Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems

Question 1.
Find the rank of the matrix A = \(\left(\begin{array}{rrrr}
1 & -3 & 4 & 7 \\
9 & 1 & 2 & 0
\end{array}\right)\)
Solution:
A = \(\left(\begin{array}{rrrr}
1 & -3 & 4 & 7 \\
9 & 1 & 2 & 0
\end{array}\right)\)
Order of A is 2 × 4. So ρ(A) ≤ 2
Consider the second order minor
\(\left|\begin{array}{cc}
1 & -3 \\
9 & 1
\end{array}\right|\) = 1 + 27 = 28 ≠ 0
There is a minor of order 2 which is not zero.
So ρ(A) = 2

Question 2.
Find the rank of the matrix A = \(\left(\begin{array}{rrrr}
-2 & 1 & 3 & 4 \\
0 & 1 & 1 & 2 \\
1 & 3 & 4 & 7
\end{array}\right)\)
Solution:
Given A = \(\left(\begin{array}{rrrr}
-2 & 1 & 3 & 4 \\
0 & 1 & 1 & 2 \\
1 & 3 & 4 & 7
\end{array}\right)\)
Order of A is 3 × 4. So ρ(A) ≤ 3
Consider the third order minor
\(\left|\begin{array}{ccc}
-2 & 1 & 3 \\
0 & 1 & 1 \\
1 & 3 & 4
\end{array}\right|\)
= -2(4 – 3) – 1(0 – 1) + 3(0 – 1)
= -2 + 1 – 3 =
= -4 ≠ 0
There exists a minor of order 3 which is not zero. So ρ(A) = 3

Question 3.
Find the rank of the matrix A = \(\left(\begin{array}{llll}
4 & 5 & 2 & 2 \\
3 & 2 & 1 & 6 \\
4 & 4 & 8 & 0
\end{array}\right)\)
Solution:
Given A = \(\left(\begin{array}{llll}
4 & 5 & 2 & 2 \\
3 & 2 & 1 & 6 \\
4 & 4 & 8 & 0
\end{array}\right)\)
Order of A is 3 × 4. So ρ(A) ≤ 3
Consider the third order minor,
\(\left|\begin{array}{lll}
4 & 5 & 2 \\
3 & 2 & 1 \\
4 & 4 & 8
\end{array}\right|\)
= 4(16 – 4) – 5(24 – 4) + 2 (12 – 8)
= 48 – 100 + 8
= -44 ≠ 0
There is a minor of order 3 which is not zero.
ρ(A) = 3

Question 4.
Examine the consistency of the system of equations:
x + y + z = 7, x + 2y + 3z = 18, y + 2z = 6
Solution:
The given system is x + y + z = 7 , x + 2y + 3z = 18, y + 2z = 6.
The matrix equation corresponding to the given system
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 1
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 2
The last equivalent matrix is in the echelon form. [A, B] has 3 non-zero rows and [A] has 2 non- zero rows.
ρ([A, B]) = 3, ρ(A) = 2
ρ(A) ≠ ρ([A, B])
Hence the given system is inconsistent and has no solution.

Question 5.
Find k if the equations 2x + 3y – z = 5, 3x – y + 4z = 2, x + 7y – 6z = k are consistent.
Solution:
The given system is 2x + 3y – z = 5, 3x – y + 4z = 2, x + 7y – 6z = k
It is also given that the system is consistent.
The matrix equation corresponding to the given system is
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 3
ρ(A) = 2; ρ([A, B]) = 2 or 3
For the equations to be consistent, ρ([A, B]) = ρ(A) = 2
k – 8 = 0 ⇒ k = 8

Question 6.
Find k if the equations x + y + z = 1, 3x – y – z = 4, x + 5y + 5z = k are inconsistent.
Solution:
The given system is x + y + z = 1, 3x – y – z = 4, x + 5y + 5z = k and it is inconsistent.
The matrix equation corresponding to the given system is
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 4
ρ(A) = 2, since the equivalent matrix has 2 non zero rows.
For the equations to be inconsistent
ρ([A, B]) ≠ ρ(A)
ρ([A, B]) ≠ 2 ⇒ k ≠ 0
So k can assume any real value other than 0.

Question 7.
Solve the equations x + 2y + z = 7, 2x – y + 2z = 4, x + y – 2z = -1 by using Cramer’s rule.
Solution:
The given system is x + 2y + z = 7, 2x – y + 2z = 4, x + y – 2z = -1
Here ∆ = \(\left|\begin{array}{ccc}
1 & 2 & 1 \\
2 & -1 & 2 \\
1 & 1 & -2
\end{array}\right|\)
= 1(2 – 2) -2(4 – 2) + 1(2 + 1)
= 12 + 3
= 15 ≠ 0
We can apply Cramer’s ruleand the system is consistent and it has unique solution.
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 5
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 6

Question 8.
The cost of 2kg. of wheat and 1kg. of sugar is ₹ 100. The cost of 1kg. of wheat and 1kg. of rice is ₹ 80. The cost of 3kg. of wheat, 2kg. of sugar and 1kg of rice is ₹ 220. Find the cost of each per kg., using Cramer’s rule.
Solution:
Let the cost of wheat per kg be ₹ x, cost of sugar per kg be ₹ y and cost of rice per kg be ₹ z, respectively.
It is given that 2x + y = 100, x + z = 80, 3x + 2y + z = 220
Here ∆ = \(\left|\begin{array}{lll}
2 & 1 & 0 \\
1 & 0 & 1 \\
3 & 2 & 1
\end{array}\right|\)
= 2 (-2) – 1(-2 )
= -2 ≠ 0
we can apply Cramer’s rule and the system is consistent and its has unique solution.
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 7
Hence the cost of wheat is ₹ 30/kg, cost of sugar is ₹ 40/kg and the cost of rice is ₹ 50/kg.

Question 9.
A salesman has the following record of sales for three months for three items A,B and C, which have different rates of commission.
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 8
Find out the rate of commission on the items A,B and C by using Cramer’s rule.
Solution:
Let the rate of commission on items A, B and C be ₹ x, ₹ y and ₹ z per unit respectively.
According to the problem we have,
January, 90x + 100y + 20z = 800
February, 130x + 50y + 40z = 900
March, 60x + 100y + 30z = 850
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 9
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 10

Question 10.
The subscription department of a magazine sends out a letter to a large mailing list inviting subscriptions for the magazine. Some of the people receiving this letter already subscribe to the magazine while others do not. From this mailing list, 60% of those who already subscribe will subscribe again while 25% of those who do not now subscribe will subscribe. On the last letter it was found that 40% of those receiving it ordered a subscription. What percent of those receiving the current letter can be expected to order a subscription?
Solution:
Let X represent people who subscribe for the magazine and Y represent people who do not subscribe for the magazine. Then
(X X) ⇒ those who already subscribed will do it again.
(X Y) ⇒ those who already subscribed will not do it again.
(Y X) ⇒ those who have not subscribed will do it now.
(Y Y) ⇒ those who have not subscribed already will not do it now also.
From the question,
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 11
Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems 12
Thus 39% of those receiving the current letter can be expected to order a subscription.