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# Properties of Determinant(With Formulas & Examples)

February 15, 2023

In math, a determinant is a unique number $\text{K}$ associated with a square matrix represented as $\text{det} (\text{K})$ or,  $|\text{K}|$. The determinant and its properties are useful as they help us to obtain the same outcomes with distinct and simpler configurations of elements.

Letâ€™s understand the properties of determinants with their formula and examples.

## Properties of Determinants

The properties of determinants can be summarized into seven broad categories. These are as discussed below.

• Interchange Property: The value of a determinant remains unchanged if the rows or columns of a determinant are interchanged.
• Sign Property: The sign of the value of a determinant changes if any two rows or any two columns are interchanged.
• Zero Property: The value of a determinant is equal to zero if any two rows or any two columns have the same elements.
• Multiplication Property: The value of the determinant becomes $k$ times the previous value of the determinant if each of the elements of a particular row or column is multiplied with a constant $k$.
• Sum Property: If a few elements of a row or column are expressed as a sum of terms, then the determinant can be expressed as a sum of two or more determinants.
• Invariance Property: If each element of a row and column of a determinant is added with the equimultiples of the elements of another row or column of a determinant, then the value of the determinant remains unchanged. This can be expressed in the form of a formula as $\text{R}_i \rightarrow \text{R}_i + k \text{R}_j$, or $\text{C}_i \rightarrow \text{C}_{i} + k \text{C}_j$.
• Triangular Property: If the elements above or below the main diagonal are equal to zero, then the value of the determinant is equal to the product of the elements of the diagonal of the matrix.

### 1. Interchange Property of Determinant

The interchange property of determinant states that the value of a determinant remains unchanged if the rows and columns of a determinant are interchanged.

If $|\text{A}| = \begin {vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}$ is a determinant of a matrix $\text{A}$, and $|\text{A}^{‘}| = \begin {vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3\\ \end{vmatrix}$ is a determinant of a matrix $\text{A}^{‘}$, then $\begin {vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix} = \begin {vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3\\ \end{vmatrix}$.

In short, we can write as $\text{det}(\text{A}) = \text{det}(\text{A}^{‘})$.

Thus from this property, it follows that the value of determinants of a matrix and its transpose is equal.

#### Examples on Interchange Property of Determinant

Example 1: Consider a matrix $\text{A} =\begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}$. Verify the interchange property of the determinant.

For matrix $\text{A} =\begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}$, $\text{A}^{‘} =\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix}$ $|\text{A}| = 1 \times 4 – 2 \times 3 = 4 – 6 = -2$ and $|\text{A}^{‘}| = 1 \times 4 – 3 \times 2 = 4 – 6 = -2$

Therefore, $\text{det}(\text{A}) = \text{det}(\text{A}^{‘})$.

Example 2: Consider a matrix $\text{B} =\begin{bmatrix} 1 & 5 & 0\\ -2 & 4 & 2\\ 1 & 3 & 6\\ \end{bmatrix}$. Verify the interchange property of the determinant.

For $\text{B} =\begin{bmatrix} 1 & 5 & 0\\ -2 & 4 & 2\\ 1 & 3 & 6\\ \end{bmatrix}, \text{B}^{‘} \begin{bmatrix} 1 & -2 & 1\\ 5 & 4 & 3\\ 0 & 2 & 6\\ \end{bmatrix}$. $|\text{B}| = \begin{vmatrix} 1 & 5 & 0\\ -2 & 4 & 2\\ 1 & 3 & 6\\ \end{vmatrix} = 1 \times \begin{vmatrix} 4 & 2\\ 3 & 6\\ \end{vmatrix} – 5 \times \begin{vmatrix} -2 & 2\\ 1 & 6\\ \end{vmatrix} + 0 \times \begin{vmatrix} -2 & 4\\ 1 & 3\\ \end{vmatrix}$

$= 1 \times (4 \times 6 – 2 \times 3) – 5 \times (-2 \times 6 – 2 \times 1) + 0$

$= 1 \times (24 – 6) – 5 \times (-12 – 2) + 0$

$= 1 \times 18 – 5 \times (-14) + 0$

$= 18 + 70 + 0 = 88$.

