Matrix Fundamentals Operations - ECE Board Exam

Matrix Fundamentals and Operations — PinoyBIX ECE EE CE ME Board Exam Reviewer

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Matrices show up in every engineering board exam in the Philippines, and they always start the same way: a notation question, a multiplication problem, or a dimension check that separates students who actually studied from those who guessed. This post covers matrix operations for the ECE, EE, CE, and ME board exam — the notation, the types, the rules for multiplication, and the one property that most students get backwards. Work through the 10 problems below using the same step-by-step format you will use on exam day. This is Part 1 of the Matrices and Determinants Series.

📋 BOARD EXAM RELEVANCE

Matrices and determinants appear across all nine Philippine engineering boards. EE and CE examinees see matrix equations in circuit mesh analysis and structural stiffness problems. ME encounters them in vibration analysis and kinematics. ECE uses them in signal processing and control systems. ChE, GeE, MetE, MinE, and Naval Architecture boards test matrix operations as part of general Engineering Mathematics. Expect at least two to four items on this topic in any full engineering board exam.

What is a Matrix

A matrix is a rectangular array of numbers arranged in rows and columns. You write it inside brackets and describe its size by stating rows first, then columns. A matrix with $m$ rows and $n$ columns is called an $m \times n$ matrix.

KEY FORMULA — ELEMENT NOTATION

$A = [a_{ij}]_{m \times n}$

$i$ is the row position, $j$ is the column position. The element $a_{23}$ sits in row 2, column 3. Row index always comes first.

Matrix Types You Must Recognize

The board exam tests matrix types through identification questions. These are not difficult, but students who never memorize the definitions lose easy points.

Square matrix: number of rows equals number of columns. All determinant problems use square matrices.

Identity matrix $I$ : a square matrix with 1’s on the main diagonal and 0’s everywhere else. Multiplying any matrix by $I$ gives the original matrix back: $AI = IA = A$ .

Zero matrix: every entry is 0.

Diagonal matrix: square matrix with nonzero entries only on the main diagonal. All other entries are 0.

Symmetric matrix: a square matrix equal to its own transpose. $A = A^T$ . Every entry $a_{ij} = a_{ji}$ .

Transpose $A^T$ : swap rows and columns. Row $i$ of $A$ becomes column $i$ of $A^T$ .

Matrix Addition and Scalar Multiplication

KEY FORMULAS — ADDITION AND SCALAR MULTIPLICATION

$A + B = [a_{ij} + b_{ij}] \qquad \text{(same size only)}$

$kA = [k \cdot a_{ij}]$

Addition requires matrices of identical size. Add the elements in matching positions. There is no shortcut here and no polar form equivalent. You go entry by entry. Scalar multiplication is simpler: multiply every element by the scalar.

Matrix Multiplication

KEY FORMULA — DIMENSION RULE

$(m \times n)(n \times p) = (m \times p)$

Inner dimensions must match. The product has the outer dimensions. Each entry of the result is the dot product of a row from the left matrix with a column from the right matrix.

For each entry $c_{ij}$ of the product matrix $C = AB$ :

$c_{ij} = \sum_{k=1}^{n} a_{ik}\, b_{kj}$

This is not element-by-element multiplication. Row $i$ of $A$ times column $j$ of $B$ , summed up, gives you one entry. You repeat this for every row-column pair in the product.

Critical rule: matrix multiplication is not commutative. $AB \neq BA$ in general. Reversing the order can produce a completely different result, or make the multiplication undefined if the new inner dimensions no longer match. This fact earns dedicated exam items almost every testing cycle.

Transpose Properties

KEY FORMULAS — TRANSPOSE RULES

$(A^T)^T = A \qquad (A+B)^T = A^T + B^T$

$(kA)^T = kA^T \qquad (AB)^T = B^T A^T$

The last rule catches most students: the transpose of a product reverses the order. $(AB)^T = B^T A^T$ , not $A^T B^T$ . Same reversal pattern as the inverse of a product, which you will see again in Part 2.

10 Worked Board Exam Problems

Problem 1. Read the matrix element.

Given: $A = \begin{bmatrix} 4 & 7 & 2 \\ 1 & 9 & 5 \\ 3 & 6 & 8 \end{bmatrix}$

Find: $a_{23}$ and $a_{31}$ .

Solution:

Step 1: For $a_{23}$ , go to row 2, column 3.

$a_{23} = 5$

Step 2: For $a_{31}$ , go to row 3, column 1.

