3×3 Matrix Multiplication

Linear Algebra · Self-Study Workbook Vol. 2

3×3 & Beyond
Matrix Multiplication

20 Questions · Multiple Choice · Instant Feedback ✏️

Name: ______________________ Date: __________

ROW × COL

Each entry = dot product of one row from A and one col from B

i·j RULE

Entry \(C_{ij}\) uses row i of A and col j of B — always!

SUM of 3

For 3×3: each entry = 3 products added together

AB ≠ BA

Swapping order gives a different result — order always matters!

ZERO ≠ SAFE

A×B can = zero matrix even when A and B are not zero!

DIMS FIRST

Always check dimensions before computing. Inner must match!

🔑 THE KEY FORMULA FOR 3×3

For \(C = AB\) where \(A,B\) are \(3\times3\):
\(\displaystyle C_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + a_{i3}b_{3j} = \sum_{k=1}^{3} a_{ik}b_{kj}\)
Example: \(C_{12}\) = (row 1 of A) · (col 2 of B) = \(a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32}\)

0 of 20 answered

🗺️ Section 1: Reading the 3×3 Grid Start Here

✏️ HOW TO READ AN ENTRY

Given \(A = \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}\)

\(a_{11}=1\) (row 1, col 1) · \(a_{12}=2\) (row 1, col 2)
\(a_{23}=6\) (row 2, col 3) · \(a_{31}=7\) (row 3, col 1)
Subscript: row first, column second — always!

Q1 Easy

Let \(A = \begin{pmatrix}5&2&8\\1&6&3\\9&4&7\end{pmatrix}\).
What is the value of \(a_{23}\)? (row 2, col 3)

💡 EXPLANATION \(a_{23}\) = row 2, col 3. Row 2 is \((1, 6, 3)\). The 3rd element is 3.
Trick: subscript → row first, col second. \(a_{ij}\) = row \(i\), col \(j\).

Q2 Easy

For the product \(C = AB\) of two \(3\times3\) matrices, entry \(C_{31}\) is computed using:
(Hint: use the i·j RULE)

💡 EXPLANATION \(C_{ij}\) = Row \(i\) of A · Col \(j\) of B.
So \(C_{31}\) = Row 3 of A · Col 1 of B.
Memory: i·j RULE — first subscript = row from A, second = col from B.

🎯 Section 2: Computing One Entry Core Skill

✏️ STEP-BY-STEP EXAMPLE

Find \(C_{12}\) where \(C = \begin{pmatrix}1&0&2\\-1&3&1\\0&2&1\end{pmatrix}\begin{pmatrix}1&2&0\\0&1&3\\2&0&1\end{pmatrix}\)

Row 1 of A = \((1, 0, 2)\) Col 2 of B = \((2, 1, 0)\)

\(C_{12} = (1)(2)+(0)(1)+(2)(0) = 2+0+0 = \mathbf{2}\)

Q3 Easy

Find \(C_{11}\) where \(C = AB\),
\(A = \begin{pmatrix}2&1&3\\0&4&1\\1&2&0\end{pmatrix}\), \(B = \begin{pmatrix}1\\2\\1\end{pmatrix}\) (treat as col of a 3×3).
Actually: \(A = \begin{pmatrix}2&1&3\\\cdots\\\cdots\end{pmatrix}\), \(B = \begin{pmatrix}1&0&0\\2&1&0\\1&0&1\end{pmatrix}\)
Compute \(C_{11}\) = Row 1 of A · Col 1 of B.

💡 EXPLANATION Row 1 of A = \((2,1,3)\), Col 1 of B = \((1,2,1)\)
\(C_{11} = (2)(1)+(1)(2)+(3)(1) = 2+2+3 = \mathbf{7}\)
Always 3 multiply-and-add steps for 3×3!

Q4 Easy

\(A=\begin{pmatrix}1&2&0\\3&1&4\\0&2&1\end{pmatrix}\), \(B=\begin{pmatrix}2&1&3\\0&4&1\\1&2&0\end{pmatrix}\)
What is \(C_{22}\)? (Row 2 of A · Col 2 of B)

💡 EXPLANATION Row 2 of A = \((3,1,4)\), Col 2 of B = \((1,4,2)\)
\(C_{22} = (3)(1)+(1)(4)+(4)(2) = 3+4+8 = \mathbf{15}\)
Common mistake: picking wrong row or column. Always double-check which row/col!

Q5 Easy

\(A=\begin{pmatrix}0&1&2\\3&0&1\\1&2&0\end{pmatrix}\), \(B=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\)
What is \(AB\)? (Recognize matrix B!)

