A technique for minimization of switching functions has been described in this presentation that can be easily programmed and is equally suitable for paper and pencil as well.
The technique does not generate all the prime implicants but looks for existence of a prime implicant thus reducing the search space. It allows simplification of switching functions having variables greater than 8 as well.
A technique for minimization of switching functions has been described in this presentation that can be easily programmed and is equally suitable for paper and pencil as well.
The technique does not generate all the prime implicants but looks for existence of a prime implicant thus reducing the search space. It allows simplification of switching functions having variables greater than 8 as well.
Physics numerical solutions, kinematics numericals solutions karachi board, federal board physics, SUVAT problems, metric physics solutions karachi, Physics solved numericals of Karachi board,Numericals of Physics metric (X) Federal Board,Federal Board Class X Physics SSC (9th & 10th),9th class physics numerical,Q1. A car is moving on a straight road at 5 m/s. It is accelerated at 3 m/s. What will be its velocity after 4 seconds?Q.2. A bullet train starts from rest and moves with uniform acceleration of 0.12 m/s2. Find its final velocity and distance covered after 5 minutes?Q.3. A bus is moving with a velocity of 72 km/hr. By applying brakes, a deceleration of 0.4 m/s2 is produced. Find the distance, covered by the bus before it stops?Q.4. What is the acceleration of an object which accelerates along a straight path from rest, and attains a velocity of 20 m/s after covering a distance of 50 m in 5 s?Q.5. a wicket keeper catches a ball moving at 30 m/s.
(a) if he does not move his hand the ball comes to rest in his glove over a distance of 1 cm. What is the average acceleration?(b) If he moves his hand as the ball is caught so that it comes to rest over 10 cm. What is its acceleration?
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SOLUTION OF ALL QUESTIONS FROM TEXT BOOK AND LAST 10 YEARS .
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Quadratic Programming : KKT conditions with inequality constraintsMrinmoy Majumder
In the case of Quadratic Programming for optimization, the objective function is a quadratic function. One of the techniques for solving quadratic optimization problems is KKT Conditions which is explained with an example in this tutorial.
The pattern of question paper in the subject Mathematics has been changed in CBSE,India.I am uploading the paper with marking scheme so that students will be benefitted-Pratima Nayak,KVS
1. Announcements
Quiz 2 on Wednesday Jan 27 on sections 1.4, 1.5, 1.7 and 1.8
Test 1 will be on Feb 1, Monday in class. More details later.
2. Last Class
a b
Let A = . If ad − bc = 0 then A is invertible and
c d
−1 1 d −b
A =
ad − bc −c a
.
3. Last Class
a b
Let A = . If ad − bc = 0 then A is invertible and
c d
−1 1 d −b
A =
ad − bc −c a
.
So if the determinant of A (or det A) is equal to 0, A−1 does not
exist.
4. Last Class
Steps for a 2 × 2 matrix A
1. First check whether det A=0. If so, stop. A is not invertible.
5. Last Class
Steps for a 2 × 2 matrix A
1. First check whether det A=0. If so, stop. A is not invertible.
2. If det A = 0, swap the main diagonal elements of A.
6. Last Class
Steps for a 2 × 2 matrix A
1. First check whether det A=0. If so, stop. A is not invertible.
2. If det A = 0, swap the main diagonal elements of A.
3. Then change the sign of both o diagonal elements (don't
swap these)
7. Last Class
Steps for a 2 × 2 matrix A
1. First check whether det A=0. If so, stop. A is not invertible.
2. If det A = 0, swap the main diagonal elements of A.
3. Then change the sign of both o diagonal elements (don't
swap these)
4. Divide this matrix (after steps 2 and 3) by detA to give A−1 .
(This divides each element of the resultant matrix.)
5. If you want to check your answer, you can see whether
1 0
AA−1 =
0 1
6. This method will not work for 3 × 3 or bigger matrices.
8. Example 7(a) section 2.2, One Case
1 2
Let A = . Find A−1 and use it to solve the equation
5 12
2
Ax = b3 where b3 =
6
Solution: Here detA = (1)(12) − (5)(2) = 2 = 0. So we can nd A−1 .
9. Example 7(a) section 2.2, One Case
1 2
Let A = . Find A−1 and use it to solve the equation
5 12
2
Ax = b3 where b3 =
6
Solution: Here detA = (1)(12) − (5)(2) = 2 = 0. So we can nd A−1 .
Interchange the positions of 1 and 12. Change the signs of 2 and 5.
Then we get
12 −2
−5 1
.
10. Example 7(a) section 2.2, One Case
1 2
Let A = . Find A−1 and use it to solve the equation
5 12
2
Ax = b3 where b3 =
6
Solution: Here detA = (1)(12) − (5)(2) = 2 = 0. So we can nd A−1 .
Interchange the positions of 1 and 12. Change the signs of 2 and 5.