$a_{31} = 3$

✓ ANSWER: $a_{23} = 5$

, $a_{31} = 3$

Examiner note: The subscript order is always row then column, never column then row. $a_{23}$ is row 2, column 3. Reversing that gives you a different element entirely.

Problem 2. Determine if addition is possible.

Given: $A$ is $2 \times 3$ , $B$ is $2 \times 3$ , $C$ is $3 \times 2$

Find: Which of $A + B$ , $A + C$ , $B + C$ are defined?

Solution:

Step 1: Check size match for $A + B$ . Both are $2 \times 3$ .

$A + B \text{ is defined} \quad \checkmark$

Step 2: Check size match for $A + C$ . $A$ is $2 \times 3$ , $C$ is $3 \times 2$ . Sizes differ.

$A + C \text{ is undefined} \quad \times$

Step 3: Check size match for $B + C$ . Same issue as above.

$B + C \text{ is undefined} \quad \times$

✓ ANSWER: Only $A + B$

is defined.

Examiner note: A $2 \times 3$ and a $3 \times 2$ matrix look related but they are not the same size. Rows by columns, in that order, always.

Problem 3. Add two matrices.

Given: $A = \begin{bmatrix} 3 & -1 \\ 2 & 5 \end{bmatrix}$ , $B = \begin{bmatrix} 4 & 7 \\ -3 & 1 \end{bmatrix}$

Find: $A + B$

Solution:

Step 1: Add corresponding entries.

$A + B = \begin{bmatrix} 3+4 & -1+7 \\ 2+(-3) & 5+1 \end{bmatrix}$

Step 2: Simplify.

$A + B = \begin{bmatrix} 7 & 6 \\ -1 & 6 \end{bmatrix}$

✓ ANSWER: $A + B = \begin{bmatrix} 7 & 6 \\ -1 & 6 \end{bmatrix}$

Examiner note: Watch signs on negative entries. $-1 + 7 = 6$ , not $-6$ . Arithmetic errors on negatives are the fastest way to lose an item you already knew how to solve.

Problem 4. Scalar multiplication.

Given: $A = \begin{bmatrix} 2 & -4 & 1 \\ 0 & 3 & -5 \end{bmatrix}$ , scalar $k = -3$

Find: $kA$

Solution:

Step 1: Multiply every entry by $-3$ .

$-3A = \begin{bmatrix} -3(2) & -3(-4) & -3(1) \\ -3(0) & -3(3) & -3(-5) \end{bmatrix}$

Step 2: Simplify.

$-3A = \begin{bmatrix} -6 & 12 & -3 \\ 0 & -9 & 15 \end{bmatrix}$

✓ ANSWER: $-3A = \begin{bmatrix} -6 & 12 & -3 \\ 0 & -9 & 15 \end{bmatrix}$

Examiner note: A negative scalar multiplied by a negative entry gives a positive result. $-3 \times -4 = +12$ , not $-12$ . Double check every sign when the scalar is negative.

Problem 5. Check if multiplication is defined and find the size of the product.

Given: $A$ is $3 \times 4$ , $B$ is $4 \times 2$

Find: Is $AB$ defined? If yes, what is its size? Is $BA$ defined?

Solution:

Step 1: Check $AB$ . Inner dimensions of $A$ and $B$ are both 4. They match.

$AB \text{ is defined, size } 3 \times 2$

Step 2: Check $BA$ . $B$ is $4 \times 2$ , $A$ is $3 \times 4$ . Inner dimensions are 2 and 3. They do not match.

$BA \text{ is undefined}$

✓ ANSWER: $AB$

is defined with size $3 \times 2$

. $BA$

is undefined.

Examiner note: Always check inner dimensions first. If they do not match, stop. Do not attempt the multiplication. This is a free point on the board exam if you know the rule.

Problem 6. Multiply two $2 \times 2$ matrices.

Given: $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ , $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$

Find: $AB$

Solution:

Step 1: Compute entry $c_{11}$ — row 1 of $A$ times column 1 of $B$ .

$c_{11} = (1)(5) + (2)(7) = 5 + 14 = 19$

Step 2: Compute $c_{12}$ — row 1 of $A$ times column 2 of $B$ .

$c_{12} = (1)(6) + (2)(8) = 6 + 16 = 22$

Step 3: Compute $c_{21}$ — row 2 of $A$ times column 1 of $B$ .

$c_{21} = (3)(5) + (4)(7) = 15 + 28 = 43$

Step 4: Compute $c_{22}$ — row 2 of $A$ times column 2 of $B$ .

