Linear Algebra · Self-Study Workbook
Matrix × Matrix
20 Questions · Multiple Choice · With Instant Feedback ✏️
★ ★ ★ ★ ★
Name: ______________________
Date: __________
R·C→R·C
"Row meets Column" — each entry is a dot product of a row × column
INNER = POSSIBLE
A(m×n) · B(n×p) — inner dims must match!
OUTER = SIZE
Result size = outer dims → m×p
AB ≠ BA
NOT commutative! Order always matters
ROW i × COL j
Entry C[i][j] = sum of (row i of A) × (col j of B)
DISTRIBUTE OK
A(B+C)=AB+AC ✔ but AB≠BA ✗
🔑 KEY FORMULA
If \( A \) is \( m \times n \) and \( B \) is \( n \times p \), then \( C = AB \) is \( m \times p \) where:\(\displaystyle C_{ij} = \sum_{k=1}^{n} a_{ik} \cdot b_{kj} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}\)
0 of 20 answered
📐 Section 1: Can We Multiply?
Dimension Check
✏️ EXAMPLE
Can we multiply \(A_{2\times3}\) and \(B_{3\times4}\)?
→ Inner dims: 3 = 3 ✔ Result: \(2 \times 4\)
Can we multiply \(A_{2\times3}\) and \(B_{4\times3}\)?
→ Inner dims: 3 ≠ 4 ✗ NOT DEFINED!
Q1 Easy
Matrix \(A\) is \(2 \times 3\) and matrix \(B\) is \(3 \times 5\).
What is the size of the product \(AB\)?
What is the size of the product \(AB\)?
💡 EXPLANATION
\(A\) is \(2\times\mathbf{3}\) and \(B\) is \(\mathbf{3}\times5\). Inner dims match (both 3) ✔
Result size = outer dims = \(2 \times 5\).
Memory trick: OUTER = SIZE → keep the 2 and the 5.
Result size = outer dims = \(2 \times 5\).
Memory trick: OUTER = SIZE → keep the 2 and the 5.
Q2 Easy
Which multiplication is NOT defined?
Both matrices must share the same inner dimension. Always check the "middle numbers"!
💡 EXPLANATION
Check the inner (middle) numbers:
A: 2=2 ✔ B: 5=5 ✔ C: 3≠4 ✗ NOT DEFINED D: 4=4 ✔
Memory trick: INNER = POSSIBLE
A: 2=2 ✔ B: 5=5 ✔ C: 3≠4 ✗ NOT DEFINED D: 4=4 ✔
Memory trick: INNER = POSSIBLE
Q3 Easy
\(A\) is \(3 \times 4\) and \(B\) is \(4 \times 3\). The result \(AB\) has size:
💡 EXPLANATION
\(A_{3\times\mathbf{4}} \cdot B_{\mathbf{4}\times3}\): inner 4=4 ✔
Result = outer dims = \(\mathbf{3} \times \mathbf{3}\)
Note: \(BA\) would give \(4 \times 4\) — order changes everything!
Result = outer dims = \(\mathbf{3} \times \mathbf{3}\)
Note: \(BA\) would give \(4 \times 4\) — order changes everything!
✏️ Section 2: Basic 2×2 Multiplication
Core Skill
✏️ EXAMPLE
\(\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}5&6\\7&8\end{pmatrix}\)
Top-left: \(1\cdot5 + 2\cdot7 = 5+14 = \mathbf{19}\)
Top-right: \(1\cdot6 + 2\cdot8 = 6+16 = \mathbf{22}\)
Bot-left: \(3\cdot5 + 4\cdot7 = 15+28 = \mathbf{43}\)
Bot-right: \(3\cdot6 + 4\cdot8 = 18+32 = \mathbf{50}\)
Q4 Easy
\(\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}3&7\\2&5\end{pmatrix} = ?\)
(Hint: What is special about the first matrix?)
(Hint: What is special about the first matrix?)
💡 EXPLANATION
The first matrix \(\begin{pmatrix}1&0\\0&1\end{pmatrix}\) is the Identity Matrix I.
\(I \cdot B = B\) always! It's like multiplying by 1.
Memory trick: "I" = Identity = does nothing
\(I \cdot B = B\) always! It's like multiplying by 1.
Memory trick: "I" = Identity = does nothing
Q5 Easy
What is the top-right entry of \(\begin{pmatrix}2&1\\3&4\end{pmatrix}\begin{pmatrix}1&5\\0&2\end{pmatrix}\)?
💡 EXPLANATION
Top-right = Row 1 of A · Col 2 of B
\( = (2)(5) + (1)(2) = 10 + 2 = \mathbf{12}\)
Always: entry \(C_{ij}\) = row \(i\) of A \(\cdot\) col \(j\) of B
\( = (2)(5) + (1)(2) = 10 + 2 = \mathbf{12}\)
Always: entry \(C_{ij}\) = row \(i\) of A \(\cdot\) col \(j\) of B
Q6 Easy
Compute: \(\begin{pmatrix}3&0\\0&2\end{pmatrix}\begin{pmatrix}4&1\\5&6\end{pmatrix}\)
What is the bottom-left entry?
