Q: Step 2: Multiply the main diagonal (top-left to bottom-right).

The main diagonal elements are $5$ and $8$. $$ 5 \times 8 = 40 $$

Q: Step 3: Multiply the anti-diagonal (top-right to bottom-left).

The anti-diagonal elements are $3$ and $2$. $$ 3 \times 2 = 6 $$

Q: Step 2: Find the minor for the first element ($4$).

Cross out the first row and first column. The remaining $2 \times 2$ matrix is $\begin{bmatrix} 0 & 5 \ 2 & 6 \end{bmatrix}$. Calculate its determinant: $(0 \times 6) - (5 \times 2) = 0 - 10 = -10$. Multiply by the element: $+4 \times (-10) = -40$.

Q: Step 4: Find the minor for the third element ($1$).

Cross out the first row and third column. The remaining $2 \times 2$ matrix is $\begin{bmatrix} 3 & 0 \ -1 & 2 \end{bmatrix}$. Calculate its determinant: $(3 \times 2) - (0 \times -1) = 6 - 0 = 6$. Multiply by the element: $+1 \times 6 = 6$.

Q: Step 1: Calculate the determinant.

$\det(C) = (4 \times 6) - (7 \times 2) = 24 - 14 = 10$. Since the determinant ($10$) is not zero, the inverse exists.

Q: Step 2: Find the Adjugate matrix.

For a $2 \times 2$ matrix, the rule for the adjugate is simple: swap the elements on the main diagonal, and flip the signs of the elements on the anti-diagonal. $$ \text{Adj}(C) = \begin{bmatrix} 6 & -7 \ -2 & 4 \end{bmatrix} $$

Question 1

Step 1: Identify the elements.

Accepted Answer

Here, $a = 5$, $b = 3$, $c = 2$, and $d = 8$.

Question 2

Step 2: Multiply the main diagonal (top-left to bottom-right).

Accepted Answer

The main diagonal elements are $5$ and $8$.
$$ 5 	imes 8 = 40 $$

Question 3

Step 3: Multiply the anti-diagonal (top-right to bottom-left).

Accepted Answer

The anti-diagonal elements are $3$ and $2$.
$$ 3 	imes 2 = 6 $$

Question 4

Step 4: Subtract the anti-diagonal product from the main diagonal product.

Accepted Answer

$$ \det(A) = 40 - 6 = 34 $$

Question 5

Step 1: Set up the expansion along the first row.

Accepted Answer

The elements of the first row are $4$, $-2$, and $1$. We will multiply each of these by the determinant of the $2 	imes 2$ matrix that remains when we cross out their respective row and column. Remember to alternate the signs: $(+) ightarrow (-) ightarrow (+)$.

Question 6

Step 2: Find the minor for the first element ($4$).

Accepted Answer

Cross out the first row and first column. The remaining $2 	imes 2$ matrix is $\begin{bmatrix} 0 & 5 \ 2 & 6 \end{bmatrix}$.
Calculate its determinant: $(0 	imes 6) - (5 	imes 2) = 0 - 10 = -10$.
Multiply by the element: $+4 	imes (-10) = -40$.

Question 7

Step 3: Find the minor for the second element ($-2$).

Accepted Answer

Cross out the first row and second column. The remaining $2 	imes 2$ matrix is $\begin{bmatrix} 3 & 5 \ -1 & 6 \end{bmatrix}$.
Calculate its determinant: $(3 	imes 6) - (5 	imes -1) = 18 - (-5) = 23$.
Multiply by the element, remembering the formula requires a minus sign here: $-(-2) 	imes 23 = +2 	imes 23 = 46$.

Question 8

Step 4: Find the minor for the third element ($1$).

Accepted Answer

Cross out the first row and third column. The remaining $2 	imes 2$ matrix is $\begin{bmatrix} 3 & 0 \ -1 & 2 \end{bmatrix}$.
Calculate its determinant: $(3 	imes 2) - (0 	imes -1) = 6 - 0 = 6$.
Multiply by the element: $+1 	imes 6 = 6$.

Question 9

Step 5: Add the results together.

Accepted Answer

$$ \det(B) = -40 + 46 + 6 = 12 $$

Question 10

Step 1: Calculate the determinant.

Accepted Answer

$\det(C) = (4 	imes 6) - (7 	imes 2) = 24 - 14 = 10$.
Since the determinant ($10$) is not zero, the inverse exists.

Question 11

Step 2: Find the Adjugate matrix.

Accepted Answer

For a $2 	imes 2$ matrix, the rule for the adjugate is simple: swap the elements on the main diagonal, and flip the signs of the elements on the anti-diagonal.
$$ 	ext{Adj}(C) = \begin{bmatrix} 6 & -7 \ -2 & 4 \end{bmatrix} $$

Question 12

Step 3: Multiply the Adjugate by $1/\det(C)$.

