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# Properties of Determinants Last updated date: 01st Dec 2023
Total views: 21.3k
Views today: 0.21k     ## Definition of Determinants

Let a$_{1}$x + b$_{1}$ = 0 …..(1) and a$_{2}$x + b$_{2}$ = 0 ……(2)

Be two simple equations in x.

From equation (1) we get,

a$_{1}$x = -b$_{1}$ or, x = - $\frac{b_{1}}{a_{1}}$;

Similarly, from equation (2) we get,

x = - $\frac{b_{2}}{a_{2}}$;

Hence, the x-eliminate of the equations (1) and (2) is

- $\frac{b_{1}}{a_{1}}$ = - $\frac{b_{2}}{a_{2}}$ or a$_{1}$b$_{2}$ - a$_{2}$b$_{1}$ = 0

The left-hand expression a$_{1}$b$_{2}$ - a$_{2}$b$_{1}$ of equation (3) is called the determinant of the second-order and is denoted by the symbol

$\begin{vmatrix} a_{1} & b_{1} \end{vmatrix}$

$\begin{vmatrix} a_{2} & b_{2} \end{vmatrix}$

The four letters  a$_{1}$, b$_{1}$, a$_{2}$, b$_{2}$ are called the elements or constituents of the determinant. The elements  a$_{1}$, b$_{1}$ form the first row, and a$_{2}$, b$_{2}$ form the second row while the elements  a$_{1}$, a$_{2}$ and  b$_{1}$, b$_{2}$ constitute the first and the second columns respectively. a$_{1}$b$_{2}$ are the elements along the leading or the principal diagonal and b$_{1}$, a$_{2}$ are the elements along the secondary diagonal. Thus, an arrangement of four letters (or numbers) of the second-order and its value is the expression a$_{1}$b$_{2}$ - a$_{2}$b$_{1}$.

Hence by definition, we have,

$\begin{vmatrix} a_{1} & b_{1} \end{vmatrix}$

$\begin{vmatrix} a_{2} & b_{2} \end{vmatrix}$ = a$_{1}$b$_{2}$ - a$_{2}$b$_{1}$.

### Properties of a Determinant

Property 1: The value of the determinant remains unaltered by changing its rows into columns and columns into rows.

Property 2: If two adjacent rows (or columns) of a determinant are interchanged, the numerical value of the determinant remains the same but its sign is altered.

Property 3: If two rows (or columns) of a determinant are identical, the value of the determinant is zero.

Property 4: If each element of a row (or a column) of a determinant be multiplied by the same quantity, then the determinant is multiplied by that quantity.

Property 5: The value of the determinant remains unaltered if each element of a row (or column) is increased or decreased by a constant multiple of the corresponding elements of another row (or column).

The adjoint or adjugate of a given determinant D is the determinant whose elements are co-factors of the corresponding elements D and is denoted by the symbol D’ .

The reciprocal or inverse of a given determinant D is the determinant which is formed by dividing every element of D’ , the adjoint of D, by D and is denoted by D’’ .

### Solved Examples

Example 1) Without expanding show that,

$\begin{vmatrix} 5 & 2 & 3 \end{vmatrix}$

$\begin{vmatrix} 7 & 3 & 4 \end{vmatrix}$ = 0

$\begin{vmatrix} 9 & 4 & 5 \end{vmatrix}$

Solution 1) We have

$\begin{vmatrix} 5 & 2 & 3 \end{vmatrix}$ $\begin{vmatrix} 5 & 2+3 & 3 \end{vmatrix}$

$\begin{vmatrix} 7 & 3 & 4 \end{vmatrix}$ = $\begin{vmatrix} 7 & 3+4 & 4 \end{vmatrix}$

$\begin{vmatrix} 9 & 4 & 5 \end{vmatrix}$ $\begin{vmatrix} 9 & 4+5 & 5 \end{vmatrix}$

