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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 1.
Solve the following systems of linear equations by Gaussian elimination method:
(i) 2x – 2y + 3z = 2, x + 2y – z = 3, 3x – y + 2z = 1
(ii) 2x + 4y + 6z = 22, 3x + 8y + 5z = 27, -x + y + 2z = 2
Solution:
(i) 2x – 2y + 3z = 2, x + 2y – z = 3 and 3x – y + 2z = 1
The matrix form of the above equations is \(\left(\begin{array}{rrr}{2} & {-2} & {3} \\ {1} & {2} & {-1} \\ {3} & {-1} & {2}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{2} \\ {3} \\ {1}\end{array}\right)\)
(i.e) AX = B
The augment matrix (A, B) is
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Q1
The above matrix is in echelon form.
Now writing the equivalent equations
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Q1.1
Substituting z = 4 in (2) we get
-6y + 20 = -4
⇒ -6y = -4 – 20 = -24
⇒ y = 4
Substituting z = 4 and y = 4 in (1) we get
x + 8 – 4 = 3
⇒ x + 4 = 3
⇒ x = 3 – 4 = -1
So, x = -1; y = 4; z = 4
(ii) 2x + 4y + 6z = 22 …… (1)
3x + 8y + 5z = 27 ……. (2)
-x + y + 2z = 2 ……. (3)
Divide equation (1) by 2 we get
x + 2y + 3z = 11 ……. (1)
3x + 8y + 5z = 27 …….. (2)
-x + y + 2z = 2 ……. (3)
The matrix form of the above equations is \(\left(\begin{array}{ccc}{1} & {2} & {3} \\ {3} & {8} & {5} \\ {-1} & {1} & {2}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{c}{11} \\ {27} \\ {2}\end{array}\right)\)
(i.e) AX = B
The augment matrix (A, B) is
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Q1.2
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Q1.3
The above matrix is in echelon form.
Now writing the equivalent equations.
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Q1.4
Substituting z = 2 in (2) we get
y – 4 = -3
⇒ y = -3 + 4 = 1
Substituting z = 2, y = 1 in (1) we get
x + 2(1) + 3(2) = 11
⇒ x + 2 + 6 = 11
⇒ x + 8 = 11
⇒ x = 11 – 8 = 3
x = 3, y = 1, z = 2

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 2.
If ax2 + bx + c is divided by x + 3, x – 5, and x – 1, the remainders are 21, 61 and 9 respectively. Find a, b and c. (Use Gaussian elimination method.)
Solution:
P(x) = ax2 + bx + c. When P(x) is divided by x + 3, x – 5 and x – 1.
The remainders are respectively P(-3), P (5) and P (1).
We are given that P(-3) = 21; P(5) = 61; P(1) = 9
Now P(-3) = 21
⇒ a(-3)2 + b(-3) + c = 21
⇒ 9a – 3b + c = 21 ……. (1)
P(5) = 61
⇒ a(5)2 + b(5) + c = 61
⇒ 25a + 5b + c = 61 ……. (2)
P(1) = 9
⇒ a(1)2 + b(1) + c = 9
⇒ a + b + c = 9 …… (3)
Now the matrix form of the above three equations is \(\left(\begin{array}{ccc}{9} & {-3} & {1} \\ {25} & {5} & {1} \\ {1} & {1} & {1}\end{array}\right)\left(\begin{array}{l}{a} \\ {b} \\ {c}\end{array}\right)=\left(\begin{array}{c}{21} \\ {61} \\ {9}\end{array}\right)\)
(i.e) AX = B
The augmented matrix (A, B) is
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Q2
The above matrix is in echelon form now writing the equivalent equations.
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Q2.1
Substituting c = 6 in (2) we get
-20b – 24(6) = -164
⇒ -20b = -164 + 144 = -20
⇒ b = 1
Substituting c = 6, b = 1 in (1) we get
a + 1 + 6 = 9
⇒ a = 9 – 7 = 2
So a = 2, b = 1, c = 6

Question 3.
An amount of ₹ 65,000 is invested in three bonds at the rates of 6%, 8% and 10% per annum respectively. The total annual income is ₹ 5,000. The income from the third bond is ₹ 800 more than that from the second bond. Determine the price of each bond. (Use Gaussian elimination method.)
Solution:
Let the amount invested in 6% bond be ₹ x
and the amount invested in 8% bond be ₹ y
and the amount invested in 10% bond be ₹ z
Now x + y + z = 65000 ……. (1)
\(\frac{6}{100} x+\frac{8}{100} y+\frac{10}{100} z=5000\)
(i.e) 6x + 8y + 10z = 500000
(÷ by 2) 3x + 4y + 5z = 250000 ……. (2)
Also given that \(\frac{10}{100} z-\frac{8}{100} y=800\)
(i.e) -8y + 10z = 80000
(÷ by 2) -4y + 5z = 40000 …… (3)
Now the matrix form of the above three equations is \(\left(\begin{array}{ccc}{1} & {1} & {1} \\ {3} & {4} & {5} \\ {0} & {-4} & {5}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{c}{65000} \\ {250000} \\ {40000}\end{array}\right)\)
(i.e) AX = B
Now, the augmented matrix (A, B) is
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Q3
Now the above matrix is in echelon form. Writing the equivalent equations.
\(\left(\begin{array}{lll}{1} & {1} & {1} \\ {0} & {1} & {2} \\ {0} & {0} & {13}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{r}{65000} \\ {55000} \\ {260000}\end{array}\right)\)
(i.e) x + y + z = 65000
⇒ y + 2z = 55000
⇒ 13z = 260000
⇒ z = ₹ 20000
Substituting z = ₹ 20000 in (2) we get
y + 40000 = 55000
⇒ y = 55000 – 40000 = ₹ 15000
Substituting z = ₹ 20000, y = ₹ 15000 in (1) we get
x + 15000 + 20000 = 65000
⇒ x = 65000 – 35000 = ₹ 30000
So the amount invested in
6% bond x = ₹ 30000
8% bond y = ₹ 15000
and 10% bond z = ₹ 20000