Accepted Answer

$$ C^{-1} = \frac{1}{10} \begin{bmatrix} 6 & -7 \ -2 & 4 \end{bmatrix} = \begin{bmatrix} 0.6 & -0.7 \ -0.2 & 0.4 \end{bmatrix} $$
This resulting matrix is the exact inverse of matrix $C$.

Question 13

The Truth:

Accepted Answer

Determinants are strictly defined only for square matrices ($2 	imes 2$, $3 	imes 3$, $n 	imes n$). It is a fundamental error to attempt to calculate the determinant of a rectangular matrix like a $3 	imes 4$. If a system of equations has 3 variables but 4 equations, you cannot use a determinant to solve it directly; you must use other methods like least-squares approximation.

Question 14

The Truth:

Accepted Answer

Many beginners assume that $\det(A + B) = \det(A) + \det(B)$. This is entirely false. Matrix determinants do not distribute over addition. 
For example, let $A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ (determinant is $1$). Let $B = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$ (determinant is $1$). 
The sum of their determinants is $1 + 1 = 2$. 
However, the matrix $(A + B) = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$. The determinant of this zero matrix is $0$. Therefore, $0 
eq 2$.

Question 15

The Truth:

Accepted Answer

If you multiply a matrix $A$ by a scalar number $k$, the determinant does not simply become $k 	imes \det(A)$. Because the scalar multiplies every element in the matrix, it factors into the determinant once for every row. The correct rule is $\det(kA) = k^n 	imes \det(A)$, where $n$ is the dimension of the matrix. 
If you have a $3 	imes 3$ matrix with a determinant of $5$, and you multiply the entire matrix by $2$, the new determinant is NOT $10$. It is $2^3 	imes 5 = 8 	imes 5 = 40$. Failing to raise the scalar to the power of $n$ is the single most common algebraic mistake made on linear algebra exams.

Question 16

Can the determinant of a matrix be a negative number?

Accepted Answer

Yes, absolutely. A negative determinant simply indicates that the geometric transformation represented by the matrix flips the orientation of space. For example, in two dimensions, if you have a shape defined by a standard X and Y axis, a matrix with a negative determinant will reflect that shape as if it were looking in a mirror. The absolute value of the number still dictates the scaling of the area or volume.

Question 17

What does it mean if the determinant is exactly zero?

Accepted Answer

If a determinant is exactly zero, the matrix is called "singular." Geometrically, this means the matrix transformation squashes the space into a lower dimension—for example, flattening a 3D cube into a 2D square, or a 2D plane into a 1D line. Mathematically, it means the matrix has no inverse, and any system of linear equations represented by that matrix will either have no solutions at all, or an infinite number of solutions, but never one unique solution.

Question 18

Is it possible to calculate the determinant of a non-square matrix?

Accepted Answer

No, it is mathematically impossible. The determinant is a property strictly defined for square matrices (where the number of rows equals the number of columns, such as $2 	imes 2$ or $3 	imes 3$). If you have a rectangular matrix (like $2 	imes 3$), it represents a transformation from one dimension to a different dimension (like 3D space to 2D space). Because the input and output dimensions don't match, the concept of a single "volume scaling factor" (the determinant) cannot exist.

Question 19

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

Accepted Answer

The determinant can be used to easily calculate the area of any triangle on a 2D Cartesian coordinate plane. If you know the $(x, y)$ coordinates of the triangle's three vertices, you can place them into a $3 	imes 3$ matrix where the first column is the x-coordinates, the second column is the y-coordinates, and the third column is filled with the number $1$. The area of the triangle is exactly one-half of the absolute value of the determinant of that matrix.

Question 20

Why is Laplace expansion considered inefficient for large matrices?

Accepted Answer

Laplace expansion is a recursive algorithm that breaks a matrix down into smaller and smaller pieces. To calculate a $4 	imes 4$ matrix, you must calculate four $3 	imes 3$ matrices. To calculate a $10 	imes 10$ matrix, you must calculate over 3.6 million $2 	imes 2$ matrices. This scales at a factorial rate ($O(n!)$). For a $20 	imes 20$ matrix, calculating the determinant via Laplace expansion would require more operations than there are atoms in the universe, making it entirely useless for modern computing compared to LU decomposition.

Question 21

What is the determinant of the Identity Matrix?

Accepted Answer

The determinant of an Identity Matrix of any size is always exactly $1$. The Identity Matrix is a square matrix with $1$s on the main diagonal and $0$s everywhere else. Because it represents a transformation that does absolutely nothing to the space (it does not stretch, shrink, or rotate it), the scaling factor is $1$. This holds true whether it is a $2 	imes 2$ Identity Matrix or a $1,000 	imes 1,000$ Identity Matrix.

Question 22

If I transpose a matrix, does its determinant change?

Accepted Answer

No, transposing a matrix (flipping it over its main diagonal so that rows become columns and columns become rows) does not change the determinant at all. Mathematically, $\det(A) = \det(A^T)$. This is a highly useful property because it means that any rule or shortcut that applies to the rows of a determinant applies equally to the columns. For instance, expanding Laplace cofactors along a column is mathematically identical to expanding along a row.

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