Replacing the 2nd column by C$_{2}$ + C$_{3}$

$\begin{vmatrix} 5 & 2 & 3 \end{vmatrix}$

$\begin{vmatrix} 7 & 3 & 4 \end{vmatrix}$ = 0

$\begin{vmatrix} 9 & 4 & 5 \end{vmatrix}$

The first and the second column are identical

(Proved)

Example 2) Without expanding show that,

$\begin{vmatrix} a+b & 2a+b & 3a+b \end{vmatrix}$

$\begin{vmatrix} 2a+b & 3a+b & 4a+b \end{vmatrix}$ = 0

$\begin{vmatrix} 4a+b & 5a+b & 6a+b \end{vmatrix}$

Solution 2) We have

$\begin{vmatrix} a+b & 2a+b & 3a+b \end{vmatrix}$

L.H.S.=   $\begin{vmatrix} 2a+b & 3a+b & 4a+b \end{vmatrix}$

$\begin{vmatrix} 4a+b & 5a+b & 6a+b \end{vmatrix}$

$\begin{vmatrix} a+b & a & a \end{vmatrix}$

$\begin{vmatrix} 2a+b & a & a \end{vmatrix}$ [C’$_{2}$ = C$_{2}$ - C$_{1}$] and [C’$_{3}$ = C$_{3}$ - C$_{2}$]

$\begin{vmatrix} 4a+b & a & a \end{vmatrix}$

= 0 [2nd and 3rd column are identical]

(Proved)

Example 3) If x + y + z = 0 then show that $\begin{vmatrix} 1 & 1 & 1 \end{vmatrix}$

$\begin{vmatrix} x & y & z \end{vmatrix}$ = 0

$\begin{vmatrix} x^{3} & y^{3} & z^{3} \end{vmatrix}$

Solution 3) We have,

$\begin{vmatrix} 1 & 1 & 1 \end{vmatrix}$ $\begin{vmatrix} 0 & 0 & 1 \end{vmatrix}$

$\begin{vmatrix} x & y & z \end{vmatrix}$ = $\begin{vmatrix} x - y & y - z & y \end{vmatrix}$ [C’$_{1}$ = C$_{1}$ - C$_{2}$] and [C’$_{2}$ = C$_{2}$ - C$_{3}$]

$\begin{vmatrix} x^{3} & y^{3} & z^{3} \end{vmatrix}$ $\begin{vmatrix} x^{3} - y^{3} & y^{3} - z^{3} & z^{3} \end{vmatrix}$

= (x - y)(y$^{3}$ - z$^{3}$) - (y - z)(x$^{3}$ - y$^{3}$) expanding by the first row

= (x - y)(y - z)[(y$^{2}$ + yz + z$^{2}$) - (x$^{2}$ + xy + y$^{2}$)]

=  (x - y)(y - z)[y(z - x) + (z$^{2}$ - x$^{2}$)]

= (x - y)(y - z)(z - x)(y + z + x)

= 0[x + y + z = 0(given)]

(Proved)

Example 4) Eliminate x,y, and z from the following equations:

$\frac{bx}{y+z}$ = a, $\frac{cy}{z+x}$ = b, $\frac{az}{x+y}$ = c

Solution 4) The given equations are,

$\frac{bx}{y+z}$ = a  or, bx - ay - az = 0

$\frac{cy}{z+x}$ = b  or, bx - cy + bz = 0

and $\frac{az}{x+y}$ = c or, cx + cy - az = 0

Eliminate x,y, and z from the above equations, we get,

$\begin{vmatrix} b & -a & -a \end{vmatrix}$

$\begin{vmatrix} b & -c & b \end{vmatrix}$ = 0

$\begin{vmatrix} c & c & -a \end{vmatrix}$

or, b(ca - bc) + a(-ab - bc) - a(bc + c$^{2}$) = 0

or, abc - b$^{2}$c - a$^{2}$b - abc - abc - ac$^{2}$ = 0

or, a$^{2}$b + b$^{2}$c + c$^{2}$a + abc = 0, which is the required result after elimination