What is the bottom-left entry?
💡 EXPLANATION
Bottom-left = Row 2 of A · Col 1 of B
\(= (0)(4) + (2)(5) = 0 + 10 = \mathbf{10}\)
The zero in row 2 of A wipes out the first product!
\(= (0)(4) + (2)(5) = 0 + 10 = \mathbf{10}\)
The zero in row 2 of A wipes out the first product!
🔁 Section 3: Commutativity — Does Order Matter?
Common Trap
\(AB \neq BA\) in general! This is the #1 mistake students make. Always check the order!
Q7 Medium
\(A = \begin{pmatrix}1&2\\0&1\end{pmatrix}\), \(B = \begin{pmatrix}1&0\\3&1\end{pmatrix}\)
Which is true about \(AB\) and \(BA\)?
Which is true about \(AB\) and \(BA\)?
💡 EXPLANATION
\(AB = \begin{pmatrix}7&2\\3&1\end{pmatrix}\) but \(BA = \begin{pmatrix}1&2\\3&7\end{pmatrix}\)
They are different! Matrix multiplication is NOT commutative.
Memory trick: AB ≠ BA — order always matters!
They are different! Matrix multiplication is NOT commutative.
Memory trick: AB ≠ BA — order always matters!
Q8 Medium
\(A\) is \(2\times3\), \(B\) is \(3\times2\). Which statement is correct?
💡 EXPLANATION
\(A_{2\times\mathbf{3}} \cdot B_{\mathbf{3}\times2}\) → \(AB\) is \(2\times2\) ✔
\(B_{3\times\mathbf{2}} \cdot A_{\mathbf{2}\times3}\) → \(BA\) is \(3\times3\) ✔
Both are defined but give different sizes!
\(B_{3\times\mathbf{2}} \cdot A_{\mathbf{2}\times3}\) → \(BA\) is \(3\times3\) ✔
Both are defined but give different sizes!
🧮 Section 4: Full Matrix Products
Core Calculation
Q9 Easy
\(\begin{pmatrix}2&3\\1&4\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix} = ?\)
💡 EXPLANATION
Top: \(2(1)+3(2) = 2+6 = \mathbf{8}\)
Bottom: \(1(1)+4(2) = 1+8 = \mathbf{9}\)
Result: \(\begin{pmatrix}8\\9\end{pmatrix}\)
Bottom: \(1(1)+4(2) = 1+8 = \mathbf{9}\)
Result: \(\begin{pmatrix}8\\9\end{pmatrix}\)
Q10 Medium
\(\begin{pmatrix}1&2\\3&4\end{pmatrix}^2 = \begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}1&2\\3&4\end{pmatrix}\)
Find the top-left entry.
Find the top-left entry.
💡 EXPLANATION
Top-left \(C_{11}\) = Row 1 · Col 1 = \((1)(1)+(2)(3) = 1+6 = \mathbf{7}\)
Note: \(A^2 \neq\) squaring each entry! Always do actual matrix multiplication.
Note: \(A^2 \neq\) squaring each entry! Always do actual matrix multiplication.
⚡ Section 5: Special Matrices
Tricky!
Q11 Medium
\(\begin{pmatrix}0&1\\0&0\end{pmatrix}^2 = ?\)
Zero matrix ≠ 0 number. A non-zero matrix squared can give the zero matrix!
💡 EXPLANATION
\(\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}\)
Entry (1,1): \((0)(0)+(1)(0)=0\) Entry (1,2): \((0)(1)+(1)(0)=0\)
Entry (2,1): \(0\) Entry (2,2): \(0\)
Result = Zero matrix! This is called a nilpotent matrix.
Entry (1,1): \((0)(0)+(1)(0)=0\) Entry (1,2): \((0)(1)+(1)(0)=0\)
Entry (2,1): \(0\) Entry (2,2): \(0\)
Result = Zero matrix! This is called a nilpotent matrix.
Q12 Medium
If \(I\) is the \(2\times2\) identity matrix and \(A = \begin{pmatrix}5&3\\1&2\end{pmatrix}\),
what is \(IA\)?
what is \(IA\)?
💡 EXPLANATION
Identity matrix \(I\) times any matrix \(A\) = \(A\).
\(IA = AI = A\) — Identity is the "1" of matrices!
\(IA = AI = A\) — Identity is the "1" of matrices!
🏗️ Section 6: Properties & Rules
Must Know
Q13 Medium
Which property is FALSE for matrix multiplication?
💡 EXPLANATION
Commutativity is FALSE! \(AB \neq BA\) in general.
The other three (associativity, distributivity, identity) ARE true.
Memory: AB ≠ BA — this is the big rule breaker!
The other three (associativity, distributivity, identity) ARE true.