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 4.
A boy is walking along the path y = ax2 + bx + c through the points (-6, 8), (-2, -12), and (3, 8). He wants to meet his friend at P(7, 60). Will he meet his friend? (Use Gaussian elimination method.)
Solution:
We are given y = ax2 + bx + c
Also we are given (-6, 8), (-2, -12) and (3, 8) are points on the path.
(i) (-6, 8) is a point on y = ax2 + bx + c
at x = -6, y = 8
(i.e) a(36) + b(-6) + c = 8
⇒ 36a – 6b + c = 8 …… (1)
(ii) (-2, -12) is a point on y = ax2 + bx + c
at x = -2, y = -12
⇒ a(-2)2 + b(-2) + c = -12
⇒ 4a – 2b + c = -12 ….. (2)
(iii) (3, 8) is a point on y = ax2 + bx + c
at x = 3, y = 8
⇒ a(3)2 + 6(3) + c = 8
⇒ 9a + 3b + c = 8 …… (3)
The matrix form of the above three equations is \(\left(\begin{array}{ccc}{36} & {-6} & {1} \\ {4} & {-2} & {1} \\ {9} & {3} & {1}\end{array}\right)\left(\begin{array}{l}{a} \\ {b} \\ {c}\end{array}\right)=\left(\begin{array}{c}{8} \\ {-12} \\ {8}\end{array}\right)\)
(i.e) AX = B
The augmented matrix (A, B) is
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Q4
The above matrix is in echelon form. Now writing the equivalent equations we get
\(\left(\begin{array}{ccc}{36} & {-6} & {1} \\ {0} & {-12} & {8} \\ {0} & {0} & {30}\end{array}\right)\left(\begin{array}{l}{a} \\ {b} \\ {c}\end{array}\right)=\left(\begin{array}{c}{8} \\ {-116} \\ {-300}\end{array}\right)\)
(i.e) 36a – 6b + c = 8
⇒ -12b + 8c = -116
⇒ 30c = -300
⇒ c = -10
Substituting c = -10 in (2) we get
-12b + 8(-10) = -116
⇒ -12b = -116 + 80 = -36
⇒ b = 3
Substituting c = -10, b = 3 in (1) we get
36a – 6(3) + (-10) = 8
⇒ 36a – 18 – 10 = 8
⇒ 36a = 8 + 18 + 10 = 36
⇒ a = 1
a = 1, b = 3 and c = -10
y = (1)x2 + (3)x + (-10)
y = x2 + 3x – 10
Now at x = 7, y= (7)2 + 3(7)- 10 = 49 + 21 – 10 = 60
(7, 60) is a point on the path so he will meet his friend.

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 Additional Problems

Question 1.
Examine the consistency of the following system of equations. If it is consistent then solve the same.
(i) 4x + 3y + 6z = 25,
x + 5y + 7z = 13,
2x + 9y + z = 1
Solution:
The augmented matrix is
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 1
The last equivalent matrix is in the echelon form. It has three non-zero rows.
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 2
∴ The given system is consistent and has a unique solution
4x + 3y + 6z = 25 …… (1)
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 3
∴ The unique solution is x = 4, y = -1, z = 2.

Question 2.
Verify whether the given system of equations is consistent. If it is consistent, solve them. 2x + 5y + 7z = 52, x + 7 + z = 9, 2x + y – z = 0
Solution:
The given system of equations is equivalent to the single matrix equation.
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 4
AX = B
The augmented matrix is
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 5
The last equivalent matrix is in the echelon form. It has three non-zero rows.
∴ \(\rho(\mathrm{A}, \mathrm{B})=3\)
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 6
Since there are three non-zero rows, \(\rho(\mathrm{A})\) = 3
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 7
The given system is consistent and has a unique solution.
To find the solution, we see that the given system of equations is equivalent to the matrix equation.
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 8
x + y + z = 9 ……. (1)
-y – 3z = -18 ……… (2)
-4z = -20 ……… (3)
(3) ⇒ z = 5, (2) ⇒ y = 18 – 3z = 13, (1) ⇒ x = 9 – y – z ⇒ x = 9 – 3 – 5 = 1
∴ Solution is x = 1,
y = 3,
z = 5

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 3.
Examine the consistency of the equations.
2x – 3y + 3z = 5,
3x + y – 3z = 13,
2x + 19y – 47z = 32
Solution:
The given system of equations can be written in the form of a matrix equation as
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 9
The augmented matrix is
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 100
The last equivalent matrix is in the echelon form. It has three non-zero rows.
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 10
∴ The given system is inconsistent and hence has no solution.

Question 4.
Show that the equations: x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30 are consistent and solve them.
Solution:
The matrix equation corresponding to the given system is.
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 11
The augmented matrix is
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 111
In the last equivalent matrix, there are two non-zero rows.
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 12
∴ The given system is consistent. But the value of the common rank is less than the number of unknowns. The given system has an infinite number of solutions.
The given system is equivalent to the matrix equation
Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5 13
x + y + z = 6 ……(1)
y + 2z = 8 ….(2)
(2) ⇒ y = 8 – 2z ; (1) ⇒ x = 6 – y – z = 6 – (8 – 2z) – z = z – 2
Taking z = k, we get x = k – 2, y = 8 – 2k; k ∈ R
Putting k = 1, we have one solution as x = – 1, y = 6, z = 1. Thus by giving different values for k we get different solutions. Hence the given system has infinite number of solutions.

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