Matrix Determinants and Inverse - ECE Board Exam Reviewer

Matrix Determinants and Inverse - ECE Board Exam Reviewer | PinoyBIX

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Determinants are where matrices stop being abstract notation and start answering a real question: does this system have one solution, many solutions, or none at all? On the Philippine engineering board exam, determinant problems show up in three forms — compute it directly, use it to test whether a matrix is invertible, or apply it inside Cramer’s rule to solve a system. This post covers the $2 \times 2$ formula, the $3 \times 3$ Sarrus rule, cofactor expansion for anything larger, and then moves into the inverse matrix itself using two different methods: the adjugate method and Gauss-Jordan elimination. Work through all 10 problems before checking your answers. This is Part 2 of the Matrices and Determinants Series.

📋 BOARD EXAM RELEVANCE — DETERMINANTS AND THE INVERSE

ECE (Electronics Engineer) — Tested in Engineering Mathematics. Determinants appear in Cramer’s rule applications and in testing system consistency for circuit analysis problems.
EE (Electrical Engineer) — High frequency. The determinant of the impedance matrix determines whether a mesh system has a unique solution, and the inverse matrix solves that system directly. $3 \times 3$ determinants appear regularly in three-mesh circuit problems.
ME (Mechanical Engineer) — Tested in Engineering Mathematics. Determinants of stiffness matrices and force coefficient matrices appear in structural analysis problems.
CE (Civil Engineer) — Tested in Engineering Mathematics. Cramer’s rule for truss and force systems requires determinant computation. Area of a triangle using determinants also appears in coordinate geometry items.
ChE (Chemical Engineer) — Tested in Engineering Mathematics. Determinants determine whether a mass balance system is well-posed, and matrix inversion solves multi-component mixing problems directly.
GeE (Geodetic Engineer) — Moderate frequency. Determinants appear in coordinate transformation problems and in testing linear independence of survey control equations.
MetE and MinE — Low to moderate frequency. Basic determinant and inverse computation in Engineering Mathematics.
Naval Architect and Marine Engineer — Moderate frequency. Stability and structural analysis problems use determinants to test matrix conditions.

Bottom line: EE and ECE examinees will use the inverse matrix directly in mesh and nodal circuit analysis, so both methods in this post need to be second nature. CE, ME, and ChE need determinant evaluation solid for Cramer’s rule and structural or mass-balance systems. All nine boards test determinant properties as standalone recall items, so the properties list below is worth memorizing regardless of which board you are taking.

What a Determinant Is

A determinant is a single scalar number computed from the elements of a square matrix. It is written as $\det(A)$ or $|A|$ . Only square matrices have determinants — you cannot compute the determinant of a $2 \times 3$ matrix, no matter how hard you try.

The determinant tells you the one fact that matters most: whether the matrix is invertible. If $\det(A) \neq 0$ , the matrix has an inverse and any linear system it represents has a unique solution. If $\det(A) = 0$ , the matrix is singular — no inverse exists, and the system either has infinitely many solutions or no solution at all, depending on the constants involved.

KEY FORMULA — Singular vs Invertible

$\det(A) \neq 0 \quad \Rightarrow \quad A \text{ is invertible (nonsingular)} \quad \Rightarrow \quad \text{unique solution exists}$

$\det(A) = 0 \quad \Rightarrow \quad A \text{ is singular} \quad \Rightarrow \quad A^{-1} \text{ does not exist}$

The $2 \times 2$ Determinant

For a $2 \times 2$ matrix, the determinant is one of the most useful shortcuts in engineering mathematics. You subtract the product of the cross diagonal from the product of the main diagonal.

KEY FORMULA — $2 \times 2$ Determinant

$\det\begin{bmatrix}a&b\\c&d\end{bmatrix} = ad - bc$

Main diagonal product minus cross diagonal product. Memorize this exactly as written — it also appears inside every cofactor expansion for larger matrices.

The $3 \times 3$ Determinant — Two Methods

For $3 \times 3$ matrices you have two options. The Sarrus rule is fast but works only for $3 \times 3$ . Cofactor expansion works for any size square matrix. Learn both — reach for Sarrus when the problem is exactly $3 \times 3$ , and use cofactor expansion for everything else, including $3 \times 3$ problems where a row or column full of zeros makes expansion faster than Sarrus.

