Matrix times vector

Day: 4 of 40
Length: 2:28
Textbook: Lay §1.4
Syllabus: Groundwork before Lesson 10

Ax is a linear combination of the columns of A, using the entries of x as the weights.

On Day 2, asking whether $\teal{\mathbf b}$ is in the span of some vectors turned into a linear system, and yesterday you learned to solve that system by row reduction. Writing the question out takes a line per equation, so today we write it as one short equation, $A\mathbf x = \teal{\mathbf b}$ . The vectors stand side by side as the columns of a matrix $A$ , and the weights stack into a vector $\mathbf x$ . This notation also sets up Day 5, where $A\mathbf x$ becomes a rule that moves every point of the plane.

Each section below opens with a problem. Try it before you open the walkthrough, even if you only get partway.

Matrix times vector

Let $\yellow{\mathbf a_1} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$ and $\blue{\mathbf a_2} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$ . Compute the linear combination $2\yellow{\mathbf a_1} + 1\blue{\mathbf a_2}$ .

Walk through the solution

Scale each vector by its weight, then add entry by entry, exactly as on Day 1:

2\begin{bmatrix} 3 \\ 2 \end{bmatrix} + 1\begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 6 \\ 4 \end{bmatrix} + \begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}.

The whole calculation depends on two lists: the vectors $\yellow{\mathbf a_1}, \blue{\mathbf a_2}$ and the weights $2, 1$ . Matrix notation writes those two lists next to each other. Put the vectors side by side as columns and stack the weights into a column:

\begin{bmatrix} \yellow{3} & \blue{-1} \\ \yellow{2} & \blue{2} \end{bmatrix}\begin{bmatrix} \yellow{2} \\ \blue{1} \end{bmatrix} = \teal{\begin{bmatrix} 5 \\ 6 \end{bmatrix}}.

The left side is a new name for the same combination.

Matrix times vector

If $A$ is an $m\times n$ matrix with columns $\mathbf a_1, \dots, \mathbf a_n$ and $\mathbf x$ is in $\mathbb{R}^n$ , then the product $A\mathbf x$ is the linear combination of the columns of $A$ that uses the entries of $\mathbf x$ as weights:

A\mathbf x = \begin{bmatrix} \mathbf a_1 & \mathbf a_2 & \cdots & \mathbf a_n \end{bmatrix}\begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = x_1\mathbf a_1 + x_2\mathbf a_2 + \cdots + x_n\mathbf a_n.

The product is defined only when $A$ has as many columns as $\mathbf x$ has entries, because each column needs exactly one weight. The answer has $m$ entries, one for each row of $A$ .

The yellow column a1 = (3, 2) stretched to 2a1 = (6, 4), the blue column a2 = (−1, 2) placed at its tip, and the teal arrow Ax ending at (5, 6), marked with an orange dot.

\mathbf x

The translation also runs backward. Any linear combination can be written as a matrix times a vector.

Write $2\begin{bmatrix} 1 \\ 4 \end{bmatrix} - 3\begin{bmatrix} 2 \\ -1 \end{bmatrix} + \begin{bmatrix} 0 \\ 5 \end{bmatrix}$ as $A\mathbf x$ . What are $A$ and $\mathbf x$ ? Enter $\mathbf x$ .

Walk through the solution

The vectors become the columns of $A$ , in order, and the weights become the entries of $\mathbf x$ . The last vector has no number in front, so its weight is $1$ , and the minus sign in front of the $3$ makes the second weight $-3$ :

\begin{bmatrix} 1 & 2 & 0 \\ 4 & -1 & 5 \end{bmatrix}\begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}.

$A$ is $2 \times 3$ because there are three vectors with two entries each. The product equals $(2 - 6 + 0,\ 8 + 3 + 5) = (-4, 16)$ .

Drag the weights until $A\mathbf x$ lands on the target. Look for a pair of weights that you can read straight off the columns.

Set the weights in

\mathbf x

so that

A\mathbf x

lands on the ringed point

\begin{bmatrix} -1 \\ 2 \end{bmatrix}

x_1 = 1

x_2 = 1

\begin{bmatrix} \textcolor{#FFD447}{3} & \textcolor{#5BC0EB}{-1} \\ \textcolor{#FFD447}{2} & \textcolor{#5BC0EB}{2} \end{bmatrix}\begin{bmatrix} \textcolor{#FFD447}{1} \\ \textcolor{#5BC0EB}{1} \end{bmatrix} = \textcolor{#3DD6B5}{\begin{bmatrix} 2 \\ 4 \end{bmatrix}}

= 1\textcolor{#FFD447}{\begin{bmatrix} 3 \\ 2 \end{bmatrix}} + 1\textcolor{#5BC0EB}{\begin{bmatrix} -1 \\ 2 \end{bmatrix}}

Which product is defined?

The row rule

Scaling whole columns explains what $A\mathbf x$ means, but it is slow to write out by hand. This section finds a faster way to compute one entry at a time.

Compute $\begin{bmatrix} 3 & -1 \\ 2 & 2 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix}$ by columns, but do not simplify the entries. Which numbers from $A$ does the top entry use?

Walk through the solution

By columns, the product is

2\begin{bmatrix} 3 \\ 2 \end{bmatrix} + 1\begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \cdot 3 + 1 \cdot (-1) \\ 2 \cdot 2 + 1 \cdot 2 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}.

The top entry is $3 \cdot 2 + (-1) \cdot 1$ . The numbers $3$ and $-1$ are exactly the top row of $A$ , and they are paired with the entries $2$ and $1$ of $\mathbf x$ in order. The bottom entry, $2 \cdot 2 + 2 \cdot 1$ , uses only the bottom row. Each entry of the answer comes from one row of $A$ .

Top line: 2 times (3, 2) plus 1 times (−1, 2) written as one column with entries 2·3 + 1·(−1) and 2·2 + 1·2, equal to (5, 6). Bottom line: the same product computed by rows, with row 1 of A, the vector x and the entry 3·2 + (−1)·1 boxed in orange.

The row rule

Entry $i$ of $A\mathbf x$ is the sum of the products of matching entries from row $i$ of $A$ and from $\mathbf x$ . For a $2 \times 2$ matrix,

\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} a_{11}x_1 + a_{12}x_2 \\ a_{21}x_1 + a_{22}x_2 \end{bmatrix}.

The row rule is usually faster by hand, and the column picture explains what the answer means. They always agree, so each one checks the other.

The identity matrix $I$ has $1$ s down its diagonal and $0$ s elsewhere. By the row rule, $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 4 \\ -7 \end{bmatrix} = \begin{bmatrix} 4 + 0 \\ 0 - 7 \end{bmatrix} = \begin{bmatrix} 4 \\ -7 \end{bmatrix}$ , and in general $I\mathbf x = \mathbf x$ .

The row rule gives a second picture of $A\mathbf x = \mathbf b$ . Each row says one entry of $A\mathbf x$ must match one entry of $\mathbf b$ , and the inputs $\mathbf x$ that make it match form a line, just like the equations on Day 3.

Drag

\mathbf x

so that both rows of

A\mathbf x

match

\mathbf b = \begin{bmatrix} 7 \\ 2 \end{bmatrix}

. Each colored line shows where one row matches.

\begin{aligned} \textcolor{#7BC96F}{3}(0) + (\textcolor{#7BC96F}{-1})(0) &= 0 & \phantom{\checkmark} \\ \textcolor{#F2665E}{2}(0) + (\textcolor{#F2665E}{2})(0) &= 0 & \phantom{\checkmark} \end{aligned}

A\mathbf x = \begin{bmatrix} \textcolor{#7BC96F}{0} \\ \textcolor{#F2665E}{0} \end{bmatrix}

One problem in three forms

Reading $A\mathbf x$ backwards turns a question into an equation. The widget above solved $A\mathbf x = (7, 2)$ , and the next problem writes that same question three ways.

Write the system $3x_1 - x_2 = 7$ , $\ 2x_1 + 2x_2 = 2$ as a vector equation and as a matrix equation. Then solve it.

Walk through the solution

Collect the coefficients of $x_1$ into one vector and the coefficients of $x_2$ into another, as on Day 2:

x_1\begin{bmatrix} 3 \\ 2 \end{bmatrix} + x_2\begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 7 \\ 2 \end{bmatrix}.

By the definition of $A\mathbf x$ , the left side is the matrix with those columns times $\mathbf x$ :

\begin{bmatrix} 3 & -1 \\ 2 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 7 \\ 2 \end{bmatrix}.