💡 EXPLANATION \(B\) is the Identity Matrix \(I_3\)! It has 1s on the diagonal, 0s elsewhere.
\(A \cdot I = A\) always. The identity matrix does nothing — like multiplying by 1!
Memory: "I" = Identity = invisible multiplier

🧮 Section 3: Full 3×3 Products Calculate All 9

✏️ FULL WORKED EXAMPLE

\(\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\begin{pmatrix}4&1&0\\2&3&1\\0&1&2\end{pmatrix}\)

C₁₁\(1·4+0·2+0·0=4\)

C₁₂\(1·1+0·3+0·1=1\)

C₁₃\(1·0+0·1+0·2=0\)

C₂₁\(0·4+2·2+0·0=4\)

C₂₂\(0·1+2·3+0·1=6\)

C₂₃\(0·0+2·1+0·2=2\)

C₃₁\(0·4+0·2+3·0=0\)

C₃₂\(0·1+0·3+3·1=3\)

C₃₃\(0·0+0·1+3·2=6\)

Result: \(\begin{pmatrix}4&1&0\\4&6&2\\0&3&6\end{pmatrix}\)

Q6 Medium

\(\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}3&1&4\\1&5&9\\2&6&5\end{pmatrix}\)

💡 EXPLANATION Left matrix is \(I_3\) (Identity). \(I \cdot A = A\) always!
Diagonal matrix with all 1s = identity = multiplication by 1.

Q7 Medium

\(A=\begin{pmatrix}1&2&1\\0&1&0\\0&0&1\end{pmatrix}\), \(B=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\)
Find \(C_{13}\) of \(AB\).

💡 EXPLANATION Row 1 of A = \((1,2,1)\), Col 3 of B = \((0,0,1)\) (it's Identity!)
\(C_{13}=(1)(0)+(2)(0)+(1)(1)=0+0+1=\mathbf{1}\)
Since B=I, AB=A, so \(C_{13}=a_{13}=1\). Confirmed!

Q8 Medium

\(A=\begin{pmatrix}2&0&0\\0&3&0\\0&0&5\end{pmatrix}\), \(B=\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}\)
What is the entire first row of \(AB\)?
(Hint: A is a diagonal matrix — there's a shortcut!)

💡 EXPLANATION Row 1 of A = \((2,0,0)\). Each \(C_{1j} = 2b_{1j}+0+0 = 2b_{1j}\).
So row 1 of C = 2 × row 1 of B = \((2,4,6)\).
Diagonal matrix shortcut: each row \(i\) of B gets scaled by \(a_{ii}\)!

🔄 Section 4: Order Matters! Most Missed

AB and BA are usually DIFFERENT matrices. Never swap the order unless you can prove they're equal!

Q9 Medium

\(A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\), \(B=\begin{pmatrix}0&0&0\\1&0&0\\0&1&0\end{pmatrix}\)
Find the (1,1) entry of \(AB\).
(Row 1 of A = (0,1,0), Col 1 of B = (0,1,0))

💡 EXPLANATION Row 1 of A = \((0,1,0)\), Col 1 of B = \((0,1,0)\)
\((AB)_{11} = (0)(0)+(1)(1)+(0)(0) = 0+1+0 = \mathbf{1}\)
Now check \((BA)_{11}\): Row 1 of B=\((0,0,0)\) → gives 0. So \(AB\neq BA\)!

Q10 Medium

\(A=\begin{pmatrix}1&2&3\\0&0&0\\0&0&0\end{pmatrix}\), \(B=\begin{pmatrix}0&0&0\\0&0&0\\1&2&3\end{pmatrix}\)
Compute \(AB\). What do you get?

💡 EXPLANATION Row 1 of A = \((1,2,3)\). \(C_{11}=(1)(0)+(2)(0)+(3)(1)=3\), \(C_{12}=1·0+2·0+3·2=6\), \(C_{13}=9\)
Rows 2 and 3 of A are all zeros → rows 2 and 3 of C are zeros.
Result: \(\begin{pmatrix}3&6&9\\0&0&0\\0&0&0\end{pmatrix}\)

⚡ Section 5: Tricky Special Cases Trap Zone

A non-zero matrix times a non-zero matrix CAN equal the zero matrix. Never assume AB=0 means A=0 or B=0!

Q11 Medium

\(A=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}\), \(B=\begin{pmatrix}0&0&0\\0&1&0\\0&0&1\end{pmatrix}\)
Compute \(AB\).

💡 EXPLANATION Row 1 of A = \((1,0,0)\). Dot with Col 1 of B = \((0,0,0)\): gives 0.
Dot with Col 2 of B = \((0,1,0)\): gives 0. Col 3 of B = \((0,0,1)\): gives 0.
Rows 2,3 of A are all zeros → entire result is zeros. AB = 0, yet A≠0 and B≠0!

Q12 Medium

\(A=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}\). What is \(A^2 = A \cdot A\)?
(Don't square entry-by-entry! Actually multiply.)

A² does NOT mean "square each entry." It means A × A!

