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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