1 thought on “determine s={[a b c] , [b+1 c+1 a+1], [c-1 a-1 b-1]} is basis for r^3. if it is write in vector z=[1 2 3] as a linear”

Answer:

Thus, the dimension of the subspace S is .

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Step-by-step explanation:

Consider the following subspace: S = {(a, b, c)|a – 3b + c = 0, b – 2c = 0, 2b – c = 0} The objective of the problem is to find the basis for subspace S. Write the subspace as, {(a, b, c)|Ax = 0 where A = [1 0 0 -3 1 2 1 -2 -1], x = [a b c]^T} Find the basis of Null A, because it is equivalent to find the basis for the subspace. The augmented matrix for the system Ax = 0 is [A:0] = [1 0 0 -3 1 2 1 -2 -1 0 0 0] Apply row operations to augmented matrix to reduce the matrix to echelon matrix. [1 0 0 -3 1 2 1 -2 -1 0 0 0] ~[1 0 0 -3 1 0 1 -2 3 0 0 0] leftarrow Row 3 + (-2) Row 2 ~[1 0 0 0 1 0 -5 -2 1 0 0 0] leftarrow Row 1+(3) Row 2 1/3 Row 3 ~[1 0 0 0 1 0 0 0 1 0 0 0] leftarrow 1+(5)Row 3 leftarrow Row 2 + (2) Row 3 Therefore, the solution of the system Ax = 0 is (a, b, c) = (0, 0, 0). Hence, the subspace S contains only (0, 0, 0) vector. S = {(0, 0, 0)} By the definition of basis, there are no vectors in S such that, those spans the subspace S and linearly independent. Hence, the subspace S has no basis Consider the following subspace, S = {[a-4b-2c 2a + 5b – 4c -a + 2c -3a + 7b + 6c]: a, b, c in R}. The objective is to find the basis for subspace S. For any vector x in S, x = [a-4b-2c 2a + 5b – 4c -a + 2c -3a + 7b + 6c] for some a, b, c R x = [a 2a -a -3a] + [-4b 5b 0 7b]+[-2c -4c 2c 6c] x = [1 2 -1 -3]a + [-4 5 0 7]b + [-2 -4 2 6]c Let V_1 = [1 2 -1 -3], v_2 = [-4 5 0 7], and v_3 = [-2 -4 2 6]. Thus, every vector in S is a linear combination of vectors v_1,v_2, and v_3. Therefore, the set of vectors {v_1,v_2,v_3} spans the set S. Check the linear independence of the set as follows: Clearly the vector v_1, is a non-zero vector. The vector v_2, is not a scalar multiple of the vector v_1. But the vector v_3 is a linear combination of v_1 and v_2 as v_3 = -2v_1 + 0 v_2. So {v_1,v_2,v_3} is linearly dependent. Therefore by the spanning set theorem, the vector v, can be discarded with no change in the span of the set. So, S = Span{v_1,v_2}. Since v_1 and v_2 are not multiples of each other, {v_1, v_2} is linearly independent and hence, a basis for S. The objective of the problem is to find the dimension of the basis. The dimension of the vector space is the number of non-zero vectors present in its basis. Consider the basis for S from part (a), B = {[1 2 -1 -3],[-4 5 0 7]} Thus, the dimension of the subspace S is .

Answer:Thus, the dimension of the subspace S is .

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Step-by-step explanation:Consider the following subspace: S = {(a, b, c)|a – 3b + c = 0, b – 2c = 0, 2b – c = 0} The objective of the problem is to find the basis for subspace S. Write the subspace as, {(a, b, c)|Ax = 0 where A = [1 0 0 -3 1 2 1 -2 -1], x = [a b c]^T} Find the basis of Null A, because it is equivalent to find the basis for the subspace. The augmented matrix for the system Ax = 0 is [A:0] = [1 0 0 -3 1 2 1 -2 -1 0 0 0] Apply row operations to augmented matrix to reduce the matrix to echelon matrix. [1 0 0 -3 1 2 1 -2 -1 0 0 0] ~[1 0 0 -3 1 0 1 -2 3 0 0 0] leftarrow Row 3 + (-2) Row 2 ~[1 0 0 0 1 0 -5 -2 1 0 0 0] leftarrow Row 1+(3) Row 2 1/3 Row 3 ~[1 0 0 0 1 0 0 0 1 0 0 0] leftarrow 1+(5)Row 3 leftarrow Row 2 + (2) Row 3 Therefore, the solution of the system Ax = 0 is (a, b, c) = (0, 0, 0). Hence, the subspace S contains only (0, 0, 0) vector. S = {(0, 0, 0)} By the definition of basis, there are no vectors in S such that, those spans the subspace S and linearly independent. Hence, the subspace S has no basis Consider the following subspace, S = {[a-4b-2c 2a + 5b – 4c -a + 2c -3a + 7b + 6c]: a, b, c in R}. The objective is to find the basis for subspace S. For any vector x in S, x = [a-4b-2c 2a + 5b – 4c -a + 2c -3a + 7b + 6c] for some a, b, c R x = [a 2a -a -3a] + [-4b 5b 0 7b]+[-2c -4c 2c 6c] x = [1 2 -1 -3]a + [-4 5 0 7]b + [-2 -4 2 6]c Let V_1 = [1 2 -1 -3], v_2 = [-4 5 0 7], and v_3 = [-2 -4 2 6]. Thus, every vector in S is a linear combination of vectors v_1,v_2, and v_3. Therefore, the set of vectors {v_1,v_2,v_3} spans the set S. Check the linear independence of the set as follows: Clearly the vector v_1, is a non-zero vector. The vector v_2, is not a scalar multiple of the vector v_1. But the vector v_3 is a linear combination of v_1 and v_2 as v_3 = -2v_1 + 0 v_2. So {v_1,v_2,v_3} is linearly dependent. Therefore by the spanning set theorem, the vector v, can be discarded with no change in the span of the set. So, S = Span{v_1,v_2}. Since v_1 and v_2 are not multiples of each other, {v_1, v_2} is linearly independent and hence, a basis for S. The objective of the problem is to find the dimension of the basis. The dimension of the vector space is the number of non-zero vectors present in its basis. Consider the basis for S from part (a), B = {[1 2 -1 -3],[-4 5 0 7]} Thus, the dimension of the subspace S is .