All three forms have the same augmented matrix, so one row reduction solves them all. Halve row 2 to get $x_1 + x_2 = 1$ , then add it to row 1 to cancel $x_2$ : $4x_1 = 8$ , so $x_1 = 2$ and $x_2 = -1$ .

The system 3x1 − x2 = 7, 2x1 + 2x2 = 2, the vector equation x1(3, 2) + x2(−1, 2) = (7, 2) and the matrix equation, stacked with the same colors. Beside them the augmented matrix row reduces to the identity with last column (2, −1), giving x = (2, −1).

Theorem 3

If $A$ is an $m\times n$ matrix with columns $\mathbf a_1, \dots, \mathbf a_n$ and $\mathbf b$ is in $\mathbb{R}^m$ , then the matrix equation $A\mathbf x = \mathbf b$ has the same solution set as the vector equation $x_1\mathbf a_1 + \cdots + x_n\mathbf a_n = \mathbf b$ . Both have the same solution set as the linear system whose augmented matrix is $\begin{bmatrix} \mathbf a_1 & \cdots & \mathbf a_n & \mathbf b \end{bmatrix}$ .

In particular, $A\mathbf x = \mathbf b$ has a solution exactly when $\mathbf b$ is a linear combination of the columns of $A$ .

Solve $\begin{bmatrix} 3 & -1 \\ 2 & 2 \end{bmatrix}\mathbf x = \begin{bmatrix} 1 \\ 6 \end{bmatrix}$ by row reducing the augmented matrix.

Walk through the solution

1
Write $\begin{bmatrix} A & \mathbf b \end{bmatrix}$ . Row 2 has all even entries, so halve it, and then swap the rows so the pivot $1$ sits on top. This avoids fractions.
$\begin{bmatrix} 3 & -1 & 1 \\ 2 & 2 & 6 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 3 \\ 3 & -1 & 1 \end{bmatrix}$
2
3

Check the answer on the column picture.

Set the weights in

\mathbf x

so that

A\mathbf x

lands on the ringed point

\begin{bmatrix} 1 \\ 6 \end{bmatrix}

x_1 = 1

x_2 = 1

\begin{bmatrix} \textcolor{#FFD447}{3} & \textcolor{#5BC0EB}{-1} \\ \textcolor{#FFD447}{2} & \textcolor{#5BC0EB}{2} \end{bmatrix}\begin{bmatrix} \textcolor{#FFD447}{1} \\ \textcolor{#5BC0EB}{1} \end{bmatrix} = \textcolor{#3DD6B5}{\begin{bmatrix} 2 \\ 4 \end{bmatrix}}

= 1\textcolor{#FFD447}{\begin{bmatrix} 3 \\ 2 \end{bmatrix}} + 1\textcolor{#5BC0EB}{\begin{bmatrix} -1 \\ 2 \end{bmatrix}}

Spanning every target

Sometimes no weights work. The next problem finds out exactly which targets a matrix can reach.

Let $A = \begin{bmatrix} 1 & 2 \\ -3 & -6 \end{bmatrix}$ . For which $\mathbf b = (b_1, b_2)$ does $A\mathbf x = \mathbf b$ have a solution?

Walk through the solution

We cannot row reduce numbers we do not know, but we can carry $b_1$ and $b_2$ along as symbols. Clear the $-3$ with $R_2 \leftarrow R_2 + 3R_1$ . Row 2 becomes $(-3 + 3,\ -6 + 6,\ b_2 + 3b_1)$ :

\begin{bmatrix} 1 & 2 & b_1 \\ -3 & -6 & b_2 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & b_1 \\ 0 & 0 & 3b_1 + b_2 \end{bmatrix}

Row 2 now says $0 = 3b_1 + b_2$ . If $3b_1 + b_2 \neq 0$ , that is false and there is no solution. If $3b_1 + b_2 = 0$ , it is true, and row 1 can be solved. So $A\mathbf x = \mathbf b$ has a solution exactly when $b_2 = -3b_1$ .

The points with $b_2 = -3b_1$ form a line through the origin. That makes sense, because the second column $(2, -6)$ is twice the first column $(1, -3)$ , so every combination is a multiple of $(1, -3)$ .

Drag $\mathbf b$ and watch the last entry of the reduced matrix. It is zero exactly when $\mathbf b$ is on the line.