And

$|\text{B}^{‘}| = \begin{vmatrix} 1 & -2 & 1\\ 5 & 4 & 3\\ 0 & 2 & 6\\ \end{vmatrix} = 1 \times \begin{vmatrix} 4 & 3\\ 2 & 6\\ \end{vmatrix} – (-2) \times \begin{vmatrix} 5 & 3\\ 0 & 6\\ \end{vmatrix} + 1 \times \begin{vmatrix} 5 & 4\\ 0 & 2\\ \end{vmatrix}$

$= 1 \times (4 \times 6 – 3 \times 2) – (-2) \times (5 \times 6 – 3 \times 0) + 1 \times (5 \times 2 – 4 \times 0)$

$= 1 \times (24 – 6) + 2 \times (30 – 0) + 1 \times (10 – 0)$

$= 1 \times 18 + 2 \times 30 + 1 \times 10$

$= 18 + 60 + 10 = 88$

Therefore, $\text{det}(\text{B}) = \text{det}(\text{B}^{‘})$.

### 2. Sign Property of Determinant

The sign property of the determinant states that the sign of the value of the determinant changes if any two rows or any two columns are interchanged.

If $|\text{A}| = \begin {vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}$ is a determinant of a matrix $\text{A},$ then $\begin {vmatrix} b_1 & b_2 & b_3\\ a_1 & a_2 & a_3\\ c_1 & c_2 & c_3\\ \end{vmatrix} = -\begin {vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}$ (first and second rows interchanged), and similarly, $\begin {vmatrix} a_3 & a_2 & a_1\\ b_3 & b_2 & b_1\\ c_3 & c_2 & c_1\\ \end{vmatrix} = -\begin {vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}$ (first and third columns interchanged).

Note:

• The value of the determinant only changes the sign if the row or the column is swapped once.
• If the value of the determinant is $\text{D}$, and the rows or columns are swapped $n$ times, then the new value of the determinant will be $(-1)^n \text{D}$.

#### Examples of Sign Property of Determinant

Example 1: Consider a matrix $\text{A} =\begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}$. Verify the sign property of the determinant.

$| \text{A}| = 1 \times 4 – 2 \times 3 = 4 – 6 = -2$.

Letâ€™s interchange the rows $\text{R}_1$ and $\text{R}_2$. The determinant now becomes $\begin{vmatrix} 3 & 4\\ 1 & 2 \end{vmatrix}$.

Solving it, we get $3 \times 2 – 4 \times 1 = 6 – 4 = 2$, which is negative of the previous determinant value.

Letâ€™s now interchange the columns $\text{C}_1$ and $\text{C}_2$.

Solving it, we get $2 \times 3 – 1 \times 4 = 6 – 4 = 2$, which is again negative of the previous determinant value.

Example 2: Consider a matrix $\text{B} =\begin{bmatrix} 1 & 5 & 0\\ -2 & 4 & 2\\ 1 & 3 & 6\\ \end{bmatrix}$. Verify the sign property of the determinant.

$|\text{B}| = \begin{vmatrix} 1 & 5 & 0\\ -2 & 4 & 2\\ 1 & 3 & 6\\ \end{vmatrix} = 1 \times \begin{vmatrix} 4 & 2\\ 3 & 6\\ \end{vmatrix} – 5 \times \begin{vmatrix} -2 & 2\\ 1 & 6\\ \end{vmatrix} + 0 \times \begin{vmatrix} -2 & 4\\ 1 & 3\\ \end{vmatrix}$

$= 1 \times (4 \times 6 – 2 \times 3) – 5 \times (-2 \times 6 – 2 \times 1) + 0$

$= 1 \times (24 – 6) – 5 \times (-12 – 2) + 0$

$= 1 \times 18 – 5 \times (-14) + 0$

$= 18 + 70 + 0 = 88$.

Letâ€™s interchange the rows $\text{R}_2$ and $\text{R}_3$. The determinant now becomes $\begin{vmatrix} 1 & 5 & 0\\ 1 & 3 & 6\\ -2 & 4 & 2\\ \end{vmatrix}$.

