Properties of Determinants

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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).


Adjugate And Reciprocal Determinants

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