Linear-Algebra-Done-Right · Unit 6 · Video 3 · Interactive Practice

Rows, Columns and Linear Combinations in Matrix Products

IKey Formulas

Formula	Name	Sizes
$(AB)_{j,k} = A_{j,\cdot}\,B_{\cdot,k}$	Entry = row times column	$A$ is $m \times n$ , $B$ is $n \times p$
$(AB)_{\cdot,k} = A\,B_{\cdot,k}$ and $(AB)_{j,\cdot} = A_{j,\cdot}\,B$	Column / row of a product	$m \times 1$ and $1 \times p$
$Ab = b_1 A_{\cdot,1} + \cdots + b_n A_{\cdot,n}$	Matrix times column	$b$ is $n \times 1$
$aB = a_1 B_{1,\cdot} + \cdots + a_n B_{n,\cdot}$	Row times matrix	$a$ is $1 \times n$

Key Insight: Column $k$ of $CR$ is a linear combination of the columns of $C$ with coefficients from column $k$ of $R$ ; row $j$ of $CR$ is a linear combination of the rows of $R$ with coefficients from row $j$ of $C$ .

IIEntry = Row Times Column

Each entry $(AB)_{j,k}$ is the $1 \times 1$ product of row $j$ of $A$ and column $k$ of $B$ .

Row j 2 Column k 1

IIIMatrix Times Column, Row Times Matrix

With $A = \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix}$ , $Ab$ combines the columns of $A$ and $aA$ combines its rows.

Product b₁ 2 b₂ 1

IVColumns and Rows of CR

Column $k$ of $CR$ uses only column $k$ of $R$ , as coefficients on the columns of $C$ ; rows mirror this.

View k 2 R₁,₂ 5 R₂,₂ 1

💡 So every column of $CR$ lies in the span of the columns of $C$ , and every row of $CR$ lies in the span of the rows of $R$ — the fact the next video builds on.

VQuiz Questions

Problem 1 · Change One Column of B

Given: $A = \begin{pmatrix} 1 & 2 \\ 0 & -1 \\ 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 & 0 \\ 1 & -1 & 4 \end{pmatrix}$ . The entry $B_{2,2}$ is changed from $-1$ to $2$ , so column $2$ of $B$ becomes $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ — decide which entries of $AB$ can change and find the new column $2$ of $AB$ .

Which entries of $AB$ can change?

The new column $2$ of $AB$ is…

✅ Correct!

(AB)_{\cdot,2} = A\,B_{\cdot,2} = 1 \cdot A_{\cdot,1} + 2 \cdot A_{\cdot,2}

, and columns

1

and

3

AB

never use column

2

B

❌ Rows of $B$ are not rows of $AB$ . Row

j

AB

A_{j,\cdot}\,B

, which uses all of

B

. The changed entry sits in column

2

B

, and only column

2

AB

, which is

A\,B_{\cdot,2}

, uses that column.

❌ Too narrow. Every entry of column

2

(AB)_{j,2} = A_{j,\cdot}\,B_{\cdot,2}

, contains the term

A_{j,2}\,B_{2,2}

, and here

A_{1,2} = 2

A_{2,2} = -1

A_{3,2} = 1

are all nonzero.

❌ Too broad. Column

k

AB

A\,B_{\cdot,k}

, which uses only column

k

B

. Columns

1

and

3

B

are unchanged, so columns

1

and

3

AB

are too.

❌ Coefficients on the wrong columns. That is

2A_{\cdot,1} + A_{\cdot,2}

, which is column

1

AB

. The first entry of the new

B_{\cdot,2}

, which is

1

, multiplies the first column of

A

❌ That is the old column.

A\begin{pmatrix} 1 \\ -1 \end{pmatrix} = A_{\cdot,1} - A_{\cdot,2}

is column

2

AB

before the change; use the new column

\begin{pmatrix} 1 \\ 2 \end{pmatrix}

❌ Only the middle entry was updated. All three entries of column

2

use

B_{2,2}

, so all three change, not just

(AB)_{2,2}

Show solution

Column $k$ of $AB$ is $A\,B_{\cdot,k}$ , so it depends on column $k$ of $B$ and on nothing else in $B$ . Changing $B_{2,2}$ changes column $2$ of $B$ only, so only column $2$ of $AB$ can change. Its new value is a linear combination of the columns of $A$ with coefficients $1$ and $2$ :

A\begin{pmatrix} 1 \\ 2 \end{pmatrix} = 1\begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix} + 2\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix} + \begin{pmatrix} 4 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -2 \\ 5 \end{pmatrix}.

Before the change it was $A_{\cdot,1} - A_{\cdot,2} = \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}$ , so all three entries of column $2$ did change, while columns $1$ and $3$ of $AB$ , $\begin{pmatrix} 4 \\ -1 \\ 7 \end{pmatrix}$ and $\begin{pmatrix} 8 \\ -4 \\ 4 \end{pmatrix}$ , stay the same.

Problem 2 · A Column of a Product

Given: $A = \begin{pmatrix} 1 & -2 \\ 0 & 3 \\ 2 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 4 & 1 & -1 \\ 2 & 0 & 3 \end{pmatrix}$ — find $(AB)_{\cdot,3}$ , column $3$ of $AB$ .

✅ Correct!

