★ Unit 1 checkpoint

Exam 1

The Unit 1 exam covers everything from sessions 1 through 13 — solving Ax = b, elimination and the LU factorization, inverses, vector spaces and subspaces, and the column space and nullspace. Four multi-part problems test rank, the elimination matrices, complete solutions with a parameter, and reasoning about rref and matrix spaces.

▶ Review lecture Exam on OCW ↗

Taking this exam

This exam covers all of Unit 1 (sessions 1–13): elimination, matrix operations and inverses, $LU$ factorization, vector spaces and subspaces, and solving $A\mathbf{x} = \mathbf{b}$ via the column space and nullspace. Attempt it closed-book under exam conditions before revealing any solutions — each problem rewards careful reasoning, not memorized formulas. If you get stuck, or once you’re done, watch the unit review video linked above, where a course TA works through problems of exactly this kind.

Problems

Problem 1 Exam 1

This question is about an $m$ by $n$ matrix $A$ for which

A\mathbf{x} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \text{ has \textbf{no solutions}} \qquad \text{and} \qquad A\mathbf{x} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \text{ has \textbf{exactly one solution}.}

(a) Give all possible information about $m$ and $n$ and the rank $r$ of $A$ .

(b) Find all solutions to $A\mathbf{x} = \mathbf{0}$ and explain your answer.

(c) Write down an example of a matrix $A$ that fits the description in part (a).

Show solution

(a) Since $A\mathbf{x} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ has exactly one solution, the nullspace contains only the zero vector: $N(A) = \lbrace \mathbf{0} \rbrace$ , so $r = n$ (no free columns). Also $m = 3$ , since $A\mathbf{x}$ lives in $\mathbb{R}^3$ .

Since $A\mathbf{x} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ has no solution, the column space is not all of $\mathbb{R}^3$ : $C(A) \neq \mathbb{R}^3$ , so $r < m$ .

Putting these together, there are two possibilities:

m = 3, \quad r = n = 1 \qquad \text{and} \qquad m = 3, \quad r = n = 2.

(b) Since $N(A) = \lbrace \mathbf{0} \rbrace$ (because $A\mathbf{x} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ has exactly one solution), there is a unique solution to $A\mathbf{x} = \mathbf{0}$ , which is clearly $\mathbf{x} = \mathbf{0}$ . (It can be either $\mathbf{x} = \begin{bmatrix} 0 \end{bmatrix}$ or $\mathbf{x} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ , depending on whether $n = 1$ or $n = 2$ .)

(c) $A$ could be

A = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \qquad \text{or} \qquad A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}

(many more possibilities). Both have full column rank, so $A\mathbf{x} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ has exactly one solution, and neither has $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ in its column space.

Problem 2 Exam 1

The 3 by 3 matrix $A$ reduces to the identity matrix $I$ by the following three row operations (in order):

\begin{aligned} E_{21}&: \quad \text{Subtract } 4 \,(\text{row } 1) \text{ from row } 2. \\ E_{31}&: \quad \text{Subtract } 3 \,(\text{row } 1) \text{ from row } 3. \\ E_{23}&: \quad \text{Subtract row } 3 \text{ from row } 2. \end{aligned}

(a) Write the inverse matrix $A^{-1}$ in terms of the $E$ ‘s. Then compute $A^{-1}$ .

(b) What is the original matrix $A$ ?

(c) What is the lower triangular factor $L$ in $A = LU$ ?

Show solution

(a) The three operations, applied in order, carry $A$ to $I$ : $E_{23}E_{31}E_{21}A = I$ . So $A^{-1} = E_{23}E_{31}E_{21}$ — apply the three operations to $I$ :

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 \\ -4 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 \\ -4 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & -1 \\ -3 & 0 & 1 \end{bmatrix} = A^{-1}

(b) $A = E_{21}^{-1}E_{31}^{-1}E_{23}^{-1}$ — apply the inverse operations in reverse order to $I$ (add row 3 to row 2, add $3\,$ (row 1) to row 3, add $4\,$ (row 1) to row 2):

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 3 & 0 & 1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 1 \\ 3 & 0 & 1 \end{bmatrix} = A

Check:

\begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 1 \\ 3 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & -1 \\ -3 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(c) $L$ collects the inverses of the two elimination steps below the diagonal:

L = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix}

The multipliers $4$ and $3$ drop straight into place — and then $U = L^{-1}A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ is upper triangular, so $A = LU$ .

