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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.