$c_{22} = (3)(6) + (4)(8) = 18 + 32 = 50$

Step 5: Assemble the result.

$AB = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}$

✓ ANSWER: $AB = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}$

Examiner note: Each entry requires a dot product, not a single multiplication. Two multiplications and one addition for a $2 \times 2$ product. Scale that up carefully for larger matrices.

Problem 7. Multiply a $2 \times 3$ by a $3 \times 2$ matrix.

Given: $A = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \end{bmatrix}$ , $B = \begin{bmatrix} 2 & 1 \\ 0 & -1 \\ 3 & 2 \end{bmatrix}$

Find: $AB$

Solution:

Step 1: Confirm size. $(2 \times 3)(3 \times 2) = 2 \times 2$ product.

Step 2: Compute each of the four entries.

$c_{11} = (1)(2)+(0)(0)+(2)(3) = 2+0+6 = 8$

$c_{12} = (1)(1)+(0)(-1)+(2)(2) = 1+0+4 = 5$

$c_{21} = (-1)(2)+(3)(0)+(1)(3) = -2+0+3 = 1$

$c_{22} = (-1)(1)+(3)(-1)+(1)(2) = -1-3+2 = -2$

Step 3: Write the result.

$AB = \begin{bmatrix} 8 & 5 \\ 1 & -2 \end{bmatrix}$

✓ ANSWER: $AB = \begin{bmatrix} 8 & 5 \\ 1 & -2 \end{bmatrix}$

Examiner note: Three-term dot products are where arithmetic errors pile up. Write out all three multiplications before adding. Do not try to combine steps in your head under exam pressure.

Problem 8. Find the transpose of a matrix.

Given: $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$

Find: $A^T$

Solution:

Step 1: The original matrix is $2 \times 3$ . The transpose will be $3 \times 2$ .

Step 2: Convert each row of $A$ into a column of $A^T$ .

$A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}$

✓ ANSWER: $A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}$

Examiner note: The main diagonal entries never move when you transpose. Everything else flips across that diagonal. A quick sanity check is to verify that $a_{12}$ in $A$ became $a_{21}$ in $A^T$ .

Problem 9. Demonstrate that $AB \neq BA$ .

Given: $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}$

Find: $AB$ and $BA$ , then compare.

Solution:

Step 1: Compute $AB$ .

$AB = \begin{bmatrix}(1)(1)+(2)(3) & (1)(0)+(2)(1) \\ (0)(1)+(1)(3) & (0)(0)+(1)(1)\end{bmatrix} = \begin{bmatrix} 7 & 2 \\ 3 & 1 \end{bmatrix}$

Step 2: Compute $BA$ .

$BA = \begin{bmatrix}(1)(1)+(0)(0) & (1)(2)+(0)(1) \\ (3)(1)+(1)(0) & (3)(2)+(1)(1)\end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}$

Step 3: Compare. $AB \neq BA$ .

✓ ANSWER: $AB = \begin{bmatrix} 7 & 2 \\ 3 & 1 \end{bmatrix} \neq BA = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}$

Examiner note: Even when both $AB$ and $BA$ are defined and the same size, they are usually not equal. The only guaranteed case where $AB = BA$ is when one of the matrices is the identity $I$ .

Problem 10. Verify $(AB)^T = B^T A^T$ .

Given: $A = \begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}$

Find: $(AB)^T$ and $B^T A^T$ , then verify they are equal.

Solution:

Step 1: Compute $AB$ .

$AB = \begin{bmatrix}(2)(1)+(1)(2) & (2)(4)+(1)(3) \\ (3)(1)+(0)(2) & (3)(4)+(0)(3)\end{bmatrix} = \begin{bmatrix} 4 & 11 \\ 3 & 12 \end{bmatrix}$

Step 2: Transpose the result.

$(AB)^T = \begin{bmatrix} 4 & 3 \\ 11 & 12 \end{bmatrix}$

Step 3: Find $A^T$ and $B^T$ .

$A^T = \begin{bmatrix} 2 & 3 \\ 1 & 0 \end{bmatrix} \qquad B^T = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$

Step 4: Compute $B^T A^T$ .

$B^T A^T = \begin{bmatrix}(1)(2)+(2)(1) & (1)(3)+(2)(0) \\ (4)(2)+(3)(1) & (4)(3)+(3)(0)\end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 11 & 12 \end{bmatrix}$

Step 5: Compare. Both give $\begin{bmatrix} 4 & 3 \\ 11 & 12 \end{bmatrix}$ .

