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## 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.

Solution:

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

5 & 6 \\

7 & 8

\end{array}\right)\)

Order of A is 2 × 2.

ρ(A) ≤ 2

Consider the second order minor

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

5 & 6 \\

7 & 8

\end{array}\right|\) = 40 – 42 = -2 ≠ 0

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

ρ(A) = 2

(ii) 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 = -3 ≠ 0

ρ(A) = 2

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

1 & 4 \\

2 & 8

\end{array}\right)\)

Since A is of order 2 × 2, ρ(A) ≤ 2

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

1 & 4 \\

2 & 8

\end{array}\right|\) = 8 – 8 = 0

Since second order minor vanishes ρ(A) ≠ 2

But first order minors, |1|, |4|, |2|, |8| are non zero.

ρ(A) = 1

(iv) 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) ≤ 3

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) + 8 + 2

= 22 ≠ 0

There is a minor of order 3, which is non zero.

ρ(A) = 3

(v) Let A = \(\left(\begin{array}{ccc}

-1 & 2 & -2 \\

4 & -3 & 4 \\

-2 & 4 & -4

\end{array}\right)\)

Since order of A is 3 × 3, ρ(A) ≤ 3

Now,

\(\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)

= 4 + 16 – 20

= 0

Since the third order minor vanishes, ρ(A) ≠ 3

Consider \(\left|\begin{array}{cc}

-1 & 2 \\

4 & -3

\end{array}\right|\) = 3 – 8 = -5 ≠ 0

There is a minor order 2, which is not zero

ρ(A) = 2

(vi) Let A = \(\left(\begin{array}{cccc}

1 & 2 & -1 & 3 \\

2 & 4 & 1 & -2 \\

3 & 6 & 3 & -7

\end{array}\right)\)

Let us transform the matrix A to an echelon form by using elementary transformations.

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)\), then find the rank of AB and the rank of BA.

Solution:

Given 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)\)

Question 3.

Solve the following system of equations by rank method.

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

Solution:

The given equations are x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0

The matrix equation corresponding to the given system is

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:

The given equations are,

5x + 3y + 7 = 4

3x + 26y + 2z = 9

7x + 2y + 10z = 5

The matrix equation corresponding to the given system is

Question 5.

Show that the following system of equations have unique solution:

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

Solution:

The given system of equations can be written in matrix equation,

The last matrix is in echelon form. It has 3 non – zero rows,

ρ(A) = ρ([A, B]) = 3 = number of unknowns.

The given system is consistent and has a unique solution.

To find the solution, we write the echelon form into matrix form.

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

1 & 1 & 1 \\

0 & 1 & 2 \\

0 & 0 & 2

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

x \\

y \\

z

\end{array}\right)=\left(\begin{array}{l}

3 \\

1 \\

0

\end{array}\right)\)

x + y + z = 3 …… (1)

y + 2z = 1 …… (2)

2z = 0 …… (3)

(3) ⇒ z = 0

(2) ⇒ y = 1

(1) ⇒ x = 2

So the unique solution is x = 2, y = 1, z = 0

Question 6.

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

Solution:

The given system can be written in matrix equation form as given below:

λ = \(\frac{-7}{2}\)

So when λ = \(\frac{-7}{2}\), the equations fail to have unique solution.

Note: The system cannot have an infinite number of solutions, since ρ([A, B]) = 3 = a number of unknowns.

Question 7.

The price of three commodities, X,Y and Z are and z respectively Mr. Anand purchases 6 units of Z and sells 2 units of X and 3 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 ₹ 5,000/-, ₹2,000/- and ₹ 5,500/- respectively. Find the prices per unit of three commodities by rank method.

Solution:

The price of three commodities X, Y, Z are given as x, y, z.

We form the following system of equations from the given conditions.

Anand → 2x + 3y – 6z = 5000

Amar → 3x – y + 2z = 2000

Amit → -x + 3y + z = 5500

The matrix equation is given by

The prices per unit of the three commodities are Rs.1000, Rs.2000 and Rs.500

Question 8.

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

Solution:

Let the amount of investment in the three different bonds be Rs. x, Rs. y and Rs. z respectively.

We get the following equations according to the given conditions,

x + y + z = 5000

The above equivalent matrix is in echelon form with 3 non-zero rows.

So ρ(A) = ρ([A, B]) = 3 = number of unknowns.

the system has a unique solution.

The matrix equation is given by

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

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)

(3) ⇒ z = 1800

(2) ⇒ y = 5800 – 2(1800) = 2200

(1) ⇒ x = 5000 – 2200 – 1800 = 1000

The amount invested in the three bonds are ₹ 1000, ₹ 2200 and ₹ 1800.