Properties of Determinants

Properties of determinants are required to find the value of the determinant with least calculations. The properties of determinants are based on the elements, the row, and column operations, and it helps to easily find the value of the determinant.

Properties of Determinants

Interchange Property

The value of a determinant remains unchanged if the rows and the columns of a determinant are interchanged.

Before the rows and the columns are interchanged

$\bigtriangleup ={\begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$

After the rows and the columns are interchanged

$\bigtriangleup _{1}={\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}}$

$\bigtriangleup =\bigtriangleup _{1}$

Verification

$\bigtriangleup =a_{1}{\begin{vmatrix}b_{2}&b_{3}\\c_{2}&c_{3}\end{vmatrix}}-a_{2}{\begin{vmatrix}b_{1}&b_{3}\\c_{1}&c_{3}\end{vmatrix}}+a_{3}{\begin{vmatrix}b_{1}&b_{2}\\c_{1}&c_{2}\end{vmatrix}}$

$\bigtriangleup =a_{1}(b_{2}c_{3}-b_{3}c_{2})-a_{2}(b_{1}c_{3}-b_{3}c_{1})+a_{3}(b_{1}c_{2}-b_{2}c_{1})$

$\bigtriangleup =a_{1}b_{2}c_{3}-a_{1}b_{3}c_{2}-a_{2}b_{1}c_{3}+a_{2}b_{3}c_{1}+a_{3}b_{1}c_{2}-a_{3}b_{2}c_{1}$

$\bigtriangleup _{1}=a_{1}{\begin{vmatrix}b_{2}&c_{2}\\b_{3}&c_{3}\end{vmatrix}}-b_{1}{\begin{vmatrix}a_{2}&c_{2}\\a_{3}&c_{3}\end{vmatrix}}+c_{1}{\begin{vmatrix}a_{2}&b_{2}\\a_{3}&b_{3}\end{vmatrix}}$

$\bigtriangleup _{1}=a_{1}(b_{2}c_{3}-c_{2}b_{3})-b_{1}(a_{2}c_{3}-c_{2}a_{3})+c_{1}(a_{2}b_{3}-b_{2}a_{3})$

$\bigtriangleup _{1}=a_{1}b_{2}c_{3}-a_{1}b_{3}c_{2}-a_{2}b_{1}c_{3}+a_{3}b_{1}c_{2}+a_{2}b_{3}c_{1}-a_{3}b_{2}c_{1}$

Hence $\bigtriangleup =\bigtriangleup _{1}$

If the rows and columns of the matrix are interchanged, then the transpose of the matrix is obtained and the determinant value and the determinant of the transpose are equal.

Sign Property

If any two rows or any two columns are interchanged, the sign of the value of the determinant changes.

$\bigtriangleup ={\begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$

After changing any two rows

$\bigtriangleup _{1}={\begin{vmatrix}a_{1}&a_{2}&a_{3}\\c_{1}&c_{2}&c_{3}\\b_{1}&b_{2}&b_{3}\end{vmatrix}}$

$\bigtriangleup =\bigtriangleup _{1}$

Verification

$\bigtriangleup =a_{1}{\begin{vmatrix}b_{2}&b_{3}\\c_{2}&c_{3}\end{vmatrix}}-a_{2}{\begin{vmatrix}b_{1}&b_{3}\\c_{1}&c_{3}\end{vmatrix}}+a_{3}{\begin{vmatrix}b_{1}&b_{2}\\c_{1}&c_{2}\end{vmatrix}}$

$\bigtriangleup =a_{1}(b_{2}c_{3}-b_{3}c_{2})-a_{2}(b_{1}c_{3}-b_{3}c_{1})+a_{3}(b_{1}c_{2}-b_{2}c_{1})$

$\bigtriangleup =a_{1}b_{2}c_{3}-a_{1}b_{3}c_{2}-a_{2}b_{1}c_{3}+a_{2}b_{3}c_{1}+a_{3}b_{1}c_{2}-a_{3}b_{2}c_{1}$

$\bigtriangleup =a_{1}{\begin{vmatrix}b_{2}&b_{3}\\c_{2}&c_{3}\end{vmatrix}}-a_{2}{\begin{vmatrix}b_{1}&b_{3}\\c_{1}&c_{3}\end{vmatrix}}+a_{3}{\begin{vmatrix}b_{1}&b_{2}\\c_{1}&c_{2}\end{vmatrix}}$

Zero Property

If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then value of determinant is zero.

