## Linear Algebra and Its Applications, Review Exercise 2.16

Review exercise 2.16. For each of the cases in review exercise 2.15, describe the relationship among the rank $r$ of $A$, the number of rows $m$, and the number of columns $n$. $A = \begin{bmatrix} 1&0 \\ 0&1 \\ 1&1 \end{bmatrix}$

we have $m = 3$, $n = 2$, and $r = 2$ (since the two columns are linearly independent). Per box 2Q on page 96, since $r = n$ we know that there is at most one solution to $Ax = b$. However since $r = n < m$ we are not guaranteed that a solution exists. So $Ax = b$ has either 0 or 1  solutions.

ii) For the matrix $A = \begin{bmatrix} 1&0&1 \\ 0&1&1 \end{bmatrix}$

we have $m = 2$, $n = 3$, and $r = 2$ (since the matrix is in echelon form with two pivots). Per box 2Q on page 96, since $r = m$ we know that there is at least one solution to $Ax = b$ and since $r = m < n$ we know that there is more than one solution.

iii) For the matrix $A = \begin{bmatrix} 1&0&1 \\ 0&1&1 \\ 1&-1&0 \end{bmatrix}$

we have $m = 3$, $n = 3$, and $r = 2$ (since the third row of the matrix is a linear combination of the first two rows). Per box 2Q on page 96, since $r < m$ we know that there may not be a solution to $Ax = b$ and since $r < n$ we know that there if there is a solution then it is not guaranteed to be unique.

iv) For the matrix $A = \begin{bmatrix} 1&0&1 \\ 0&1&1 \\ 0&0&1 \end{bmatrix}$

we have $m = 3$, $n = 3$, and $r = 3$ (since the matrix is in echelon form with three pivots). Per box 2Q on page 96, since $r = m$ we know that there is at least one solution to $Ax = b$ and since $r = n$ we know that the solution is unique.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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