0.2 } \\ 0.7 & 0.3 \end{matrix} \right) \) =\(\left( \overset { M }{ 76\quad } \overset { E }{ 24 } \right) \)
So in the very next study period, there will be 76 students do maths work and 24 students do the English work.
After two study periods,
\(\left( \overset { M }{ 76\quad } \overset { E }{ 24 } \right) \) \(_{ E }^{ M }\left( \begin{matrix} \overset { M }{ 0.8 } & \overset { E }{ 0.2 } \\ 0.7 & 0.3 \end{matrix} \right) \)
= (60.8+16.8 15.2+7.2)
= (77.6 22.4)
After two study periods there will be 78 (approx) students do maths work and 22 (approx) students do English work.

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.

Here \(\triangle =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{matrix} \right| =13\neq 0\)
\(\therefore \) We can apply Cramer’s Rule and the system is consistent and it has unique solution.
\({ \triangle }_{ x }=\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{matrix} \right| =-13\)
\( { \triangle }_{ y }=\left| \begin{matrix} 1 & 4 & 1 \\ 2 & 1 & 3 \\ 3 & 1 & -1 \end{matrix} \right| =39\)
\( { \triangle }_{ z }=\left| \begin{matrix} 1 & 1 & 4 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{matrix} \right| =26\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -13 }{ 13 } =-1\)
\( y=\frac { { \triangle }y }{ { \triangle } } =\frac { 39 }{ 13 } =3\)
\( z=\frac { { \triangle }z }{ { \triangle } } =\frac { 26 }{ 13 } =2\)
\(\therefore \) The solution is (x, y, z) = (−1, 3, 2)

Generic 3

2 x 3 = 6

The Transition probability matrix is T = \(\left( \begin{matrix} 0.25 & 0.75 \\ 0.35 & 0.65 \end{matrix} \right) \)
At equilibrium, (S F) \(\left( \begin{matrix} 0.25 & 0.75 \\ 0.35 & 0.65 \end{matrix} \right) \) = (S F)
where S + F = 1
0.25 S + 0.35 F = S
0.25 S + 0.35 (1 – S) = S
On solving this, we get S = \(\frac { 0.35 }{ 1.10 } \)
⇒ S = 0.318 and F = 0.682
\(\therefore \) Akash’s batting average is 31.8%

The transition porbability matrix is T = \(\left( \begin{matrix} \frac { 1 }{ 3 } & \frac { 2 }{ 3 } \\ \frac { 4 }{ 5 } & \frac { 1 }{ 5 } \end{matrix} \right) \)
At equilibrium, (S H) \(\left( \begin{matrix} \frac { 1 }{ 3 } & \frac { 2 }{ 3 } \\ \frac { 4 }{ 5 } & \frac { 1 }{ 5 } \end{matrix} \right) \) = (S H)
where S + H = 1
\(\frac { 4 }{ 5 } S+\frac { 1 }{ 3 } H=S\)
\(\frac { 1 }{ 3 } (1 - H)+\frac { 4 }{ 5 } H= 1- H\)
On solving this, we get S = \(\frac { 6 }{ 11 } \) and H = \(\frac { 5 }{ 11 } \)
In the long run, on a randomly selected day, his chances of being happy is \(\frac { 5 }{ 11 } \)