Solving it we get, $1 \times \begin{vmatrix} 3 & 6\\ 4 & 2\\ \end{vmatrix} – 5 \times \begin{vmatrix} 1 & 6\\ -2 & 2\\ \end{vmatrix} + 0 \times \begin{vmatrix} 1 & 3\\ -2 & 4\\ \end{vmatrix}$

$= 1 \times (3 \times 2 – 6 \times 4) – 5 \times (1 \times 2 – 6 \times (-2)) + 0$

$= 1 \times (6 – 24) – 5 \times (2 + 12) + 0$

$= 1 \times (-18) – 5 \times 14 + 0$

$= -18 – 70 + 0 = -88$, which is negative of the value of the previous determinant.

Letâ€™s now interchange the columns $\text{C}_1$ and $\text{C}_2$.

Solving it we get, $0 \times \begin{bmatrix} 4 & -2\\ 3 & 1\\ \end{bmatrix} – 5 \times \begin{bmatrix} 2 & -2\\ 6 & 1\\ \end{bmatrix} + 1 \times \begin{bmatrix} 2 & 4\\ 6 & 3\\ \end{bmatrix}$

$= 0 \times (4 \times 1 – (-2) \times 3) – 5 \times (2 \times 1 – (-2) \times 6) + 1 \times (2 \times 3 – 4 \times 6)$

$= 0 – 5 \times (2 + 12) + 1 \times (6 – 24)$

$= 0 – 5 \times 14 + 1 \times (-18)$

$= 0 – 70 – 18 = -88$, which is again negative of the value of the previous determinant.

### 3. Zero Property of Determinant

The zero property of determinant states that the value of a determinant is equal to zero if any two rows or any two columns have the same elements.

If $|\text{A}| = \begin {vmatrix} a_1 & a_2 & a_3\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ \end{vmatrix}$ is a determinant of a matrix $\text{A},$ then its value will be $0$(Rows $\text{R}_1$ and $\text{R}_2$ are identical).

Similarly, If $|\text{A}| = \begin {vmatrix} a_1 & a_2 & a_2\\ b_1 & b_2 & b_2\\ c_1 & c_2 & c_2\\ \end{vmatrix}$ is a determinant of a matrix $\text{A},$ then its value will be $0$(Columns $\text{C}_2$ and $\text{C}_3$ are identical).

#### Examples of Zero Property of Determinant

Example 1: Find the value of the determinant $\begin {vmatrix} -5 & 6\\ -5 & 6\\ \end{vmatrix}$

In the given determinant, the columns $\text{C}_1$ and $\text{C}_2$ are identical, and hence the value is $0$.

Letâ€™s check it by expansion.

$\begin {vmatrix} -5 & 6\\ -5 & 6\\ \end{vmatrix} = -5 \times 6 – 6 \times (-5) = -30 + 30 = 0$.

Example 2: Find the value of the determinant $\begin {vmatrix} 2 & -3 & 6\\ 4 & 1 & 3\\ 2 & -3 & 6\\ \end{vmatrix}$.

In the given determinant, the rows $\text{R}_1$ and $\text{R}_2$ are identical, and hence the value is $0$.

Letâ€™s check it by expansion.

$\begin {vmatrix} 2 & -3 & 6\\ 4 & 1 & 3\\ 2 & -3 & 6\\ \end{vmatrix} = 2 \times \begin {vmatrix} 1 & 3\\ -3 & 6\\ \end{vmatrix} – (-3) \times \begin {vmatrix} 4 & 3\\ 2 & 6\\ \end{vmatrix} + 6 \times \begin {vmatrix} 4 & 1\\ 2 & -3\\ \end{vmatrix}$

$= 2 \times (1 \times 6 – 3 \times (-3)) + 3 \times (4 \times 6 – 3 \times 2) + 6 \times (4 \times (-3) – 1 \times 2)$

$= 2 \times (6 + 9) + 3 \times (24 – 6) + 6 \times (-12 – 2)$

$= 2 \times 15 + 3 \times 18 + 6 \times (-14)$

$= 30 + 54 – 84 = 0$

### 4. Multiplication Property of Determinant

The multiplication property of the determinant states that the value of the determinant becomes $k$ times the previous value of the determinant if each of the elements of a particular row or column is multiplied with a constant $k$.