💡 EXPLANATION Each entry of \(A^2\): e.g. \(C_{11} = (1)(1)+(1)(1)+(1)(1) = 3\). Same for all 9 entries.
Result: all 3s. Key: each entry = sum of 3 ones = 3, not \(1^2=1\)!

📐 Section 6: Mixed-Size Matrices Non-Square

✏️ EXAMPLE — 3×3 times 3×1

\(\begin{pmatrix}2&1&0\\0&3&1\\1&0&2\end{pmatrix}\begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}2(1)+1(2)+0(3)\\0(1)+3(2)+1(3)\\1(1)+0(2)+2(3)\end{pmatrix}=\begin{pmatrix}4\\9\\7\end{pmatrix}\)

Result is a 3×1 column vector — makes sense! \(3\times3\) · \(3\times1\) → \(3\times1\)

Q13 Easy

\(\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix} = ?\)
(This is the first column of A!)

💡 EXPLANATION Row 1 · \((1,0,0)\) = \(1·1+2·0+3·0 = 1\)
Row 2 · \((1,0,0)\) = \(4·1+5·0+6·0 = 4\)
Row 3 · \((1,0,0)\) = \(7·1+8·0+9·0 = 7\)
Multiplying by \((1,0,0)^T\) picks out the first column of A!

Q14 Medium

\(A = \begin{pmatrix}1&2&3\end{pmatrix}\) (1×3 row), \(B = \begin{pmatrix}1\\2\\3\end{pmatrix}\) (3×1 col)
What is \(AB\)? (size = 1×1, a single number)

💡 EXPLANATION \(1\times3\) · \(3\times1\) → result is 1×1 (a scalar/dot product).
\(AB = 1·1+2·2+3·3 = 1+4+9 = \mathbf{14}\)
This is the dot product of the two vectors!

📜 Section 7: Properties to Know Rules

Q15 Medium

Which property is always TRUE for 3×3 matrix multiplication?

💡 EXPLANATION Only Associativity is always true!
B: False — \(AB\neq BA\) in general.
C: False — \((AB)^2 = ABAB \neq A^2B^2 = AABB\) (order matters!)
D: False — a non-zero matrix can square to zero (nilpotent)!

Q16 Tricky

\((A+B)^2 = ?\) in matrix algebra.
(Remember: AB ≠ BA !)

💡 EXPLANATION \((A+B)^2 = (A+B)(A+B) = A^2+AB+BA+B^2\)
Since \(AB \neq BA\), you cannot combine them into \(2AB\)!
Normal algebra: \((a+b)^2=a^2+2ab+b^2\). Matrix algebra: keep AB and BA separate!

🏆 Section 8: Challenge Round Hardest!

Q17 Tricky

\(A=\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}\) (called a permutation matrix). What is \(A^2\)?
(Hint: this matrix swaps rows. What happens when you swap twice?)

💡 EXPLANATION \(C_{11}=(0)(0)+(0)(0)+(1)(1)=1\), \(C_{12}=0\), \(C_{13}=0\).
Similarly all diagonal = 1, all off-diagonal = 0.
\(A^2 = I\). This means \(A = A^{-1}\) (its own inverse)!
Swapping rows twice = back to original = identity transformation.

Q18 Tricky

\(A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\). What is \(A^3\)?
(First find A², then multiply by A again.)

💡 EXPLANATION \(A^2 = A\cdot A\): entry \((1,3)\): row 1 of A=\((0,1,0)\) · col 3 of A=\((0,1,0)\)=1. Others work out to \(\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\).
\(A^3 = A^2 \cdot A\): row 1 of \(A^2\)=\((0,0,1)\) · col 1 of A=\((0,0,0)\)=0. All entries → 0.
\(A^3 = 0\). This is a nilpotent matrix of order 3!

Q19 Tricky

\(A=\begin{pmatrix}1&0&0\\2&1&0\\3&4&1\end{pmatrix}\) (lower triangular). What is \(A^2\)?
Focus on the diagonal entries only.

💡 EXPLANATION Diagonal entries of \(A^2\): \(C_{11}=(1)(1)+0+0=1\), \(C_{22}=(0+1·1+0)=1\), \(C_{33}=1\).
Key rule: For triangular matrices, the diagonal of \(A^n\) = (diagonal entries of A) raised to the \(n\)th power!
All diagonal entries of A = 1, so diagonal of \(A^2\) = all 1s.

Q20 Tricky

Which statement about 3×3 matrices is ALWAYS TRUE?

💡 EXPLANATION A: False — many matrices square to I (e.g. the permutation matrix in Q17, or \(\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}\)).
B: False — zero divisors exist (Q11 showed this).
C: False — it's \(B^{-1}A^{-1}\), NOT \(A^{-1}B^{-1}\).
D: TRUE! Reverse order: Socks-Shoes Rule — put on socks then shoes; remove shoes then socks. Always reverse the order!

📝 My Scratch Work & Notes: