Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1

Question 1.

Find the rank of each of the following matrices

(i) \(\left(\begin{array}{ll}

5 & 6 \\

7 & 8

\end{array}\right)\)

Solution:

Let A = \(\left|\begin{array}{ll}

5 & 6 \\

7 & 8

\end{array}\right|\)

Order of A is 2 × 2 [∴ P(A) ≤ 2]

Consider the second order minor = \(\left|\begin{array}{ll}

5 & 6 \\

7 & 8

\end{array}\right|\)

= 40 – 42

|A| = -2 ≠ 0

There is a minor of order 2, which is not zero

ρ(A) = 2

(ii) \(\left(\begin{array}{ll}

1 & -1 \\

3 & -6

\end{array}\right)\)

Solution:

Let A = \(\left(\begin{array}{ll}

1 & -1 \\

3 & -6

\end{array}\right)\)

Order of A is 2 × 2 [∴ ρ(A) ≤ 2]

Consider the second order minor = \(\left|\begin{array}{ll}

1 & -1 \\

3 & -6

\end{array}\right|\)

= -6 – (-3)

= -6 + 3 = -3

|A| ≠ 0

There is a minor of order 2, which is not zero

∴ ρ(A) = 2

(iii) \(\left(\begin{array}{ll}

1 & 4 \\

2 & 8

\end{array}\right)\)

Solution:

Let A = \(\left(\begin{array}{ll}

1 & 4 \\

2 & 8

\end{array}\right)\)

Order of A is 2 × 2 [∴ ρ(A) ≤ 2]

Consider the second order minor = \(\left|\begin{array}{ll}

1 & 4 \\

2 & 8

\end{array}\right|\)

= 8 – 8 = 0

Since the second are minor vanishes, ρ(A) ≠ 2

Consider a first order minor |1| ≠ 0

There is a minor of order 1, which is not zero

∴ ρ(A) = 1

(iv) \(\left[\begin{array}{ccc}

2 & -1 & 1 \\

3 & 1 & -5 \\

1 & 1 & 1

\end{array}\right]\)

Solution:

Let A = \(\left[\begin{array}{ccc}

2 & -1 & 1 \\

3 & 1 & -5 \\

1 & 1 & 1

\end{array}\right]\)

Order of A is 3 × 3

∴ ρ(A) ≤ 2

Consider the third order minor \(\left|\begin{array}{ccc}

2 & -1 & 1 \\

3 & 1 & -5 \\

1 & 1 & 1

\end{array}\right|\)

= 2(1 + 5) – (-1) (3 + 5) + 1 (3 – 1)

= 2 (6) + 1(8) + 1(2)

= 12 + 8 + 2

= 22 ≠ 0

There is a minor of order 3, which is not zero

∴ ρ(A) = 3

(v) \(\left[\begin{array}{ccc}

-1 & 2 & -2 \\

4 & -3 & 4 \\

-2 & 4 & -4

\end{array}\right]\)

Solution:

Let A = \(\left[\begin{array}{ccc}

-1 & 2 & -2 \\

4 & -3 & 4 \\

-2 & 4 & -4

\end{array}\right]\)

Order of A is 3 × 3

∴ ρ(A) ≤ 3

Consider the third order minor \(\left|\begin{array}{ccc}

-1 & 2 & -2 \\

4 & -3 & 4 \\

-2 & 4 & -4

\end{array}\right|\)

= -1(12 – 16) – 2 (-16 + 8) – 2 (16 – 6)

= -1(-4) – 2 (-8) – 2 (10)

= 4 + 16 – 20

= 0

Since the third order minor vanishes, therefore

∴ ρ(A) ≠ 3

Consider a second order minor \(\left|\begin{array}{ll}

-1 & 2 \\

4 & -3

\end{array}\right|\)

= 3 – 8

= -5 ≠ 0

There is a minor of order 2, which is not zero.

∴ ρ(A) = 2

The order of A is 3 × 4

∴ ρ(A) ≤ 3

The number of non zero rows is 2

∴ ρ(A) = 2

The order of A is 3 × 4

∴ ρ(A) ≤ 3

Consider the third order minor \(\left|\begin{array}{ccc}

3 & 1 & -5 \\

1 & -2 & 1 \\

1 & 5 & -7

\end{array}\right|\)

= 3 (14 – 5) – 1 (- 7 – 1) – 5 (5 + 2)

= 3 (9) – 1 (-8) – 5 (7)

= 27 + 8 – 35

= 0

\(\left|\begin{array}{ccc}

1 & -5 & 1 \\

-2 & 1 & -5 \\

5 & -7 & 2

\end{array}\right|\)

= 1 (2 – 35) + 5 (-4 + 25) – 1 (14 – 5)

= 1 (-33) + 5(21) – 1 (9)

= -33 + 105 – 9

= 63 ≠ 0

There is a minor of order 3, which is not zero.

∴ ρ(A) = 3

Order of A is 3 × 4

∴ ρ(A) ≤ 3

Consider the third order minor \(\left|\begin{array}{ccc}

1 & -2 & 3 \\

-2 & 4 & -1 \\

-1 & 2 & 7

\end{array}\right|\)

= 1 (28 + 2)+ 2 (-14 – 1) + 3 (-4 + 4)

= 1 (30) + 2 (- 15) + 3 (0)

= 0

\(\left|\begin{array}{ccc}

-2 & 3 & 4 \\

4 & -1 & -3 \\

2 & 7 & 6

\end{array}\right|\)

= -2 (-6 + 21) – 3 (24 + 6) + 4 (28 + 2)

= -2(15) – 3 (30)+ 4 (30)

= 0

\(\left|\begin{array}{ccc}

1 & 3 & 4 \\

-2 & -1 & -3 \\

-1 & 7 & 6

\end{array}\right|\)

= 1 (-6 + 21) – 3 (-12 – 3) + 4 (-14 – 1)

= 1 (15) -3 (-15) + 4 (-15)

= 15 + 45 – 60

= 0

\(\left|\begin{array}{ccc}

1 & -2 & 4 \\

-2 & 4 & -3 \\

-1 & 2 & 6

\end{array}\right|\)

= 1 (24 + 6) + 2 (-12 – 3) + 4 (-4 + 4)

= 1 (30) + 2 (-15) + 4(0)

= 30 – 30

= 0

Since all third order minors vanishes, ρ(A) ≠ 3

Now, let us consider the second order minors.

Consider one of the second order minors

\(\left|\begin{array}{ll}

-2 & 3 \\

4 & -1

\end{array}\right|\) = 2 – 12 = -10 ≠ 0

There is a minor of order 2 which is not zero

∴ ρ(A) = 2

Question 2.

If A = \(\left(\begin{array}{ccc}

1 & 1 & -1 \\

2 & -3 & 4 \\

3 & -2 & 3

\end{array}\right)\) and B = \(\left(\begin{array}{ccc}

1 & -2 & 3 \\

-2 & 4 & -6 \\

5 & 1 & -1

\end{array}\right)\)

find the rank of A B and the rank of B A.

Solution:

To find the rank of AB

Order of AB is 3 × 3

∴ ρ(AB) ≤ 3

Consider the third order minor |AB| = \(\left|\begin{array}{ccc}

-6 & 3 & -2 \\

28 & -12 & 20 \\

22 & -11 & 18

\end{array}\right|\)

= -6(-216 + 220) -3(504 – 440) – 2(-308 + 264)

= – 6(4) – 3(64) – 2(-44)

= -24 – 192 + 88

= 128 ≠ 0

There is a minor of order 3, which is not zero.

∴ ρ(AB) = 3

To find the rank of BA

Order of BA is 3 × 3

∴ ρ(BA) ≤ 3

Consider the third order minor |BA| = \(\left|\begin{array}{ccc}

6 & 1 & 0 \\

-12 & -2 & 0 \\

4 & 4 & -4

\end{array}\right|\)

= 6(8 – 0) – 1(48 – 0) + 0(-48 + 8)

= 6(8) – 1(48) + 0(-40)

= 48 – 48 + 0

= 0

Since the third order minor vanishes, therefore P(BA) ≠ 3

Consider a second order minor \(\left|\begin{array}{ll}

-12 & -2 \\

4 & 4

\end{array}\right|\) = -48 + 8 = -40 ≠ 0

There is a minor of order 2 which is not zero

∴ ρ(BA) = 2

Question 3.

Solve the following system of equations by rank method

x + y + z = 9, 2x + 5y + 7z = 52, 2x – y – z = 0

Solution:

x + y + z = 9

2x + 5y + 7z = 52

2x – y – z = 0

The matrix equation corresponding to the given system is

Obviously the last equivalent matrix is in the echelon form. It has three non-zero rows.

P(A) = P (A, B) = 3 = Number of unknowns

The given system is consistent and has unique solution.

To find the solution, let us rewrite the above ech-elon form into the matrix form.

\(\left[\begin{array}{lll}

1 & 1 & 1 \\

0 & 3 & 5 \\

0 & 0 & 2

\end{array}\right]\) \(\left[\begin{array}{l}

x \\

y \\

z

\end{array}\right] =\left[\begin{array}{c}

9 \\

34 \\

16

\end{array}\right]\)

x + y + z = 9 ……..(1)

3y + 5z = 34 ……… (2)

2z = 16 ……… (3)

z = \(\frac { 16 }{ 2 }\) = 8

z = 8

Substitute z = 8 in eqn (2)

3y + 5 (8) = 34

3y + 40 = 34

3y = 34 – 40

3y = -6

y = -2

Substitute y = -2 and z = 8 in eqn (1)

x = (-2) + 8 = 9

x + 6 = 9

x = 9 – 6

x = 3

∴ x = 3, y = -2, z = 8

Question 4.

Show that the equations 5x + 3y + 7z = 4, 3x + 26y + 2z = 9, 7x + 2y + 10z = 5 are consistent and solve them by rank method.