If $|\text{A}| = \begin {vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}$ is a determinant of a matrix $\text{A},$ then $\begin {vmatrix} a_1 & a_2 & a_3\\ kb_1 & kb_2 & kb_3\\ c_1 & c_2 & c_3\\ \end{vmatrix} = k|\text{A}|$ (Multiplying each element of row $\text{R}_2$ by $k$).

Similarly, if $|\text{A}| = \begin {vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}$ is a determinant of a matrix $\text{A},$ then $\begin {vmatrix} a_1 & a_2 & ka_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & kc_3\\ \end{vmatrix} = k|\text{A}|$ (Multiplying each element of column $\text{C}_3$ by $k$).

This property helps in taking a common factor from each row or column of the determinant. Also if the corresponding elements of any two rows or columns are equal then the value of the determinant is equal to zero.

#### Examples of Multiplication Property of Determinant

Example 1: Given that the value of the determinant $\begin {vmatrix} 2 & 5\\ 7 & 21\\ \end{vmatrix} = 7$, then find the value of $\begin {vmatrix} 6 & 15\\ 7 & 21\\ \end{vmatrix}$.

$\begin {vmatrix} 6 & 15\\ 7 & 21\\ \end{vmatrix} = \begin {vmatrix} 3 \times 2 & 3 \times 5\\ 7 & 21\\ \end{vmatrix} = 3 \times \begin {vmatrix} 2 & 5\\ 7 & 21\\ \end{vmatrix} = 3 \times 7 = 21$.

Letâ€™s now find the value of the determinant by expansion.

$\begin {vmatrix} 6 & 15\\ 7 & 21\\ \end{vmatrix} = 6 \times 21 – 15 \times 7 = 126 – 105 = 21$ which is equal to $3 \times \begin {vmatrix} 2 & 5\\ 7 & 21\\ \end{vmatrix}$.

### 5. Sum Property of Determinant

This property states that if a few elements of a row or column are expressed as a sum of terms, then the determinant can be expressed as a sum of two or more determinants.

If $|\text{A}| = \begin {vmatrix} x_1 + y_1 & x_2 + y_2 & x_3 + y_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}$ is a determinant of a matrix $\text{A},$ then $\begin {vmatrix} x_1 + y_1 & x_2 + y_2 & x_3 + y_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix} = \begin {vmatrix} x_1 & x_2 & x_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix} + \begin {vmatrix} y_1 & y_2 & y_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}$.

You can see above that the elements of the first row represent the sum of terms, which can be split into two different determinants. Further, the new determinants also have the same second and third row, as the earlier determinant.

#### Examples of Sum Property of Determinant

Example 1: Find the value of the determinant $\begin {vmatrix} 25 & 28 & 2\\ 42 & 47 & -1\\ 68 & 72 & 1\\ \end{vmatrix}$.

The values present in the given determinant are large numbers and if we expand it, the expansion will become a bit lengthy and complicated. Therefore, weâ€™ll split the numbers into smaller numbers so that the calculation becomes easier.

$\begin {vmatrix} 25 & 28 & 2\\ 42 & 47 & -1\\ 68 & 72 & 1\\ \end{vmatrix} = \begin {vmatrix} 25 & 25 + 3 & 2\\ 42 & 42 + 5 & -1\\ 68 & 68 + 4 & 1\\ \end{vmatrix} = \begin {vmatrix} 25 & 25 & 2\\ 42 & 42 & -1\\ 68 & 68 & 1\\ \end{vmatrix} + \begin {vmatrix} 25 & 3 & 2\\ 42 & 5 & -1\\ 68 & 4 & 1\\ \end{vmatrix}$

$= 0 + \begin {vmatrix} 25 & 3 & 2\\ 42 & 5 & -1\\ 68 & 4 & 1\\ \end{vmatrix}$ (Since columns $\text{C}_1$ and $\text{C}_2$ are identical and hence the value of first determinant is $0$).