Verification

$\bigtriangleup ={\begin{vmatrix}3&2&3\\3&2&3\\2&1&2\end{vmatrix}}$

$\bigtriangleup =3{\begin{vmatrix}2&3\\1&2\end{vmatrix}}-2{\begin{vmatrix}3&3\\2&2\end{vmatrix}}+3{\begin{vmatrix}3&2\\2&1\end{vmatrix}}$

$\bigtriangleup =3(4-3)-2(6-6)+3(3-4)$

$\bigtriangleup =3(1)-2(0)+3(-1)$

$\bigtriangleup =3-0-3=0$

Multiplication Property

If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k

$\bigtriangleup ={\begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$ $\bigtriangleup _{1}={\begin{vmatrix}ka_{1}&ka_{2}&ka_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$

$\bigtriangleup _{1}=k\bigtriangleup$

Verification

$\bigtriangleup ={\begin{vmatrix}1&1&1\\2&1&4\\3&2&3\end{vmatrix}}$ $k=2$

$\bigtriangleup _{1}={\begin{vmatrix}2&2&2\\2&1&4\\3&2&3\end{vmatrix}}$

$\bigtriangleup ={\begin{vmatrix}1&1&1\\2&1&4\\3&2&3\end{vmatrix}}=1(3-8)-1(6-12)+1(4-3)=1(-5)-1(-6)+1(1)=-5+6+1=2$

$\bigtriangleup _{1}={\begin{vmatrix}2&2&2\\2&1&4\\3&2&3\end{vmatrix}}=2(3-8)-2(6-12)+2(4-3)=2(-5)-2(-6)+2(1)=-10+12+2=4$

$\bigtriangleup _{1}=2\bigtriangleup$

Sum Property

If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants.

${\begin{vmatrix}a_{1}+d_{1}&a_{2}+d_{2}&a_{3}+d_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}={\begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}+{\begin{vmatrix}d_{1}&d_{2}&d_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$

Verification

L.H.S = ${\begin{vmatrix}a_{1}+d_{1}&a_{2}+d_{2}&a_{3}+d_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$

$=(a_{1}+d_{1})(b_{2}c_{3}-b_{3}c_{2})-(a_{2}+d_{2})(b_{1}c_{3}-b_{3}c_{1})+(a_{3}+d_{3})(b_{1}c_{2}-b_{2}c_{1})$

$=a_{1}(b_{2}c_{3}-b_{3}c_{2})-a_{2}(b_{1}c_{3}-b_{3}c_{1})+a_{3}(b_{1}c_{2}-b_{2}c_{1})+d_{1}(b_{2}c_{3}-b_{3}c_{2})-d_{2}(b_{1}c_{3}-b_{3}c_{1})+d_{3}(b_{1}c_{2}-b_{2}c_{1})$

$={\begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}+{\begin{vmatrix}d_{1}&d_{2}&d_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$= R.H.S

Property of Invariance

If, to each element of any row or column of a determinant, the equimultiples of corresponding elements of other row (or column) are added, then value of determinant remains the same, i.e., the value of determinant remain same if we apply the operation $R_{i}=R_{i}+kR_{j}$ or $C_{i}=C_{i}+kC_{j}$

$\bigtriangleup ={\begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$ $\bigtriangleup _{1}={\begin{vmatrix}a_{1}+kc_{1}&a_{2}+kc_{2}&a_{3}+kc_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$

Here we have multiplied the elements of the third row ($R_{3}$) by a constant $k$ and added them to the corresponding elements of the first row ($R_{1}$). This is shown symbolically as $R_{1}=R_{1}+kR_{3}$

Using Sum property

$\bigtriangleup _{1}={\begin{vmatrix}a_{1}+kc_{1}&a_{2}+kc_{2}&a_{3}+kc_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}={\begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}+{\begin{vmatrix}kc_{1}&kc_{2}&kc_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}$

$\bigtriangleup _{1}=\bigtriangleup +0$ (as the $R_{1}$ and $R_{3}$ are proportional)

$\bigtriangleup _{1}=\bigtriangleup$

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 matrix.

$\bigtriangleup ={\begin{vmatrix}a_{1}&a_{2}&a_{3}\\0&b_{2}&b_{3}\\0&0&c_{3}\end{vmatrix}}={\begin{vmatrix}a_{1}&0&0\\b_{1}&b_{2}&0\\c_{1}&c_{2}&c_{3}\end{vmatrix}}=a_{1}b_{2}c_{3}$

Verification

L.H.S $=a_{1}(b_{2}c_{3}-0)-a_{2}(0-0)+a_{3}(0-0)=a_{1}b_{2}c_{3}$

R.H.S = $=a_{1}(b_{2}c_{3}-0)-0(b_{1}c_{3}-0)+0(b_{1}c_{2}-b_{2}c_{1})=a_{1}b_{2}c_{3}$

L.H.S = R.H.S