Problem 3 Exam 1

This 3 by 4 matrix depends on $c$ :

A = \begin{bmatrix} 1 & 1 & 2 & 4 \\ 3 & c & 2 & 8 \\ 0 & 0 & 2 & 2 \end{bmatrix}

(a) For each $c$ find a basis for the column space of $A$ .

(b) For each $c$ find a basis for the nullspace of $A$ .

(c) For each $c$ find the complete solution $\mathbf{x}$ to $A\mathbf{x} = \begin{bmatrix} 1 \\ c \\ 0 \end{bmatrix}$ .

Show solution

(a) Elimination (subtract $3\,$ (row 1) from row 2) gives

\begin{bmatrix} 1 & 1 & 2 & 4 \\ 0 & c-3 & -4 & -4 \\ 0 & 0 & 2 & 2 \end{bmatrix}

so there are two cases.

If $c \neq 3$ , then $c - 3$ is a pivot and

U = \begin{bmatrix} \boxed{1} & 1 & 2 & 4 \\ 0 & \boxed{c-3} & -4 & -4 \\ 0 & 0 & \boxed{2} & 2 \end{bmatrix} \longrightarrow R = \begin{bmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}

so a basis for $C(A)$ is the first three columns of $A$ :

\left\lbrace \begin{bmatrix} 1 \\ 3 \\ 0 \end{bmatrix},\ \begin{bmatrix} 1 \\ c \\ 0 \end{bmatrix},\ \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix} \right\rbrace

If $c = 3$ , then $c - 3 = 0$ and

U = \begin{bmatrix} \boxed{1} & 1 & 2 & 4 \\ 0 & 0 & \boxed{-4} & -4 \\ 0 & 0 & 0 & 0 \end{bmatrix} \longrightarrow R = \begin{bmatrix} 1 & 1 & 0 & 2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}

so take the first and third columns of $A$ as a basis for $C(A)$ :

\left\lbrace \begin{bmatrix} 1 \\ 3 \\ 0 \end{bmatrix},\ \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix} \right\rbrace

(b) If $c \neq 3$ , only $x_4$ is free and the special solution gives

N(A) = \left\lbrace x_4 \begin{bmatrix} -2 \\ 0 \\ -1 \\ 1 \end{bmatrix} \right\rbrace

If $c = 3$ , both $x_2$ and $x_4$ are free and the special solutions give

N(A) = \left\lbrace x_2 \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -2 \\ 0 \\ -1 \\ 1 \end{bmatrix} \right\rbrace

(c) By inspection, the right-hand side $\begin{bmatrix} 1 \\ c \\ 0 \end{bmatrix}$ is exactly the second column of $A$ , so

\mathbf{x}_p = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}

is one particular solution (other correct answers are possible). Adding the nullspace:

for $c \neq 3$ , the complete solution is

\mathbf{x} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -2 \\ 0 \\ -1 \\ 1 \end{bmatrix}

for $c = 3$ , the complete solution is

\mathbf{x} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -2 \\ 0 \\ -1 \\ 1 \end{bmatrix}

Problem 4 Exam 1

(a) If $A$ is a 3 by 5 matrix, what information do you have about the nullspace of $A$ ?

(b) Suppose row operations on $A$ lead to this matrix $R = \operatorname{rref}(A)$ :

R = \begin{bmatrix} 1 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}

Write all known information about the columns of $A$ .

(c) In the vector space $M$ of all 3 by 3 matrices (you could call this a matrix space), what subspace $S$ is spanned by all possible row reduced echelon forms $R$ ?

Show solution

(a) $N(A)$ is a subspace of $\mathbb{R}^5$ , and its dimension is $5 - r$ with $r \leq 3$ : so $N(A)$ has dimension at least $2$ (and at most $5$ ).

(b) The pivot columns of $R$ are columns $1$ , $4$ , $5$ — so columns $1$ , $4$ , $5$ of $A$ form a basis for $C(A)$ . From the free columns of $R$ :

\text{Column } 2 = 4 \times (\text{Column } 1), \qquad \text{Column } 3 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

(c) Every rref matrix is upper triangular, and the span of all of them is exactly

S = \left\lbrace \begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{bmatrix} \right\rbrace,

the set of upper triangular matrices. (A basis of six echelon forms is

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix},\ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix},\ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix},\ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix},\ \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix},\ \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}.)