KEY FORMULA — Sarrus Rule for $3 \times 3$ Determinants

$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$

$\det(A) = (aei + bfg + cdh) - (ceg + afh + bdi)$

Positive terms: three downward diagonals (left to right). Negative terms: three upward diagonals (right to left). Extend the matrix by copying the first two columns to the right to see the diagonals clearly.

KEY FORMULA — Cofactor Expansion (any size)

$\det(A) = a_{i1}C_{i1} + a_{i2}C_{i2} + a_{i3}C_{i3} \qquad \text{where } C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor — the determinant of the submatrix left after deleting row $i$ and column $j$ . Expand along any row or column, the result is always the same number. Pick the row or column with the most zeros to save arithmetic.

Sign pattern for $3 \times 3$ :

$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$

Determinant Properties

These properties appear as direct board exam questions, some pure recall, some designed to save you computation time if you recognize them fast enough.

KEY PROPERTIES — DETERMINANTS

$\text{Row/column swap: sign flips} \qquad \det(kA) = k^n \det(A) \text{ for an } n \times n \text{ matrix}$

$\det(AB) = \det(A) \cdot \det(B) \qquad \det(A^T) = \det(A) \qquad \det(A^{-1}) = \dfrac{1}{\det(A)}$

Identical rows or columns, or any row or column of all zeros, both force $\det(A) = 0$ with no computation required. The last property means you can find $\det(A^{-1})$ without ever computing the inverse itself — if $\det A = 4$ , then $\det(A^{-1}) = 0.25$ .

The Inverse Matrix — Two Methods

An inverse exists only when $\det A \neq 0$ . Always check the determinant first before attempting to find the inverse. Trying to invert a singular matrix is a guaranteed wrong answer, no matter how carefully you do the rest of the work.

KEY FORMULA — $2 \times 2$ Inverse (fastest method)

$A^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$

Swap the main diagonal entries, negate the off-diagonal entries, divide everything by the determinant. This works for $2 \times 2$ only. For $3 \times 3$ and larger, use the adjugate or Gauss-Jordan method below.

KEY FORMULA — Adjugate Method ( $3 \times 3$ and larger)

$A^{-1} = \dfrac{1}{\det A}\,\text{adj}(A)$

Build the matrix of cofactors $C_{ij}$ , then transpose it to get the adjugate: $\text{adj}(A) = [C_{ij}]^T$ . Divide every entry by $\det A$ .

GAUSS-JORDAN METHOD

$[A \mid I] \xrightarrow{\text{row reduce}} [I \mid A^{-1}]$

Augment $A$ with the identity matrix. Apply row operations until the left side becomes $I$ . Whatever appears on the right side is $A^{-1}$ . Preferred for larger matrices and for solving $Ax = b$ directly without finding the inverse first.

KEY PROPERTIES — INVERSE

$AA^{-1} = A^{-1}A = I \qquad (A^{-1})^{-1} = A$

$(AB)^{-1} = B^{-1}A^{-1}$

The product rule reverses the order. $(AB)^{-1} = B^{-1}A^{-1}$ , not $A^{-1}B^{-1}$ . Same reversal pattern you saw for the transpose of a product back in Part 1.

Six Key Determinant Properties

These properties appear as direct questions on the board exam — some are pure recall, some require you to apply them to save computation time.

Property 1: Swapping any two rows (or columns) changes the sign of the determinant: $\det = -\det(A)$ .

Property 2: If any two rows (or columns) are identical, $\det(A) = 0$ .

Property 3: If any row (or column) is all zeros, $\det(A) = 0$ .

Property 4: Multiplying one row by a scalar $k$ multiplies the determinant by $k$ : $\det(kA) = k^n \det(A)$ for an $n \times n$ matrix.

Property 5: $\det(AB) = \det(A) \cdot \det(B)$ . The determinant of a product equals the product of the determinants.

Property 6: $\det(A^T) = \det(A)$ . The transpose has the same determinant.

10 Worked Problems — Board Exam Type

Problem 1. Evaluate a $2 \times 2$ determinant.

Given: $A = \begin{bmatrix} 5 & 3 \\ 2 & 8 \end{bmatrix}$

Find: $\det A$

Solution:

Step 1: Apply the $2 \times 2$ determinant formula.

