Solution to a System of Linear equations
Solve the following system of linear equations, using matrix inversion method:
5x + 2 y = 3, 3x + 2 y = 5 .
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).
Solve the following system of equations, using matrix inversion method:
2x1 + 3x2 + 3x3 = 5,
x1 – 2x2 + x3 = -4,
3x1 – x2 – 2x3 = 3
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).
Solve, by Cramer’s rule, the system of equations
x1 − x2 = 3, 2x1 + 3x2 + 4x3 = 17, x2 + 2x3 = 7.
First we evaluate the determinants
So, the solution is (x1 = 2, x2 = - 1, x3 = 4).
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 y = ax2 + bx + c 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).)
The path y = ax2 + bx + 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 x = 70, we get y = 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.
Solve the following system of linear equations, by Gaussian elimination method :
4x + 3y + 6z = 25, x + 5 y + 7z = 13, 2x + 9 y + z = 1.
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.
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.)
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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