Now solving $\begin {vmatrix} 25 & 3 & 2\\ 42 & 5 & -1\\ 68 & 4 & 1\\ \end{vmatrix}$, we get $25 \times \begin {vmatrix} 5 & -1\\ 4 & 1\\ \end{vmatrix} – 3 \times \begin {vmatrix} 42 & -1\\ 68 & 1\\ \end{vmatrix} + 2 \times \begin {vmatrix} 42 & 5\\ 68 & 4\\ \end{vmatrix}$

$= 25 \times (5 \times 1 – (-1) \times 4) – 3 \times (42 \times 1 – (-1) \times 68) + 2 \times (42 \times 4 – 5 \times 68)$

$= 25 \times (5 + 4) – 3 \times (42 + 68) + 2 \times (168 – 340)$

$= 25 \times 9 – 3 \times 110 + 2 \times (-172)$

$= -449$

### 6. Invariance Property of Determinant

The invariance property of the determinant states that if each element of a row and column of a determinant is added with the equimultiples of the elements of another row or column of a determinant, then the value of the determinant remains unchanged. This can be expressed in the form of a formula as $\text{R}_i \rightarrow \text{R}_i + k \text{R}_j$ , or $\text{C}_i \rightarrow \text{C}_i + k \text{C}_j$.

#### Examples of Invariance Property of Determinant

Example 1: Find the value of the determinant $\begin {vmatrix} 12 & 36 & 1\\ 10 & 30 & 0\\ 7 & 21 & 2\\ \end{vmatrix}$.

In the given determinant, we observe that every element of column $\text{C}_2$ is three times the value of corresponding elements in column $\text{C}_1$, therefore, applying the rule $\text{C}_2 \rightarrow \text{C}_2 – 3 \times \text{C}_1$, we get $\begin {vmatrix} 12 & 36 & 1\\ 10 & 30 & 0\\ 7 & 21 & 2\\ \end{vmatrix} = \begin {vmatrix} 12 & 36 – 3 \times 12 & 1\\ 10 & 30 – 3 \times 10 & 0\\ 7 & 21 – 3 \times 7 & 2\\ \end{vmatrix} = \begin {vmatrix} 12 & 0 & 1\\ 10 & 0 & 0\\ 7 & 0 & 2\\ \end{vmatrix}$

Now expanding the determinant, we get

$= 12 \times (0 \times 2 – 0 \times 0) – 0 \times (10 \times 2 – 0 \times 7) + 1 \times (10 \times 0 – 0 \times 7)$

$= 12 \times (0 – 0) – 0 \times (20 – 0) + 1 \times (0 – 0)$

$= 12 \times 0 – 0 + 1 \times 0 = 0$

Example 2: Find the value of the determinant $\begin {vmatrix} 2 & 4 & 13\\ 5 & 1 & 26\\ -3 & 5 & -16\\ \end{vmatrix}$.

Applying the rule $\text{C}_3 \rightarrow \text{C}_3 – 5 \text{C}_1$

$\begin {vmatrix} 2 & 4 & 13\\ 5 & 1 & 26\\ -3 & 5 & -16\\ \end{vmatrix} = \begin {vmatrix} 2 & 4 & 13 – 5 \times 2\\ 5 & 1 & 26 – 5 \times 5\\ -3 & 5 & -16 – 5 \times (-3)\\ \end{vmatrix} = \begin {vmatrix} 2 & 4 & 3\\ 5 & 1 & 1\\ -3 & 5 & -1\\ \end{vmatrix}$

Now expanding the determinant we get $\begin {vmatrix} 2 & 4 & 3\\ 5 & 1 & 1\\ -3 & 5 & -1\\ \end{vmatrix} = 2 \times \begin {vmatrix} 1 & 1\\ 5 & -1\\ \end{vmatrix} – 4 \times \begin {vmatrix} 5 & 1\\ -3 & -1\\ \end{vmatrix} + 3 \times \begin {vmatrix} 5 & 1\\ -3 & 5\\ \end{vmatrix}$

$= 2 \times (1 \times (-1) – 1 \times 5) – 4 \times (5 \times (-1) – 1 \times (-3)) + 3 \times (5 \times 5 – 1 \times (-3))$

$= 2 \times (-1 – 5) – 4 \times (-5 + 3) + 3 \times (25 + 3)$

$= 2 \times (-6) – 4 \times (-2) + 3 \times 28$

$= -12 + 8 + 84 = 80$

### 7. Triangular Property of Determinant

The triangular property of the determinant states that if the elements above or below the main diagonal are equal to zero, then the value of the determinant is equal to the product of the elements of the diagonal matrix.