$\det A = (5)(8) - (3)(2)$

Step 2: Compute.

$\det A = 40 - 6 = 34$

✓ ANSWER: $\det A = 34$

Examiner note: Main diagonal $5 \times 8 = 40$ , cross diagonal $3 \times 2 = 6$ , subtract in that order, never add. A positive nonzero determinant means the matrix is invertible.

Problem 2. Find the value of $k$ that makes the matrix singular.

Given: $A = \begin{bmatrix} k & 4 \\ 3 & k \end{bmatrix}$

Find: value of $k$ such that $A$ is singular

Solution:

Step 1: A singular matrix has $\det A = 0$ . Set up the equation.

$\det A = k^2 - (4)(3) = 0$

Step 2: Solve for $k$ .

$k^2 = 12 \implies k = \pm 2\sqrt{3}$

✓ ANSWER: $k = \pm 2\sqrt{3}$

Examiner note: Singular matrix means zero determinant, period. Set $\det A = 0$ and solve. This type of item tests the determinant formula and the definition of singularity in a single question.

Problem 3. Evaluate a $3 \times 3$ determinant using the Sarrus Rule.

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

Find: $\det A$

Solution:

Step 1: Compute the three positive (downward) diagonals.

$(2)(-1)(1) + (1)(2)(4) + (3)(0)(0) = -2 + 8 + 0 = 6$

Step 2: Compute the three negative (upward) diagonals.

$(3)(-1)(4) + (2)(2)(0) + (1)(0)(1) = -12 + 0 + 0 = -12$

Step 3: Subtract.

$\det A = 6 - (-12) = 18$

✓ ANSWER: $\det A = 18$

Examiner note: Write out all six diagonal products before combining them. Skipping that step is the most common source of sign errors in Sarrus rule problems.

Problem 4. Evaluate a $3 \times 3$ determinant by cofactor expansion along the zero-heavy column.

Given: $B = \begin{bmatrix} 3 & 0 & 2 \\ -1 & 0 & 5 \\ 4 & 7 & 1 \end{bmatrix}$

Find: $\det B$

Solution:

Step 1: Column 2 has two zeros. Expand along column 2 to minimize work.

$\det B = (0)C_{12} + (0)C_{22} + (7)C_{32}$

Step 2: Only the $b_{32} = 7$ term survives. Compute its cofactor with sign $(-1)^{3+2} = -1$ .

$C_{32} = (-1)\det\begin{bmatrix}3&2\\-1&5\end{bmatrix} = (-1)\big[(3)(5)-(2)(-1)\big] = (-1)(17) = -17$

Step 3: Compute the determinant.

$\det B = (7)(-17) = -119$

✓ ANSWER: $\det B = -119$

Examiner note: Two zeros in one column reduced this from three separate $2 \times 2$ computations down to one. Scan for zeros before you pick a row or column — on a timed exam, that choice matters.

Problem 5. Apply the determinant product rule.

Given: $\det A = 4$ and $\det B = -3$

Find: $\det(AB)$

Solution:

Step 1: Apply the product rule directly. There is no need to compute $AB$ itself.

$\det(AB) = \det(A) \cdot \det(B)$

Step 2: Substitute and compute.

$\det(AB) = (4)(-3) = -12$

✓ ANSWER: $\det(AB) = -12$

Examiner note: Also note that $\det(AB) = \det(BA) = -12$ , even though $AB \neq BA$ in general. Determinants of products are commutative even when matrix multiplication itself is not.

Problem 6. Find the inverse of a $2 \times 2$ matrix.

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

Find: $A^{-1}$

Solution:

Step 1: Compute $\det A$ .

$\det A = (4)(2) - (3)(3) = 8 - 9 = -1$

Step 2: Since $\det A \neq 0$ , the inverse exists. Swap the main diagonal, negate the off-diagonal.

$A^{-1} = \dfrac{1}{-1}\begin{bmatrix} 2 & -3 \\ -3 & 4 \end{bmatrix}$

Step 3: Simplify.

$A^{-1} = \begin{bmatrix} -2 & 3 \\ 3 & -4 \end{bmatrix}$

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

Examiner note: Dividing by $-1$ flips every sign. Students who forget this step hand in the adjugate as their final answer and lose the item. Always divide by $\det A$ , even when the determinant is $-1$ .

