Answer:
Step-by-step explanation:
a. For a matrix equation Ax=b to the consistent then, there will be a unique solution for x and for we to have a unique solution for x, the matrix A must be linear dependent.
A matrix equation is consistent if there is at least one set of values for the unknowns that satisfies each equation in the system.
Given a set of vectors, you can determine if they are linearly independent by writing the vectors as the columns of the matrix A, and solving Ax = 0. If there are any non-zero solutions, then the vectors are linearly dependent. If the only solution is x = 0, then they are linearly independent.
So if a matrix A is linearly independent, then the solution is not consistent.
So it is false.
b. The first non-zero element in each row, called the leading entry, is 1. Each leading entry is in a column to the right of the leading entry in the previous row.
So, if the matrix A is reduce to echelon form and each has a leading entry, then the solution to the matrix is consistent because he unknowns will have a unique value.
Therefore,
This is also false.
c. This is true,
If Ax=b is inconsistent, the matrix A doesn't have a leading entries in each row,
This can only be wrong if only it was the last row that did not have a leading element., because at that case the last unknown will be assign to a variable. But, since it every row that doesn't have a leading entry, then the matrix is independent so equation Ax=b is inconsistent
This is True.
d. Every matrix equation Ax=b corresponds to a vector equation with the same solution set
This is true for the vector equation but not true for the vector solution,
Every matrix has it owns vector solution, since the element in the vectors are different.
This is false.
e. This is true since, I(m) of A is not entire in Rm, there is a vector in Rm, say b, which cannot be obtained as Ax=b, for some x which belong to Real number. Therefore the equation Ax=b has no solution i.e, it is not consistent.
Then this is true
f. The solution set of a linear system whose augmented matrix is [a1 a2 a3 | b] is the same as the solution set of Ax=b, if A=[a1 a2 a3 ].
This is true since they both have the same Matrix A
They will form the same matrix equation and the same vector solution
Then this is true