$\begin {vmatrix} a_1 & a_2 & a_3\\ 0 & b_2 & b_3\\ 0 & 0 & c_3\\ \end{vmatrix} = \begin {vmatrix} a_1 & 0 & 0\\ b_1 & b_2 & 0\\ c_1 & c_2 & c_3\\ \end{vmatrix} = a_1 b_2 c_2$

Note:

• $\begin {vmatrix} a_1 & a_2 & a_3\\ 0 & b_2 & b_3\\ 0 & 0 & c_3\\ \end{vmatrix}$ is called an upper triangular determinant and $\begin {vmatrix} a_1 & 0 & 0\\ b_1 & b_2 & 0\\ c_1 & c_2 & c_3\\ \end{vmatrix} = a_1 b_2 c_2$ is called a lower triangular determinant.

#### Examples of Triangular Property of Determinant

Example 1: Find the value of $\begin {vmatrix} 3 & 0 & -2\\ 0 & 5 & 7\\ 0 & 0 & 4\\ \end{vmatrix}$

$\begin {vmatrix} 3 & 0 & -2\\ 0 & 5 & 7\\ 0 & 0 & 4\\ \end{vmatrix} = 3 \times 5 \times 4 = 60$

Letâ€™s check the value of the determinant by expansion.

$\begin {vmatrix} 3 & 0 & -2\\ 0 & 5 & 7\\ 0 & 0 & 4\\ \end{vmatrix} = 3 \times \begin {vmatrix} 5 & 7\\ 0 & 4\\ \end{vmatrix} – 0 \times \begin {vmatrix} 0 & 7\\ 0 & 4\\ \end{vmatrix} – 2 \times \begin {vmatrix} 0 & 5\\ 0 & 0\\ \end{vmatrix}$

$= 3 \times (5 \times 4 – 7 \times 0) – 0 – 2 \times (0 \times 0 – 5 \times 0)$

$= 3 \times (20 – 0) – 0 – 2 \times (0 – 0) = 3 \times 20 – 0 – 2 \times 0 = 3 \times 20 – 0 – 0 = 60$.

Example 2: Find the value of $\begin {vmatrix} 1Â & 0Â & 0\\ 5 & 3 & 0\\ 6 & 7 & 4\\ \end{vmatrix}$.

$\begin {vmatrix} 1Â & 0Â & 0\\ 5 & 3 & 0\\ 6 & 7 & 4\\ \end{vmatrix} = 1 \times 3 \times 4 = 12$

Letâ€™s check the value of the determinant by expansion.

$\begin {vmatrix} 1Â & 0Â & 0\\ 5 & 3 & 0\\ 6 & 7 & 4\\ \end{vmatrix} = 1 \times \begin {vmatrix} 3Â & 0\\ 7 & 4\\ \end{vmatrix} – 0 \times \begin {vmatrix} 5Â & 0\\ 6 & 4\\ \end{vmatrix} + 0 \times \begin {vmatrix} 5Â & 3\\ 6 & 7\\ \end{vmatrix}$ $= 1 \times (3 \times 4 – 0 \times 7) – 0 + 0 = 1 \times (12 – 0) – 0 + 0 = 12$

## Examples of Using Properties of Determinants

Example 1: Find the value of the determinant $\begin {vmatrix} aÂ & bÂ & c\\ a + 2x & b + 2y & c + 2z\\ x & y & z\\ \end{vmatrix}$.

Applying $\text{R}_1 \rightarrow \text{R}_1 + 2 \text{R}_3$

$\begin {vmatrix} aÂ & bÂ & c\\ a + 2x & b + 2y & c + 2z\\ x & y & z\\ \end{vmatrix} = \begin {vmatrix} aÂ + 2x & b + 2y & c + 2z\\ a + 2x & b + 2y & c + 2z\\ x & y & z\\ \end{vmatrix} = 0$ (Elements of rows $\text{R}_1$ and $\text{R}_2$ are identical).