Problem 7. Verify the inverse by computing $AA^{-1}$ .

Given: $A = \begin{bmatrix} 4 & 3 \\ 3 & 2 \end{bmatrix}$ , $A^{-1} = \begin{bmatrix} -2 & 3 \\ 3 & -4 \end{bmatrix}$ from Problem 6

Find: $AA^{-1}$ , confirm it equals $I$

Solution:

Step 1: Multiply $AA^{-1}$ entry by entry.

$AA^{-1} = \begin{bmatrix}(4)(-2)+(3)(3) & (4)(3)+(3)(-4)\\(3)(-2)+(2)(3) & (3)(3)+(2)(-4)\end{bmatrix}$

Step 2: Compute each entry.

$= \begin{bmatrix}-8+9 & 12-12\\-6+6 & 9-8\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} = I$

✓ ANSWER: $AA^{-1} = I$ ✓ Verified.

Examiner note: This check takes 30 seconds and protects you from a sign error that could cost you the entire problem. Board exams rarely give partial credit, so verifying before you write your final answer is always worth the time.

Problem 8. Find the inverse of a $3 \times 3$ matrix using the adjugate method.

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

Find: $A^{-1}$

Solution:

Step 1: Compute $\det A$ by expanding along row 1.

$\det A = 1\det\begin{bmatrix}1&3\\0&1\end{bmatrix} - 2\det\begin{bmatrix}0&3\\2&1\end{bmatrix} + 0 = 1(1-0) - 2(0-6) = 1 + 12 = 13$

Step 2: Compute all nine cofactors.

$C_{11} = 1 \quad C_{12} = 6 \quad C_{13} = -2$

$C_{21} = -2 \quad C_{22} = 1 \quad C_{23} = 4$

$C_{31} = 6 \quad C_{32} = -3 \quad C_{33} = 1$

Step 3: Transpose the cofactor matrix to get the adjugate.

$\text{adj}(A) = \begin{bmatrix}1&-2&6\\6&1&-3\\-2&4&1\end{bmatrix}$

Step 4: Divide every entry by $\det A = 13$ .

$A^{-1} = \dfrac{1}{13}\begin{bmatrix}1&-2&6\\6&1&-3\\-2&4&1\end{bmatrix}$

✓ ANSWER: $A^{-1} = \dfrac{1}{13}\begin{bmatrix}1&-2&6\\6&1&-3\\-2&4&1\end{bmatrix}$

Examiner note: The adjugate is the transpose of the cofactor matrix, not the cofactor matrix itself. Skipping the transpose step gives a completely wrong inverse — write the cofactors in a grid first, then flip across the main diagonal.

Problem 9. Find the inverse of a $2 \times 2$ matrix using Gauss-Jordan elimination.

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

Find: $A^{-1}$ using row operations

Solution:

Step 1: Write the augmented matrix $[A \mid I]$ .

$\left[\begin{array}{cc|cc}2&1&1&0\\5&3&0&1\end{array}\right]$

Step 2: $R_2 \leftarrow R_2 - \dfrac{5}{2}R_1$ to eliminate the 5 in column 1.

$\left[\begin{array}{cc|cc}2&1&1&0\\0&\frac{1}{2}&-\frac{5}{2}&1\end{array}\right]$

Step 3: $R_2 \leftarrow 2R_2$ to make the leading entry 1.

$\left[\begin{array}{cc|cc}2&1&1&0\\0&1&-5&2\end{array}\right]$

Step 4: $R_1 \leftarrow R_1 - R_2$ , then $R_1 \leftarrow \dfrac{1}{2}R_1$ to finish reducing the left side to $I$ .

$\left[\begin{array}{cc|cc}1&0&3&-1\\0&1&-5&2\end{array}\right]$

Step 5: Read the inverse directly from the right side.

$A^{-1} = \begin{bmatrix}3&-1\\-5&2\end{bmatrix}$

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

Examiner note: Quick check with the direct formula: $\det A = (2)(3)-(1)(5) = 1$ , so $A^{-1} = \frac{1}{1}\begin{bmatrix}3&-1\\-5&2\end{bmatrix}$ . Matches — Gauss-Jordan and the formula agree, which confirms neither method had an arithmetic slip.

Problem 10. Applied: determinant in circuit analysis.

