 Student, in Algebra, many cases we need to know the number of vectors which are dependent and independent. For example, to find the rank of the matrix, we need to know the number of independent and dependent number of rows and column. So this, we want to find the linearly dependence of vector. So what is the linearly dependence of vector here is the set of n into 1 vectors. So we have this set into n into 1 vectors. So this is the vector 1 to up to so on k are linearly dependent if there exists real numbers alpha 1, alpha 2 up to so on alpha k, not all 0. You have real numbers alpha 1, alpha 2, alpha k, not all 0. You have their value exists, all 0 will not exist. Such that this equation is satisfied. So the equation is i varies 1 to k. We have k, yk, we have k. Vectors, alpha 1, alpha 2 up to so on. If we open this, you will have an equation. So we have equation alpha 1, y1, plus alpha 2, y2, plus up to so on alpha k, yk. So this is equals to 0. Equation we have satisfied or after satisfying the equation we have real constant or real numbers. These are alpha 1, alpha 2 which is not all 0. This is called the linearly dependent case. Otherwise the set of vectors is linearly independent. And what happens if this is not all 0? If we have any of these zeroes then we will say the set is linearly independent. Okay, further we will check this. For a linearly independent set, the only solution to the above equation is this. In the independent case we have only solution, only one solution. And the solution we have is alpha 1, alpha 2 up to so on, alpha k equals to 0. Only solution will come in the independent case. In the dependent case, the only solution is not the solution exists, it can be done. So here is the example, we have the hypothetical example. So the y1 is the vector, y2 is the vector and y3 is the vector. We have three dimensions. So three dimensions are open, alpha 1, y1, alpha 2, y2 and the alpha 3, y3 equals to 0. Now put the value of y1 in this equation. So alpha 1, y1, alpha 2, this is the value of y2 and alpha 3, this is the value of y3 equals to 0. Now consider the equation 1, up equation 1 determinant case over here. So 1 multiplied by alpha 1 plus 0 multiplied by alpha 2 plus 1 multiplied by alpha 3 equals to 0, equation 1 determinant. Similarly, equation 2 determinant, we are doing 1 into alpha 1, 1 into alpha 2, 4 into alpha 3, equation 2 determinant. Again 1 into alpha 1 minus 1 into alpha 2 minus 2 into alpha 3 equation 3 determinant. Equals to 0. So here, equation 1, we will determine the value of the alpha 1, so the alpha 1 equals to minus alpha 3. This is from equation 1 and put this value in equation 2. So, alpha 1 minus alpha 3 plus alpha 2 plus 4 alpha 3 equals to 0. From here, we are seeing alpha 2 plus 3 alpha 3 equals to 0. So, alpha 2 equals to minus 3 alpha 3. So, look at this. This is the equation 1 from equation 1 which is equals to minus alpha 3. Substituting alpha 1 into alpha 2's value we have entered and we have alpha 2's value which is minus 3 alpha 3. So now, alpha 1 equals to minus alpha 3 and alpha 2 equals to minus alpha 3. This is the satisfied for equation 3. Let us check this. This is the satisfied for equation 3. So, here is the alpha 1, alpha 1 which is equals to minus alpha 3 minus alpha 2's value plus the minus 3 alpha 3 minus 2 alpha 3 equals to 0. So, minus alpha 3 plus 3 alpha 3 minus 2 alpha 3 which is equals to 0. So, minus 3 alpha 3 plus 3 alpha 3 equals to 0. So, satisfy this equation? Satisfy equation 3. Now, here we do not have the only solution in the dependent case. Now, look at this. The solution of this equation is possible for any value of alpha 1, alpha 2 and alpha 3, not all 0. The solution will satisfy this, not all 0. Similarly, if I put alpha 3's value here as 1, what will happen to you? We can take any value of alpha 1 equals to minus 1. I will get alpha 3's value minus 3 and similarly, I will get alpha 3's value as well. So, what is our dependent? Therefore, the set of vectors is linearly dependent. Means, the only solution is not there. The equation must be satisfied. If the equation is satisfied, we will determine the solution. After the equation is satisfied, we have alpha 1, alpha 2, alpha 3's value other than 0. We will determine the value on that. So, that is called, this is the vector is linearly dependent. So, here, this is the another example. Here, y1 equals to this, y2 equals to this and y3 equals to this. For three dimensions, we have opened it three times. For equation 1, so alpha 1, y1, 0, 1, 1 plus alpha 2, y2, 1, 1, minus 2, alpha 3, y3, 3, 4, 1. From here, we will determine equation 1, 2, 3. So, 0, alpha 1, alpha 1's value is not there. 1 into alpha 2 plus 3 into alpha 3, equation 1 is determined. Similarly, equation 2 and equation 3, we have determined from here. For equation 1, we are determining equation 1 from here. In equation 1, we have alpha 2's value. What is the value of alpha 2? Look at the previous equation. What is the value of alpha 2? Alpha 2 equals to minus 3 alpha 3. This is from equation 1. Substituting equation 1 into 2. In 2, we have entered alpha 1 plus, from here, we have the value. Minus 3 alpha 3 plus 4 alpha 3 equals to 0. Alpha 1 minus 3 alpha 3 plus 4 alpha 3 equals to 0. So, alpha 1 plus alpha 3 equals to 0. Alpha 1 equals to minus alpha 3. So, alpha 1 minus alpha 3. Third value, alpha 1 and alpha 2, we have determined. Substituting alpha 1 and alpha 2 in 3. So, we get alpha 3 equals to 0. Now, alpha 2's value is minus 3 alpha 3. Substituting alpha 1's value is minus alpha 3. In equation 3, we are substituting. Minus 2 into minus 3 alpha 3 and plus alpha 3 equals to 0. So, minus alpha 3, 3 to the 6, alpha 3 plus alpha 3 equals to 0. Minus 6 alpha 3 equals to 0 and the alpha 3 equals to 0. So, this is the alpha 3 equals to 0. So, the only solution of alpha 1, alpha 2, alpha 3 which is equals to 0. Now, how do you get it? Because alpha 3's value equals to 0. I am entering it into alpha 1. 0. I entered it into alpha 2. 0. And I already have alpha 3. 0. The only solution, if I have the only solution, alpha 1, alpha 2, alpha 3 equals to 0. Therefore, the set of vectors are linearly independent. So, what do we have now? Therefore, the set of linearly independent. Here is this example. Determine, if you have a linearly dependent case. Students, from here, we have determined from a particular example that if we have a linearly dependent case, if we have vectors, then we have to satisfy the equation. After satisfying the equation, real numbers, their only solution does not exist. Their solution also comes. Other than zeros, the value can exist. But in the independent case, we have the only solution. We are checking the vectors' independence. The only solution is that you have real numbers that we have, alpha 1, alpha 2, alpha 3. The real numbers, their values become zero. So, linearly independent, we have only solution which is equals to the values which is equals to the zero because it is the only solution. And for the dependent, we do not have the only solution for the dependent. The solution can also exist. So, this is the example of the linearly dependent of vectors.