Example 2: Show that $\begin {vmatrix} xÂ & pÂ & q\\ p & x & q\\n q & q & x\\ \end{vmatrix} = (x – p) \left(x^2 + px – 2q^2 \right)$

Applying $\text{C}_1 \rightarrow \text{C}_1 – \text{C}_2$

$\begin {vmatrix} xÂ & pÂ & q\\ p & x & q\\ q & q & x\\ \end{vmatrix} = \begin {vmatrix} x – p & pÂ & q\\ p – x & x & q\\ qÂ – q & q & x\\ \end{vmatrix} = \begin {vmatrix} x – p & pÂ & q\\ p – x& x & q\\ 0 & q & x\\ \end{vmatrix} = \begin {vmatrix} x – p & pÂ & q\\ -(x – p)& x & q\\ 0 & q & x\\ \end{vmatrix} = (x – p)\begin {vmatrix} 1 & pÂ & q\\ -1& x & q\\ 0 & q & x\\ \end{vmatrix}$

Applying $\text{R}_1 \rightarrow \text{R}_1 + \text{R}_2$

$(x – p)\begin {vmatrix} 1 – 1 & p + x & q + q\\ -1& x & q\\ 0 & q & x\\ \end{vmatrix} = (x – p)\begin {vmatrix} 0 & p + x & 2q\\ -1& x & q\\ 0 & q & x\\ \end{vmatrix}$

$= (x – p)\begin {vmatrix} 0 & -1 & 0\\ p + x & x & q\\ 2q & q & x\\ \end{vmatrix}$

Expanding the determinant, we get $(x – p) \left(x^2 + px – 2q^2 \right)$

## Practice Problems

Find the value of the determinant $\begin {vmatrix} aÂ & a + bÂ & a + b + c\\ 2a & 3a + 2b & 4a + 3b + 2c\\ 3a & 6a + 3b & 10a + 6b + 3c\\ \end{vmatrix}$

Find the value of the determinant $\begin {vmatrix} aÂ & a + bÂ & a + b + c\\ 2a & 3a + 2b & 4a + 3b + 2c\\ 3a & 6a + 3b & 10a + 6b + 3c\\ \end{vmatrix}$

If $x = -4$ is a root of $\begin {vmatrix} xÂ & 2Â & 3\\ 1 & x & 1\\ 3 & 2 & x\\ \end{vmatrix} = 0$, then find the two roots.

## FAQs

### What are the properties of determinants?

The seven important properties of determinants are
1. Interchange Property: The value of a determinant remains unchanged if the rows or columns of a determinant are interchanged.
2. Sign Property: The sign of the value of a determinant changes if any two rows or any two columns are interchanged.
3. Zero Property: The value of a determinant is equal to zero if any two rows or any two columns have the same elements.
4. Multiplication Property: The value of the determinant becomes $k$ times the previous value of the determinant if each of the elements of a particular row or column is multiplied with a constant $k$.
5. Sum Property: If a few elements of a row or column are expressed as a sum of terms, then the determinant can be expressed as a sum of two or more determinants.
6. Invariance Property: If each element of a row and column of a determinant is added with the equimultiples of the elements of another row or column of a determinant, then the value of the determinant remains unchanged. This can be expressed in the form of a formula as $\text{R}_i \rightarrow \text{R}_i + k \text{R}_j$, or $\text{C}_i \rightarrow \text{C}_{i} + k \text{C}_j$.
7. Triangular Property: If the elements above or below the main diagonal are equal to zero, then the value of the determinant is equal to the product of the elements of the diagonal of the matrix.

### What are the uses of properties of determinants?

The properties of determinants are useful to easily find the value of the determinant with the least calculations. Based on the elements and the row and column operations, the value of the determinant can be easily computed.

### How do the properties of determinants differ from the properties of matrices?

The properties of determinants differed from the properties of matrices, as much as the determinant differs from the matrix. For example, in a determinant, the elements of a particular row or column can be multiplied with a constant, but in a matrix, the multiplication of a matrix with constant multiplies each element of the matrix.

## Conclusion

The determinant and its properties are useful as they help us to obtain the same outcomes with distinct and simpler configurations of elements. The seven important properties of determinants are Interchange Property, Sign Property, Zero Property, Multiplication Property, Sum Property, Invariance Property, and Triangular Property.