Given: a two-mesh circuit produces the impedance matrix $Z = \begin{bmatrix} 10 & -4 \\ -4 & 8 \end{bmatrix}$ (in ohms)

Find: whether the system has a unique solution for the mesh currents

Solution:

Step 1: Compute $\det(Z)$ .

$\det(Z) = (10)(8) - (-4)(-4) = 80 - 16 = 64$

Step 2: Since $\det(Z) \neq 0$ , the system has a unique solution.

✓ ANSWER: $\det(Z) = 64 \neq 0$ — unique solution exists for the mesh currents.

Examiner note: This is exactly how determinants connect to circuit analysis on the ECE and EE board exam. Before solving for mesh currents with Cramer’s rule or the inverse matrix, always compute the determinant of the impedance matrix first. If it comes out zero, either the circuit is genuinely ill-posed or there is an error somewhere in the mesh equations.

Common Mistakes and Examiner Traps

❌ Common Mistake	✅ Correct Approach
Computing $\det A = ac - bd$ instead of $ad - bc$ for a $2 \times 2$ matrix.	$\det\begin{bmatrix}a&b\\c&d\end{bmatrix} = ad - bc$ . Main diagonal minus cross diagonal. Write the formula before substituting numbers.
Treating every cofactor as positive, ignoring the sign pattern during expansion.	The sign of $C_{ij}$ is $(-1)^{i+j}$ . Write the checkerboard pattern next to the matrix before you expand.
Using the cofactor matrix directly as the adjugate, without transposing it first.	The adjugate is the transpose of the cofactor matrix. Compute the cofactors first, then flip the result across the main diagonal.
Writing $\det(kA) = k \cdot \det(A)$ for a $3 \times 3$ or larger matrix.	$\det(kA) = k^n \det(A)$ . The scalar gets raised to the power $n$ , the size of the matrix — for a $3 \times 3$ , that means $k^3$ , not $k$ .
Concluding that $\det(A) = 0$ automatically means the system has no solution.	$\det(A) = 0$ means no unique solution. The system could have infinitely many solutions or none at all — which one depends on the constants, not the coefficient matrix alone.
Writing $(AB)^{-1} = A^{-1}B^{-1}$ without reversing the order.	$(AB)^{-1} = B^{-1}A^{-1}$ . The order always reverses — the same reversal rule you saw for the transpose of a product in Part 1.

Board Exam Quick Tips

Check for zero-determinant conditions before computing anything. Identical rows, proportional rows, or a zero row all guarantee $\det = 0$ . Spotting this takes 5 seconds. Computing a full $3 \times 3$ determinant takes 2 to 3 minutes. Use your time wisely.
Expand along the row or column with the most zeros. Each zero eliminates one term entirely. A row or column with two zeros turns a $3 \times 3$ problem into a single $2 \times 2$ computation.
Check the determinant before you attempt to invert. One second spent computing $\det A$ saves you from wasting two minutes trying to invert a matrix that has no inverse at all.
Use the $2 \times 2$ inverse formula directly. It is faster than the adjugate method and faster than Gauss-Jordan. Save those two methods for $3 \times 3$ and larger matrices.
Verify your inverse with a quick $AA^{-1} = I$ check. It takes 30 seconds for a $2 \times 2$ matrix and catches sign errors before you commit to a wrong final answer.

Frequently Asked Questions

Q1. What is the difference between a minor and a cofactor?

A minor $M_{ij}$ is the determinant of the submatrix formed by deleting row $i$ and column $j$ from the original matrix. A cofactor $C_{ij}$ is the signed minor: $C_{ij} = (-1)^{i+j} M_{ij}$ . The sign alternates in a checkerboard pattern starting with $+$ at position $(1,1)$ . The cofactor is simply the minor with the correct sign already attached.

Q2. Can I use the Sarrus rule for $4 \times 4$ matrices?

No. The Sarrus rule only works for $3 \times 3$ matrices. For $4 \times 4$ and larger, use cofactor expansion, ideally along whichever row or column has the most zeros. That reduces the problem to a set of $3 \times 3$ determinants, which you can then finish off with Sarrus. Board exam problems rarely go past $3 \times 3$ , but if you see a $4 \times 4$ , cofactor expansion is the method to reach for.

Q3. Why does the adjugate method require transposing the cofactor matrix?

