8 PROPERTIES OF DETERMINANTS OF MATRICES

The determinant of a matrix is the scalar value that is obtained from a square matrix.

Eight properties of determinants are reflection, all-zero, proportionality, switching, scalar multiple, sum, property of invariance, and triangle properties.

Property #1: Reflection property

This means that even if a matrix's rows turn into columns or its columns turn into rows, the determinant will be unaltered.

Proof: 

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

Interchanging the rows and columns

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

The determinant of A is:

$(4×1)-(3×2)=-2$

The determinant of A' is:

$(4×1)-(2×3)=-2$

Thus, we see that $A=A'$

Property #2: All-zero property

If all elements of a column (or row) are zero, then the determinant of the matrix is zero.

Proof:

Let $A=\begin{bmatrix} 0 & 0 & 0 \\ 1 & 2 & 3 \\ 4 & 5 & 6\end{bmatrix}$

$0(2×6-3×5)-0(1×6-3×4)+0(1×5-2×4)=0$

Property #3: Proportionality (repetition) property.

If every element of a row (or column) are proportional or identical to the elements of another row (or column), then the determinant of the matrix is zero.

Proof:

Let $A=\begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 2 & 3 & 6\end{bmatrix}$

If you observed, the elements of the first row are identical to that of the second row, hence the determinant of the matrix is zero.

To prove,

$1(2×6-3×3)-2(1×6-3×2)+3(1×3-2×2)$

$1(12-9)-2(6-6)+3(3-4)$

$1(3)-2(0)+3(-1)$

$3-0-3=0$

Property #4: Switching property

The interchange of any two rows (or two columns) of a matrix changes the sign of its determinant.

This means that if the determinant of a matrix is 2, switching two of its rows( or columns) would change its determinant to -2.

Proof:

Let $A=\begin{bmatrix} 3 & 2 \\ 4 & 6 \end{bmatrix}$

Interchanging the first and second columns,

$A'=\begin{bmatrix} 2 & 3 \\ 6 & 4 \end{bmatrix}$

Determinant of  A is:

$3×6-4×2=10$

Determinant of A'

$2×4-3×6=-10$

Therefore, interchanging A column changes its determinant from 10 to -10.

Property #5: Scalar multiple property

If each element in a row (or in a column) of a matrix is multiplied by a scalar value, then the determinant of the matrix is also multiplied by the scalar value.

Proof:

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

Then, the determinant of A is

$A=(1×4-2×3)=-2$

Assuming, we multiply each element of the first row by 2 so that

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

Then, the determinant of the matrix A is two 

$A=2×4-4×3$

$A=8-12$

$A=-4$

As can be seen, when we multiply the first row by 2, the determinant of the transform ed matrix A (-4) is twice that of the original matrix A (-2).

Property #6: Sum property

A determinant can be expressed as the total of the determinants in the same order if each element in any row (or column) is expressed as the sum of two values.

That is, if

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

Then

$DetA=Det\begin{bmatrix} a & e\\c & f \end{bmatrix}+Det\begin{bmatrix} b & e \\d & f \end{bmatrix}$

Proof:

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

The determinant of A will be:

$|A|=Det\begin{bmatrix} 1 & 5 \\3 & 6 \end{bmatrix}+Det\begin{bmatrix} 2 & 5 \\4 & 6 \end{bmatrix}$

$|A|=(1×6-5×3)+(2×6-5×4)$

$|A|=6-15+12-20$

$|A|=-9-8=-17$

Property #7: Property of invariance

The value of the determinant of the transform matrix will be the same as the original matrix if we add any constant element from the other columns (or rows) to each element of any column (or row).

If $A=\begin{bmatrix} a_1 & b_1 \\a_2 & b_2 \end{bmatrix}$

Then adding the constant multiple

If $A=\begin{bmatrix} a_1+ma_2 & b_1+mb_2 \\a_2 & b_2 \end{bmatrix}$

will yield the same constant as the original value.

Proof:

Let $A=\begin{bmatrix} 3 & 9 \\4 & 10 \end{bmatrix}$

The determinant of A is:

$|A|=3×10-9×4=-6$

Now, let multiply the second column by 2 and add it to the first column so that 

$A=\begin{bmatrix} 3 +(2×9) & 9 \\4+(2×10) & 10 \end{bmatrix}$

$A=\begin{bmatrix} 3 +18 & 9 \\4+20 & 10 \end{bmatrix}$

$A=\begin{bmatrix} 21 & 9 \\24 & 10 \end{bmatrix}$

$|A|=21×10-9×24$

$|A|=210-216=-6$

Therefore, the determinant of the transformed matrix will equal the determinant of the original matrix

Property #8: Triangle property

If all elements above or below the main diagonal of a matrix are equal to zero, then the value of the determinant is equal to zero, then the value of the determinant is equal to the product of the elements of the main diagram

Proof:

Given $A=\begin{bmatrix} 1 & 6 & 9 \\ 0 & 7 & 10 \\ 0 & 0 & 3\end{bmatrix}$

If we draw a diagonal line through 1,7,  and 3, the lower part will all be zero. Hence, the determinant of matrix A is the product of the diagonal: $1×7×3=21$

To prove, 

$1(7×3-10×0)-6(0×3-10×0)+9(0×0-7×0)$

$1(21-0)-6(0-0)+8(0-0)$

$1(21)-6(0)+8(0)=21$

READ ALSO: MULTIPLICATION OF 2X2, 3X3 MATRICES

There you have it. Got questions? Feel free to ask our telegram community.

Don't forget to subscribe to our telegram channel so that you could get FREE access to our resources and quizzes

Help us grow our readership by sharing this post

Related Posts

Post a Comment

Subscribe Our Newsletter