Then we get
12 −2
−5 1
. Divide each element of the matrix by detA which is 2. This gives
−1 6 −1
A =
−5/2 1/2
x1 6 −1 2 6
x= = =
x2 −5/2 1 /2 6 −2
A−1 b
11. Algorithm to nd A−1
To nd the inverse of a square matrix A of any size,
1. Write the augmented matrix with A and I (the identity matrix
of proper size) written side by side.
12. Algorithm to nd A−1
To nd the inverse of a square matrix A of any size,
1. Write the augmented matrix with A and I (the identity matrix
of proper size) written side by side.
2. Do proper row reductions on both A and I till A is reduced to
I.
13. Algorithm to nd A−1
To nd the inverse of a square matrix A of any size,
1. Write the augmented matrix with A and I (the identity matrix
of proper size) written side by side.
2. Do proper row reductions on both A and I till A is reduced to
I.
3. This changes I to a new matrix which is A−1
14. Algorithm to nd A−1
To nd the inverse of a square matrix A of any size,
1. Write the augmented matrix with A and I (the identity matrix
of proper size) written side by side.
2. Do proper row reductions on both A and I till A is reduced to
I.
3. This changes I to a new matrix which is A−1
4. If you cannot reduce A to I (if you get a row full of zeros for
example), A−1 does not exist
15. Example 30, section 2.2
5 10
Let A = . Find A−1 using the algorithm.
4 7
Solution: Start with the augmented matrix
5 10 1 0
4 7 0 1
16. Example 30, section 2.2
5 10
Let A = . Find A−1 using the algorithm.
4 7
Solution: Start with the augmented matrix
5 10 1 0
4 7 0 1
Divide R1 by 5
1 2 1/5 0
4 7 0 1
30. Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
31. Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
32. Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
3. The equation Ax = 0 has Non-trivial solution
33. Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
3. The equation Ax = 0 has Non-trivial solution
4. The columns of A are linearly
34. Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
3. The equation Ax = 0 has Non-trivial solution
4. The columns of A are linearly Dependent
35. Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
3. The equation Ax = 0 has Non-trivial solution
4. The columns of A are linearly Dependent
5. A cannot be row-reduced to the 3 × 3 identity matrix
36. Section 2.3 Invertible Matrices
If a square matrix A of size n × n is invertible the following is true
37. Section 2.3 Invertible Matrices
If a square matrix A of size n × n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
38. Section 2.3 Invertible Matrices
If a square matrix A of size n × n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n × n identity matrix
39. Section 2.3 Invertible Matrices
If a square matrix A of size n × n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n × n identity matrix
3. The equation Ax = 0 has only the trivial solution (no free
variables)
40. Section 2.3 Invertible Matrices
If a square matrix A of size n × n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n × n identity matrix
3. The equation Ax = 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly
41. Section 2.3 Invertible Matrices
If a square matrix A of size n × n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n × n identity matrix
3. The equation Ax = 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly independent
42. Section 2.3 Invertible Matrices
If a square matrix A of size n × n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n × n identity matrix
3. The equation Ax = 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly independent
5. The equation Ax = b is consistent for every b in Rn
43. Section 2.3 Invertible Matrices
If a square matrix A of size n × n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n × n identity matrix
3. The equation Ax = 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly independent
5. The equation Ax = b is consistent for every b in Rn
6. The columns of A span Rn (because every row has a pivot and
recall theorem 4, sec 1.4)
44. Example 2, Section 2.3
−4 6
Determine whether A = is invertible. Why(not)?
6 −9
45. Example 2, Section 2.3
−4 6
Determine whether A = is invertible. Why(not)?
6 −9
Solution: Clearly the determinant is (-4)(-9)-(6)(6)=36-36=0. So
not invertible. Checking invertibility of 2 × 2 matrices are thus easy.
Not so obvious fact: The second column is -1.5 times the rst
column. Since we have linearly dependent columns, by IMT, A is
not invertible.
46. Example 4, Section 2.3
−7 0 4
Determine whether A = 3 0 −1 is invertible. Why(not)?
2 0 9
Solution: Remember what happens to a set of vectors if the zero
vector is present in that set?
47. Example 4, Section 2.3
−7 0 4
Determine whether A = 3 0 −1 is invertible. Why(not)?
2 0 9
Solution: Remember what happens to a set of vectors if the zero
vector is present in that set? The vectors are linearly dependent.
Since the second column of A is full of zeros, we have linearly
dependent columns, and so A is not invertible.
50. Example 6, Section 2.3
1 −5 −4
0 3 4
0 0 0
Since the third row (and the third column) does not have a pivot,
we have linearly dependent columns and by the IMT A is not
invertible.
52. Example 8, Section 2.3
1 3 7 4
0 5 9 6
Determine whether A =
is invertible. Why(not)?
0 0 2 8
0 0 0 10
Solution: This matrix is already in the echelon form. There are 4
pivot rows and 4 pivot columns. So A is invertible.