Drag

\mathbf b

to a point other than the origin where

A\mathbf x = \mathbf b

has a solution.

\begin{bmatrix} A & \mathbf b \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 2 \\ 0 & 0 & \textcolor{#E8ECF1}{7} \end{bmatrix}

\text{last entry} = 3b_1 + b_2

The yellow column a1 = (3, 2) and the blue column a2 = (6, 4) lie on one teal line through the origin. The point b = (7, 2), ringed in orange, sits below the line.

\yellow{(3, 2)}

Let $A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 0 & 1 & -1 \end{bmatrix}$ . Find the condition on $b_1, b_2, b_3$ for $A\mathbf x = \mathbf b$ to have a solution. Do three columns in $\mathbb{R}^3$ reach everything?

Walk through the solution

Carry the $b$ 's along again. $R_2 \leftarrow R_2 - 2R_1$ clears column 1, and then $R_3 \leftarrow R_3 - R_2$ clears below the second pivot. Row 3 becomes $(0,\ 1 - 1,\ -1 - (-1),\ b_3 - (b_2 - 2b_1))$ :

\begin{bmatrix} 1 & 0 & 2 & b_1 \\ 2 & 1 & 3 & b_2 \\ 0 & 1 & -1 & b_3 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 2 & b_1 \\ 0 & 1 & -1 & b_2 - 2b_1 \\ 0 & 1 & -1 & b_3 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 2 & b_1 \\ 0 & 1 & -1 & b_2 - 2b_1 \\ 0 & 0 & 0 & 2b_1 - b_2 + b_3 \end{bmatrix}

The system has a solution exactly when $2b_1 - b_2 + b_3 = 0$ . That equation describes a plane through the origin, so the three columns reach only a plane. For example, $\mathbf b = (1, 0, 0)$ gives $2 \neq 0$ and has no solution.

The trouble is the row of zeros in the echelon form of $A$ . Only two of its three rows have a pivot.

Theorem 4

Let $A$ be an $m\times n$ matrix. These statements are either all true or all false:

(a) For each $\mathbf b$ in $\mathbb{R}^m$ , the equation $A\mathbf x = \mathbf b$ has a solution.

(b) Each $\mathbf b$ in $\mathbb{R}^m$ is a linear combination of the columns of $A$ .

(d) $A$ has a pivot position in every row.

Statements (a), (b) and (c) say the same thing in three vocabularies. Statement (d) connects them to row reduction, and Day 12 returns to it as the standard test for whether columns span. If every row of $A$ has a pivot, the augmented column can never hold a pivot, so every system is consistent. If some row of $A$ has no pivot, Problem 7 shows how a target can be missed.

Slide

k

until some row of the echelon form has no pivot. The orange entries are the pivots.

k = 0

A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 5 & \textcolor{#5BC0EB}{0} \\ 1 & 3 & 3 \end{bmatrix}

\text{rows with a pivot: } 3 \text{ of } 3

Linearity

Because the entries of $\mathbf x$ are weights on the columns, the product treats sums and multiples of $\mathbf x$ in a simple way.

Let $A = \begin{bmatrix} 2 & 5 \\ 3 & 1 \end{bmatrix}$ , $\mathbf u = \begin{bmatrix} 4 \\ -1 \end{bmatrix}$ and $\mathbf v = \begin{bmatrix} -3 \\ 5 \end{bmatrix}$ . Compute $A\mathbf u$ , $A\mathbf v$ and $A(\mathbf u + \mathbf v)$ . Enter $A(\mathbf u + \mathbf v)$ .

Walk through the solution

Use the row rule for each product.

\begin{aligned} A\mathbf u &= \begin{bmatrix} 2 \cdot 4 + 5 \cdot (-1) \\ 3 \cdot 4 + 1 \cdot (-1) \end{bmatrix} = \begin{bmatrix} 3 \\ 11 \end{bmatrix}, \\ A\mathbf v &= \begin{bmatrix} 2 \cdot (-3) + 5 \cdot 5 \\ 3 \cdot (-3) + 1 \cdot 5 \end{bmatrix} = \begin{bmatrix} 19 \\ -4 \end{bmatrix}. \end{aligned}

For the third, first add: $\mathbf u + \mathbf v = (1, 4)$ . Then $A(\mathbf u + \mathbf v) = (2 + 20,\ 3 + 4) = (22, 7)$ .