Memory: AB ≠ BA — this is the big rule breaker!
Q14 Tricky
If \(AB = AC\) and \(A \neq 0\), can we conclude \(B = C\)?
Matrix algebra is NOT like regular number algebra. You CANNOT cancel A from both sides in general!
💡 EXPLANATION
You can only cancel \(A\) if \(A\) is invertible (has an inverse \(A^{-1}\)).
In general, \(AB = AC\) does NOT imply \(B = C\).
Matrices don't always have the "cancellation" property!
In general, \(AB = AC\) does NOT imply \(B = C\).
Matrices don't always have the "cancellation" property!
📊 Section 7: Larger Matrices
Level Up
Q15 Medium
\(\begin{pmatrix}1&2&0\\0&1&3\end{pmatrix}\begin{pmatrix}2\\1\\4\end{pmatrix} = ?\)
💡 EXPLANATION
Row 1 · Col: \(1(2)+2(1)+0(4) = 2+2+0 = \mathbf{4}\)
Row 2 · Col: \(0(2)+1(1)+3(4) = 0+1+12 = \mathbf{13}\)
Result: \(\begin{pmatrix}4\\13\end{pmatrix}\)
Row 2 · Col: \(0(2)+1(1)+3(4) = 0+1+12 = \mathbf{13}\)
Result: \(\begin{pmatrix}4\\13\end{pmatrix}\)
Q16 Medium
How many multiplications are needed to compute one entry of \(AB\) where \(A\) is \(m\times n\) and \(B\) is \(n\times p\)?
💡 EXPLANATION
Each entry \(C_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}\)
This requires exactly \(n\) multiplications (and \(n\) additions).
\(n\) is the inner dimension — the shared dimension!
This requires exactly \(n\) multiplications (and \(n\) additions).
\(n\) is the inner dimension — the shared dimension!
🎯 Section 8: Tricky Trap Questions
Most Missed!
Q17 Tricky
\(A = \begin{pmatrix}1&1\\1&1\end{pmatrix}\). What is \(A^2\)?
Don't just square each entry! You must actually multiply the matrices.
💡 EXPLANATION
\((1)(1)+(1)(1)=2\) for each entry.
So \(A^2 = \begin{pmatrix}2&2\\2&2\end{pmatrix} = 2A\)
Common mistake: thinking \(1^2=1\) entry-by-entry — WRONG!
So \(A^2 = \begin{pmatrix}2&2\\2&2\end{pmatrix} = 2A\)
Common mistake: thinking \(1^2=1\) entry-by-entry — WRONG!
Q18 Tricky
\((AB)^T = ?\) where \(T\) means transpose (flip rows↔cols).
💡 EXPLANATION
The transpose reverses the order: \((AB)^T = B^T A^T\)
This is called the "Socks-Shoes" rule — you put socks on first, shoes on second; but to take them off, shoes come off first!
Memory: REVERSE THE ORDER when transposing a product
This is called the "Socks-Shoes" rule — you put socks on first, shoes on second; but to take them off, shoes come off first!
Memory: REVERSE THE ORDER when transposing a product
Q19 Tricky
Let \(A = \begin{pmatrix}2&0\\0&3\end{pmatrix}\) (diagonal matrix).
What is \(A^3\)?
What is \(A^3\)?
💡 EXPLANATION
For diagonal matrices, you just raise each diagonal entry to the power!
\(A^3 = \begin{pmatrix}2^3&0\\0&3^3\end{pmatrix} = \begin{pmatrix}8&0\\0&27\end{pmatrix}\)
Memory: Diagonal matrices are easy to power — just power the diagonal!
\(A^3 = \begin{pmatrix}2^3&0\\0&3^3\end{pmatrix} = \begin{pmatrix}8&0\\0&27\end{pmatrix}\)
Memory: Diagonal matrices are easy to power — just power the diagonal!
Q20 Tricky
Given \(A = \begin{pmatrix}3&1\\6&2\end{pmatrix}\) and \(B = \begin{pmatrix}2&4\\-6&-12\end{pmatrix}\),
what is \(AB\)?
what is \(AB\)?
When the answer comes out as the zero matrix despite non-zero inputs, it is NOT an error!
💡 EXPLANATION
(1,1): \(3(2)+1(-6)=6-6=0\) (1,2): \(3(4)+1(-12)=12-12=0\)
(2,1): \(6(2)+2(-6)=12-12=0\) (2,2): \(6(4)+2(-12)=24-24=0\)
\(AB = \mathbf{0}\) even though \(A \neq 0\) and \(B \neq 0\)!
This is a zero divisor — unique to matrices!
(2,1): \(6(2)+2(-6)=12-12=0\) (2,2): \(6(4)+2(-12)=24-24=0\)
\(AB = \mathbf{0}\) even though \(A \neq 0\) and \(B \neq 0\)!
This is a zero divisor — unique to matrices!
📝 My Notes & Work:
_
_
_
_
_
_
_
_