The cofactor $C_{ij}$ sits at position $(i,j)$ in the cofactor matrix, but the adjugate needs $C_{ij}$ at position $(j,i)$ instead. That swap is what makes $A \cdot \text{adj}(A)$ produce $\det A$ down the diagonal and zero everywhere else. Skip the transpose and that alignment breaks, giving you a wrong inverse that still looks plausible.

Q4. Which method is faster for finding a $3 \times 3$ inverse on the board exam — adjugate or Gauss-Jordan?

Gauss-Jordan is usually faster under exam conditions, since it is mechanical, row operation by row operation, with fewer opportunities for sign errors than nine separate cofactor calculations. The adjugate method pulls ahead when you already need the cofactors for another part of the same problem, such as computing $\det A$ by expansion first.

Q5. How does the determinant relate to the area of a triangle?

Given three vertices $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ , the area works out to:

$\text{Area} = \dfrac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

This shows up in CE and GeE board exams under coordinate geometry. The absolute value bars make sure the area comes out positive no matter which order you list the vertices in.

What is Next

You can now evaluate determinants and find inverses using three different methods. Part 3 puts both of those tools to direct use, solving systems of equations with Cramer’s rule and Gaussian elimination — the two techniques tested more often than anything else across all nine Philippine engineering boards.

→ Continue to Part 3 — Systems of Linear Equations

→ Back to the Matrices and Determinants Series Index

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Matrix Determinants and Inverse – ECE Board Exam Reviewer | PinoyBIX

What a Determinant Is

The $2 \times 2$ Determinant

The $3 \times 3$ Determinant — Two Methods

Determinant Properties

The Inverse Matrix — Two Methods

Six Key Determinant Properties

10 Worked Problems — Board Exam Type

Problem 1. Evaluate a $2 \times 2$ determinant.

Problem 2. Find the value of $k$ that makes the matrix singular.

Problem 3. Evaluate a $3 \times 3$ determinant using the Sarrus Rule.

Problem 4. Evaluate a $3 \times 3$ determinant by cofactor expansion along the zero-heavy column.

Problem 5. Apply the determinant product rule.

Problem 6. Find the inverse of a $2 \times 2$ matrix.

Problem 7. Verify the inverse by computing $AA^{-1}$ .

Problem 8. Find the inverse of a $3 \times 3$ matrix using the adjugate method.

Problem 9. Find the inverse of a $2 \times 2$ matrix using Gauss-Jordan elimination.

Problem 10. Applied: determinant in circuit analysis.

Common Mistakes and Examiner Traps

Board Exam Quick Tips

Frequently Asked Questions

What is Next

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What a Determinant Is

The Determinant

The Determinant — Two Methods

Determinant Properties

The Inverse Matrix — Two Methods

Six Key Determinant Properties

10 Worked Problems — Board Exam Type

Problem 1. Evaluate a determinant.

Problem 2. Find the value of that makes the matrix singular.

Problem 3. Evaluate a determinant using the Sarrus Rule.

Problem 4. Evaluate a determinant by cofactor expansion along the zero-heavy column.

Problem 5. Apply the determinant product rule.

Problem 6. Find the inverse of a matrix.

Problem 7. Verify the inverse by computing .

Problem 8. Find the inverse of a matrix using the adjugate method.

Problem 9. Find the inverse of a matrix using Gauss-Jordan elimination.

Problem 10. Applied: determinant in circuit analysis.

Common Mistakes and Examiner Traps

Board Exam Quick Tips

Frequently Asked Questions

What is Next

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The $2 \times 2$ Determinant

The $3 \times 3$ Determinant — Two Methods

Problem 1. Evaluate a $2 \times 2$ determinant.

Problem 2. Find the value of $k$ that makes the matrix singular.

Problem 3. Evaluate a $3 \times 3$ determinant using the Sarrus Rule.

Problem 4. Evaluate a $3 \times 3$ determinant by cofactor expansion along the zero-heavy column.

Problem 6. Find the inverse of a $2 \times 2$ matrix.

Problem 7. Verify the inverse by computing $AA^{-1}$ .

Problem 8. Find the inverse of a $3 \times 3$ matrix using the adjugate method.

Problem 9. Find the inverse of a $2 \times 2$ matrix using Gauss-Jordan elimination.