Now compare. $A\mathbf u + A\mathbf v = (3 + 19,\ 11 - 4) = (22, 7)$ , the same vector. Multiplying the sum gives the same answer as adding the products.

Matrix times vector is linear

For any $A$ , any $\mathbf u, \mathbf v$ and any scalar $c$ ,

A(\mathbf u + \mathbf v) = A\mathbf u + A\mathbf v, \qquad A(c\,\mathbf u) = c\,(A\mathbf u).

The reason is that the entries of $\mathbf x$ are the weights on the columns. For two columns,

\begin{aligned} A(\mathbf u + \mathbf v) &= (u_1 + v_1)\mathbf a_1 + (u_2 + v_2)\mathbf a_2 \\ &= (u_1\mathbf a_1 + u_2\mathbf a_2) + (v_1\mathbf a_1 + v_2\mathbf a_2) = A\mathbf u + A\mathbf v, \\ A(c\,\mathbf u) &= (c\,u_1)\mathbf a_1 + (c\,u_2)\mathbf a_2 = c\,(u_1\mathbf a_1 + u_2\mathbf a_2) = c\,(A\mathbf u). \end{aligned}

The same regrouping works for any number of columns.

Tomorrow these two rules become the definition of a linear transformation, and $A\mathbf x$ becomes a way to move the whole plane at once.

Practice

Warm-up

Is $\begin{bmatrix} 2 & 1 \\ 0 & 3 \\ -1 & 4 \end{bmatrix}\begin{bmatrix} 1 \\ 5 \\ 2 \end{bmatrix}$ defined? (Modeled on Lay §1.4 #1.)

Compute $\begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix}\begin{bmatrix} 4 \\ 1 \end{bmatrix}$ .

Show the solution

By rows, the first entry is $1 \cdot 4 + (-2) \cdot 1 = 2$ and the second is $3 \cdot 4 + 0 \cdot 1 = 12$ . By columns, $4(1, 3) + 1(-2, 0) = (4 - 2,\ 12 + 0) = (2, 12)$ , which agrees.

Let $\mathbf a_1 = (2, 1)$ , $\mathbf a_2 = (0, 3)$ and $\mathbf a_3 = (-1, 1)$ . Write $4\mathbf a_1 - \mathbf a_2 + 2\mathbf a_3$ as $A\mathbf x$ with $A = \begin{bmatrix} \mathbf a_1 & \mathbf a_2 & \mathbf a_3 \end{bmatrix}$ , and enter $\mathbf x$ . (Modeled on Lay §1.4, Example 2.)

Show the solution

The weights become the entries of $\mathbf x$ , in the order of the columns, so $\mathbf x = (4, -1, 2)$ :

\begin{bmatrix} 2 & 0 & -1 \\ 1 & 3 & 1 \end{bmatrix}\begin{bmatrix} 4 \\ -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 8 - 0 - 2 \\ 4 - 3 + 2 \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \end{bmatrix}.

Core

Compute $\begin{bmatrix} 3 & -1 & 2 \\ 0 & 4 & -5 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}$ as a combination of the columns, then check it with the row rule. (Modeled on Lay §1.4 #3.)

Show a hint

There are three columns, so there are three scaled vectors to add. Watch the sign of the third weight.

Show the solution

2\begin{bmatrix} 3 \\ 0 \end{bmatrix} + 1\begin{bmatrix} -1 \\ 4 \end{bmatrix} - 1\begin{bmatrix} 2 \\ -5 \end{bmatrix} = \begin{bmatrix} 6 \\ 0 \end{bmatrix} + \begin{bmatrix} -1 \\ 4 \end{bmatrix} + \begin{bmatrix} -2 \\ 5 \end{bmatrix} = \begin{bmatrix} 3 \\ 9 \end{bmatrix}

By rows, the first entry is $3\cdot 2 + (-1)\cdot 1 + 2\cdot(-1) = 3$ and the second is $0\cdot 2 + 4\cdot 1 + (-5)\cdot(-1) = 9$ .

Solve $\begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}\mathbf x = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$ by row reducing the augmented matrix. (Modeled on Lay §1.4 #11.)

Show a hint

The top left entry is already $1$ . Clear below it first.

