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Chapter: 12th Mathematics : UNIT 1 : Applications of Matrices and Determinants

Solved Example Problems on Applications of Matrices: Solving System of Linear Equations

Solution to a System of Linear equations: (i) Matrix Inversion Method (ii) Cramer’s Rule (iii) Gaussian Elimination Method

Solution to a System of Linear equations


(i) Matrix Inversion Method


Example 1.22

Solve the following system of linear equations, using matrix inversion method:

5+ 2 = 3, 3+ 2 = 5 .

Solution

The matrix form of the system is AX = B , where 

We find |A| =   = 10 - 6= 4 ≠ 0. So, A−1  exists and A−1 = 

Then, applying the formula X = A−1B , we get


So the solution is (x = −1, y = 4).

 

Example 1.23

Solve the following system of equations, using matrix inversion method:

2x1 + 3x2 + 3x3 = 5,

x1 – 2x2 + x3 = -4,

3x1 – x2 – 2x3 = 3

Solution

The matrix form of the system is AX = B,where


So, the solution is ( x1 = 1, x2 = 2, x3 = −1) .

 

Example 1.24

If , find the products AB and BA and hence solve the system of equations x − y + z = 4, x – 2y – 2z = 9, 2x + y +3z =1.

Solution


Writing the given system of equations in matrix form, we get


Hence, the solution is (x = 3, y = - 2,  z = −1).


(ii) Cramer’s Rule


Example 1.25

Solve, by Cramer’s rule, the system of equations

x1  x2 = 3, 2x1 + 3x2 + 4x3 = 17, x2 + 2x3 = 7.

Solution

First we evaluate the determinants


So, the solution is (x1 = 2, x2 = - 1,  x3 = 4).

 

Example 1.26

In a T20 match, Chennai Super Kings needed just 6 runs to win with 1 ball left to go in the last over. The last ball was bowled and the batsman at the crease hit it high up. The ball traversed along a path in a vertical plane and the equation of the path is = ax2 + bx + with respect to a xy -coordinate system in the  vertical  plane  and  the  ball  traversed   through   the   points  (10,8), (20,16), (30,18) , can you conclude that Chennai Super Kings won the match?


Justify your answer. (All distances are measured in metres and the meeting point of the plane of the path with the farthest boundary line is (70, 0).)

Solution

The path  axbx c  passes through the points (10,8), (20,16), (40, 22) . So, we get the system of equations 100a + 10b + c = 8, 400a + 20b + c= 16,1600a + 40b + c = 22. To apply Cramer’s rule, we find


When = 70, we get = 6. 

So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder standing even just before theboundary line to jump and catch the ball.

  Hence the ball went for a super six and the Chennai Super Kings won the match.



(iii) Gaussian Elimination Method


Example 1.27

Solve the following system of linear equations, by Gaussian elimination method :

4+ 3+ 6= 25, + 5 + 7= 13, 2+ 9 + = 1.

Solution

Transforming the augmented matrix to echelon form, we get


The equivalent system is written by using the echelon form:

 x + 5y  + 7z = 13 , … (1)

 17y + 22z = 27 , … (2)

 199z = 398 . … (3)


Substituting z = 2, y = -1  in (1), we get x = 13 - 5  × (−1 ) − 7 × 2 = 4 .

So, the solution is ( x =4, y = - 1, z = 2 ).

Note. The above method of going from the last equation to the first equation is called the method of back substitution.

 

Example 1.28

The  upward  speed  v(t) of a rocket  at time t is approximated by v(t) = at2 + bt + c, 0 ≤  t ≤ 100 where a, b, and c are constants. It has been found that the speed at times t = 3, t = 6 , and t = 9 seconds are respectively, 64, 133, and 208 miles per second respectively. Find the speed at time  t = 15 seconds. (Use Gaussian elimination method.)


Solution

Since v(3) =64, v(6) = 133 and v(9) = 208 , we get the following system of linear equations

 9a +3b + c = 64 ,

 36a + 6b + c = 133,

 81a + 9b + c = 208 .

We solve the above system of linear equations by Gaussian elimination method.

Reducing the augmented matrix to an equivalent row-echelon form by using elementary row  operations, we get


Writing the equivalent equations from the row-echelon matrix, we get

9a + 3b + c =  64, 2b + c = 41, c= 1.

By back substitution, we get 


So, we get v (t) = 1/3 t2  + 20t + 1.

Hence, v(15) = 1/3 (225) + 20(15) + 1 = 75 + 300 + 1 = 376


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12th Mathematics : UNIT 1 : Applications of Matrices and Determinants


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