Assume that the matrix A is row equivalent to B. Find a basis for the row space of the matrix A.
-$$A = \left[ \begin{array} { r r r r r }
1 & 5 & 1 & 3 & 0 \
- 2 & - 4 & - 5 & - 4 & 1
\end{array} \right]$$
A) $\operatorname { dim } \mathrm { Nul } \mathrm { A } = 2 , \operatorname { dim } \operatorname { Col } \mathrm { A } = 3$
B) $\operatorname { dim } \mathrm { Nul } \mathrm { A } = 4 , \operatorname { dim } \operatorname { Col } \mathrm { A } = 1$
C) $\operatorname { dim } \operatorname { Nul } \mathrm { A } = 3 , \operatorname { dim } \operatorname { Col } \mathrm { A } = 2$
D) $\operatorname { dim } \mathrm { Nul } \mathrm { A } = 3 , \operatorname { dim } \operatorname { Col } \mathrm { A } = 3$

Question

Assume that the matrix A is row equivalent to B. Find a basis for the row space of the matrix A.
-$$A = \left[ \begin{array} { r r r r r } 
1 & 5 & 1 & 3 & 0 \
- 2 & - 4 & - 5 & - 4 & 1
\end{array} \right]$$
A) $\operatorname { dim } \mathrm { Nul } \mathrm { A } = 2 , \operatorname { dim } \operatorname { Col } \mathrm { A } = 3$
B) $\operatorname { dim } \mathrm { Nul } \mathrm { A } = 4 , \operatorname { dim } \operatorname { Col } \mathrm { A } = 1$
C) $\operatorname { dim } \operatorname { Nul } \mathrm { A } = 3 , \operatorname { dim } \operatorname { Col } \mathrm { A } = 2$
D) $\operatorname { dim } \mathrm { Nul } \mathrm { A } = 3 , \operatorname { dim } \operatorname { Col } \mathrm { A } = 3$

Quizplus · Accepted Answer

The basis for the row space of matrix A can be found by reducing it to row echelon form (REF) or reduced row echelon form (RREF). Since A is row equivalent to B, we focus on A's row operations to find its row space. The dimension of the null space (Nul A) plus the dimension of the column space (Col A) equals the number of columns in A. A has 5 columns, and since it has 2 rows, the maximum rank (dimension of the column space) it can have is 2, assuming the rows are linearly independent. This means the dimension of the null space (Nul A) is 3, because the rank-nullity theorem states that the number of columns (5) is the sum of the rank (dimension of Col A, which is 2 here) and the nullity (dimension of Nul A). Therefore, the correct answer is that the dimension of the null space of A is 3, and the dimension of the column space of A is 2.

Assume That the Matrix a Is Row Equivalent to B A=[15130−2−4−5−41]A = \left[ \begin{array} { r r r r r } 1 & 5 & 1 & 3 & 0 \\- 2 & - 4 & - 5 & - 4 & 1\end{array} \right]A=[1−2​5−4​1−5​3−4​01​]

Assume That the Matrix a Is Row Equivalent to B $A = \left[ \begin{array} { r r r r r } 1 & 5 & 1 & 3 & 0 \\- 2 & - 4 & - 5 & - 4 & 1\end{array} \right]$