Solution:

5x + 3y + 7z = 4

3x + 26y + 2z = 9

7x + 2y + 10z = 5

The matrix equation corresponding to the given systematic

Obviously, the last equivalent matrix is in the echelon form. It has two non-zero rows.

P([A, B]) = 2 ; P(A) = 2

P(A) = P ([A, B]) = 2 < Number of unknowns

The given system is consistent and has infinitely many solutions.

The given system is equivalent to the matrix equation

\(\left[\begin{array}{lll}

1 & 5 & 1 \\

0 & -11 & 1 \\

0 & 0 & 0

\end{array}\right]\) \(\left[\begin{array}{l}

x \\

y \\

z

\end{array}\right] =\left[\begin{array}{c}

2 \\

-3 \\

0

\end{array}\right]\)

x + 5y + z = 2 ……. (1)

-11y + z = -3 ……… (2)

let z = k

Where K ε R

Question 5.

Show that the following system of equations have unique solutions:

x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6 by rank method.

Solution:

x + y + z = 3

x + 2y + 3z = 4

x + 4y + 9z = 6

The matrix equation corresponding to the given system is

The last equivalent matrix is in the echelon form [A, B] has 3 non-zero rows and [A] has 3 non-zero rows.

p([A,B]) = 3; ρ(A) = 3

ρ([A, B]) = ρ(A) = No. of unknowns

∴ The system of equations have unique solution.

Question 6.

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

Solution:

3x – y + λz = 1

2x + y + z = 2

x + 2y – λz = -1

The matrix equation corresponding to the given system is

If the equations fail to have unique solution.

ρ(A) ≠ ρ(A, B)

ρ(A, B) = 3

ρ(A) ≠ 3

Therefore 7 + 2λ = 0

2λ = -7 and λ = -7/2

Question 7.

The price of three commodities. X, Y, and Z are x,y, and z respectively Mr. Anand purchases 6 units of Z and sells 2 units of Y. Mr. Amar purchases a unit of Y and sells 3 units of X and 2 units of Z. Mr. Amit purchases a unit of X and sells 3 units of Y and a unit of Z. In the process they earn Rs 5,000/-, Rs 2,000/- and Rs 5,500/- respectively. Find the prices per unit of three commodities by the rank method.

Solution:

Let the equations for Mr. Anand, Mr. Amar, and Mr. Amit are

2x + 3y – 6z = 5000

3x – y + 2z = 2000

-x + 3y + z = 5500 respectively

The matrix equation corresponding to the given system is

∴ The given system is equivalent to the matrix equation

\(\left[\begin{array}{lll}

-1 & 4 & -8 \\

0 & 1 & -2 \\

0 & 0 & 7

\end{array}\right]\) \(\left[\begin{array}{l}

x \\

y \\

z

\end{array}\right] =\left[\begin{array}{c}

3000 \\

1000 \\

3500

\end{array}\right]\)

-x + 4y – 8z = 3000 …….. (1)

y – 2z = 1000 ………. (2)

7z = 3500 ……….. (3)

Eqn (3) ⇒ z = \(\frac { 3500 }{ 7 }\)

∴ z = 500

Eqn (2) ⇒ y – 2(500) = 1000

y = 1000 + 1000

∴ y = 2000

Eqn (1) ⇒ -x + 4 (2000) – 8 (500) = 3000

-x + 8000 – 4000 = 3000

-x + 4000 = 3000

-x = 3000 – 4000

-x = – 1000

∴ x = 1000

The price of three commodities, x, y and z are Rs. 1000; Rs. 2000 and Rs. 500 respectively.

Question 8.

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 the first two investments is Rs 70/- more than the income from the third, then find the amount of investment in each bond by the rank method.

Solution:

Let the three different bonds be x, y, and z

x + y + z = 5000 …….. (1)

6x + 7y = 8z + 7000

6x + 7y – 8z = 7000 ……… (3)

The matrix equation corresponding to the given system is

∴ The given system is equivalent to the matrix equation

\(\left[\begin{array}{lll}

1 & 1 & 1 \\

0 & 1 & 2 \\

0 & 0 & -16

\end{array}\right]\) \(\left[\begin{array}{l}

x \\

y \\

z

\end{array}\right] =\left[\begin{array}{c}

5000 \\

5800 \\

-28800

\end{array}\right]\)

x + y + z = 5000 ……… (1)

y + 2z = 5800 …….. (2)

-16z = -28800 ……… (3)

eqn (3) ⇒ z = \(\frac { -28800 }{ -16 }\)

∴ z = 1800

eqn (2) ⇒ y + 2(1800) = 5800

y + 3600 = 5800

y = 5800 – 3600

∴ y = 2200

eqn (1) ⇒ x + 2200 + 1800 = 5000

x + 4000 = 5000

x = 5000 – 4000

∴ x = 1000

The amount of investment in each bond is Rs 1000, Rs 2200 and Rs 1800.