✓ ANSWER: $(AB)^T = B^T A^T = \begin{bmatrix} 4 & 3 \\ 11 & 12 \end{bmatrix}$

✓ Verified.

Examiner note: The order reverses when you transpose a product. $(AB)^T = B^T A^T$ , never $A^T B^T$ . This same reversal applies to inverses: $(AB)^{-1} = B^{-1}A^{-1}$ . One rule, two applications.

Common Mistakes and Examiner Traps

❌ Wrong Approach	✅ Correct Approach
Adding matrices of different sizes, e.g. adding a $2 \times 3$ to a $3 \times 2$	Matrices must have identical dimensions for addition. Check sizes before you start.
Multiplying entry by entry instead of row by column, e.g. $c_{11} = a_{11} \times b_{11}$	Each entry $c_{ij}$ is a dot product: row $i$ of $A$ times column $j$ of $B$ , summed.
Assuming $AB = BA$ and swapping matrix order in the middle of a problem	Matrix multiplication is not commutative. Order is fixed. Never swap unless the problem tells you to.
Writing $(AB)^T = A^T B^T$ instead of reversing the order	$(AB)^T = B^T A^T$ . The order always reverses when transposing or inverting a product.
Reading $a_{23}$ as row 3, column 2 instead of row 2, column 3	$a_{ij}$ always means row $i$ , column $j$ . Row index first, column index second, no exceptions.
Confusing $AI = A$ with $AI = I$ or thinking the identity does something to the matrix	Multiplying by $I$ on either side gives back the original matrix unchanged: $AI = IA = A$ .

Board Exam Quick Tips

Check inner dimensions before you multiply. If they do not match, the product is undefined. This is a one second check that saves you from solving a problem that has no answer.
The product size uses the outer dimensions. $(m \times n)(n \times p) = (m \times p)$ . Write this down during the problem so you know how many entries to compute.
Never swap matrix order. If a problem gives you $AB$ , compute $AB$ . Do not compute $BA$ and assume it is the same thing.
Transpose reverses the product order. $(AB)^T = B^T A^T$ . Memorize this alongside its twin: $(AB)^{-1} = B^{-1}A^{-1}$ . Both follow the same reversal rule.
Matrix addition is always rectangular. Entries are added in matching positions, row by row, column by column. There is no faster method, and no situation where you can skip to an answer without going entry by entry.

Frequently Asked Questions

Can two matrices of different sizes ever be multiplied?

Yes, but only when the inner dimensions match. A $2 \times 3$ matrix can multiply a $3 \times 4$ matrix because the inner dimensions are both 3. The product is $2 \times 4$ . A $2 \times 3$ cannot multiply a $2 \times 4$ because the inner dimensions are 3 and 2, which do not match.

What makes a matrix symmetric?

A matrix is symmetric when it equals its own transpose: $A = A^T$ . This means every entry $a_{ij}$ equals $a_{ji}$ . All symmetric matrices are square. Not all square matrices are symmetric.

Is there any case where $AB = BA$ ?

Yes. Any matrix multiplied by the identity matrix $I$ satisfies $AI = IA = A$ . Also, any matrix multiplied by its own inverse satisfies $AA^{-1} = A^{-1}A = I$ . Outside of these special cases, assume $AB \neq BA$ unless the problem proves otherwise.

How do you find the size of a matrix product without computing it?

Write the two sizes next to each other: $(m \times n)(n \times p)$ . The inner dimensions must match. The product size is always the two outer dimensions: $m \times p$ . This rule works for any chain of matrix multiplications.

What is the difference between a zero matrix and a zero determinant?

A zero matrix has all entries equal to zero. A zero determinant means the matrix is singular, but the matrix itself may have nonzero entries. These are two different concepts that appear in different types of board exam problems. A zero matrix always has a zero determinant, but a zero determinant does not mean the matrix is a zero matrix.

What is Next

Part 1 gave you the vocabulary and the operation rules. Part 2 builds directly on that foundation. In Part 2 — Determinants and the Inverse Matrix, you will learn how to evaluate determinants for 2×2, 3×3, and 4×4 matrices, expand using cofactors, and find the inverse using two different methods. Determinants appear in almost every board exam problem involving linear systems, so Part 2 is the pivot point of the whole series.

Return to the Matrices and Determinants Series Index for the full formula sheet, the 30-item practice exam, and the study schedule.

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Matrix Fundamentals Operations – ECE Board Exam | PinoyBIX

What is a Matrix

Matrix Types You Must Recognize

Matrix Addition and Scalar Multiplication

Matrix Multiplication

Transpose Properties