(AB)_{\cdot,3} = A\,B_{\cdot,3} = -1 \cdot A_{\cdot,1} + 3 \cdot A_{\cdot,2}

❌ Coefficients on the wrong columns. That is

3A_{\cdot,1} - A_{\cdot,2}

. The first entry of

B_{\cdot,3}

, which is

-1

, multiplies the first column of

A

❌ Wrong column of $B$ . That is

A\,B_{\cdot,1}

, column

1

AB

. Column

3

AB

needs column

3

B

, which is

\begin{pmatrix} -1 \\ 3 \end{pmatrix}

❌ Check the sign. That is

A_{\cdot,1} + 3A_{\cdot,2}

; the coefficient on

A_{\cdot,1}

B_{1,3} = -1

, not

1

Show solution

Column $3$ of $AB$ is $A$ times column $3$ of $B$ , and a matrix times a column is a linear combination of the matrix's columns, with coefficients from that column:

(AB)_{\cdot,3} = A\begin{pmatrix} -1 \\ 3 \end{pmatrix} = -1\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + 3\begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ -2 \end{pmatrix} + \begin{pmatrix} -6 \\ 9 \\ 3 \end{pmatrix} = \begin{pmatrix} -7 \\ 9 \\ 1 \end{pmatrix}.

Only column $3$ of $B$ is needed; the rest of $B$ never enters.

Problem 3 · A Row Times a Matrix

Given: $a = \begin{pmatrix} 2 & -1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ 0 & 2 \\ -1 & 1 \end{pmatrix}$ — describe $aB$ and compute it.

$aB$ is a linear combination of…

$aB =$

✅ Correct!

aB = 2B_{1,\cdot} - B_{2,\cdot} + 3B_{3,\cdot} = \begin{pmatrix} -1 & 9 \end{pmatrix}

❌ Not quite.

a

has three entries and

B

has three rows (but only two columns): in

aB = a_1 B_{1,\cdot} + a_2 B_{2,\cdot} + a_3 B_{3,\cdot}

, entry

i

of the row

a

multiplies row

i

B

❌ Check the sign.

a_2 = -1

, so row

2

B

is subtracted:

-\begin{pmatrix} 0 & 2 \end{pmatrix}

, not

+\begin{pmatrix} 0 & 2 \end{pmatrix}

❌ Coefficients reversed. That is

3B_{1,\cdot} - B_{2,\cdot} + 2B_{3,\cdot}

. The first entry of

a

multiplies the first row of

B

❌ Check $B_{3,1}$ . It is

-1

, so

3B_{3,\cdot} = \begin{pmatrix} -3 & 3 \end{pmatrix}

and the first entry is

2 - 0 - 3 = -1

Show solution

A $1 \times 3$ row times a $3 \times 2$ matrix is a linear combination of the three rows of the matrix, with coefficients from the row:

aB = 2\begin{pmatrix} 1 & 4 \end{pmatrix} - 1\begin{pmatrix} 0 & 2 \end{pmatrix} + 3\begin{pmatrix} -1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 8 \end{pmatrix} + \begin{pmatrix} 0 & -2 \end{pmatrix} + \begin{pmatrix} -3 & 3 \end{pmatrix} = \begin{pmatrix} -1 & 9 \end{pmatrix}.

Check one entry with row times column: entry $2$ of $aB$ is $a\,B_{\cdot,2} = 2 \cdot 4 + (-1) \cdot 2 + 3 \cdot 1 = 9$ .

Problem 4 · Columns and Rows of CR

Given: $C = \begin{pmatrix} 1 & 2 \\ 0 & -1 \\ 3 & 1 \\ 2 & 0 \end{pmatrix}$ with columns $u = C_{\cdot,1}$ , $v = C_{\cdot,2}$ , and $R = \begin{pmatrix} 2 & 0 & -1 \\ 1 & 3 & 4 \end{pmatrix}$ with rows $r_1 = R_{1,\cdot}$ , $r_2 = R_{2,\cdot}$ — express column $3$ and row $3$ of $CR$ .

Column $3$ of $CR$ is…

Row $3$ of $CR$ is…

✅ Correct! Column

3

takes its coefficients

-1, 4

from column

3

R

; row

3

takes its coefficients

3, 1

from row

3

C

❌ Not quite. Column

k

CR

R_{1,k}\,u + R_{2,k}\,v

: the coefficients are column

k

R

. That option uses a different column of

R

❌ Wrong column of $R$ .

2u + v

uses column

1

R

, so it is column

1

CR

❌ Coefficients swapped.

R_{1,3} = -1

multiplies

u = C_{\cdot,1}

and

R_{2,3} = 4

multiplies

v = C_{\cdot,2}

❌ Not quite. Row

j

CR

C_{j,1}\,r_1 + C_{j,2}\,r_2

: the coefficients are row

j

C

. That option uses row

2

C

, which is

\begin{pmatrix} 0 & -1 \end{pmatrix}

❌ Coefficients swapped.

C_{3,1} = 3

multiplies

r_1

and

C_{3,2} = 1

multiplies

r_2

❌ Wrong row of $C$ .

r_1 + 2r_2

uses row

1

C

, so it is row

1

CR

Show solution

Column 3. $(CR)_{\cdot,3} = C\,R_{\cdot,3} = R_{1,3}\,C_{\cdot,1} + R_{2,3}\,C_{\cdot,2} = -u + 4v$ :

-\begin{pmatrix} 1 \\ 0 \\ 3 \\ 2 \end{pmatrix} + 4\begin{pmatrix} 2 \\ -1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 7 \\ -4 \\ 1 \\ -2 \end{pmatrix}.

Row 3. $(CR)_{3,\cdot} = C_{3,\cdot}\,R = C_{3,1}\,r_1 + C_{3,2}\,r_2 = 3r_1 + r_2$ :

3\begin{pmatrix} 2 & 0 & -1 \end{pmatrix} + \begin{pmatrix} 1 & 3 & 4 \end{pmatrix} = \begin{pmatrix} 7 & 3 & 1 \end{pmatrix}.

Both agree with entry $(CR)_{3,3} = C_{3,\cdot}\,R_{\cdot,3} = 3 \cdot (-1) + 1 \cdot 4 = 1$ .

Solved: 0 / 4