Show the solution

\begin{aligned} \begin{bmatrix} 1 & 2 & 4 \\ 3 & -1 & 5 \end{bmatrix} &\sim \begin{bmatrix} 1 & 2 & 4 \\ 0 & -7 & -7 \end{bmatrix} && R_2 - 3R_1 \\ &\sim \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \end{bmatrix} && R_2 \div (-7),\ R_1 - 2R_2 \end{aligned}

So $\mathbf x = (2, 1)$ . Check: $2\,(1, 3) + 1\,(2, -1) = (4, 5)$ .

You are told that $\begin{bmatrix} 2 & 1 & 0 \\ 1 & -1 & 3 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} = \begin{bmatrix} 4 \\ -4 \end{bmatrix}$ . Without any row operations, find weights $c_1, c_2, c_3$ with $\begin{bmatrix} 4 \\ -4 \end{bmatrix} = c_1\begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_3\begin{bmatrix} 0 \\ 3 \end{bmatrix}$ . (Modeled on Lay §1.4 #25.)

Show a hint

Read the given product with the definition of $A\mathbf x$ . What are the weights on the columns?

Show the solution

The product $A\mathbf x$ is the combination of the columns of $A$ with the entries of $\mathbf x$ as weights, so $c_1 = 1$ , $c_2 = 2$ and $c_3 = -1$ . Check: $(2, 1) + (2, -2) + (0, -3) = (4, -4)$ .

Let $A = \begin{bmatrix} 2 & -1 \\ -4 & 2 \end{bmatrix}$ . For which $h$ does $A\mathbf x = \begin{bmatrix} 3 \\ h \end{bmatrix}$ have a solution? (Modeled on Lay §1.4 #15.)

Show a hint

Row reduce $\begin{bmatrix} A & \mathbf b \end{bmatrix}$ with $h$ carried along, and look at the last entry of row 2.

Show the solution

$R_2 \leftarrow R_2 + 2R_1$ gives row 2 as $(-4 + 4,\ 2 - 2,\ h + 6) = (0, 0, h + 6)$ . That row says $0 = h + 6$ , which holds only when $h = -6$ . The columns are parallel, so they reach only the line $b_2 = -2b_1$ , and $(3, -6)$ is on it.

A $4 \times 4$ matrix $A$ has an echelon form with pivots in only three rows. Do the columns of $A$ span $\mathbb{R}^4$ ? (Modeled on Lay §1.4 #17.)

Challenge

True or false: if $\begin{bmatrix} A & \mathbf b \end{bmatrix}$ has a pivot position in every row, then $A\mathbf x = \mathbf b$ has a solution. (Modeled on Lay §1.4 #23.)

Let $A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 0 & 1 & -1 \end{bmatrix}$ from Problem 7. Check that $\mathbf b = (3, 7, 1)$ satisfies the condition, and then find the solution of $A\mathbf x = \mathbf b$ with $x_3 = 0$ .

Show a hint

Put $b_1 = 3$ , $b_2 = 7$ , $b_3 = 1$ into the reduced matrix from Problem 7. Column 3 has no pivot, so $x_3$ is free.

Show the solution

The condition is $2b_1 - b_2 + b_3 = 6 - 7 + 1 = 0$ , so a solution exists. The reduced rows become $x_1 + 2x_3 = 3$ and $x_2 - x_3 = 7 - 6 = 1$ . With $x_3 = 0$ they give $x_1 = 3$ and $x_2 = 1$ .

Check: $3(1, 2, 0) + 1(0, 1, 1) + 0(2, 3, -1) = (3, 7, 1)$ . Every other value of $x_3$ gives another solution, $x_1 = 3 - 2x_3$ and $x_2 = 1 + x_3$ .

Suppose $A\mathbf u = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $A\mathbf v = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$ . You do not know $A$ , $\mathbf u$ or $\mathbf v$ . Find $A(2\mathbf u - \mathbf v)$ .

Show a hint

Split $A(2\mathbf u - \mathbf v)$ with the two linearity rules.

Show the solution

By linearity, $A(2\mathbf u - \mathbf v) = A(2\mathbf u) + A(-\mathbf v) = 2A\mathbf u - A\mathbf v$ . So

2\begin{bmatrix} 1 \\ 2 \end{bmatrix} - \begin{bmatrix} 3 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 - 3 \\ 4 + 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 5 \end{bmatrix}.

Knowing what $A$ does to $\mathbf u$ and $\mathbf v$ is enough to know what it does to every combination of them. Tomorrow's lesson builds on exactly this fact.