4.2 Properties of Determinant

Maths > Matrices and Determinants > 4.0 Determinant of a square matrix

  Matrices and Determinants
    1.0 Introduction
    2.0 Algebra of Matrices
    3.0 Special Matrices
    4.0 Determinant of a square matrix
    5.0 Adjoint of a square Matrix
    6.0 Inverse of a Matrix
    7.0 Types of Equations Homogenous & Non-Homogenous
    8.0 Cramer's rule
    9.0 Types of Linear Equations
4.2 Properties of Determinant
  • |$A$|=|${A^T}$| for any square matrix $A$. i.e. the value of a determinant remains unaltered, if the rows and clumns are interchanged
Example:

$\left| {\begin{array}{c} {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = \left| {\begin{array}{c} {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|$


  • If any two rows (or columns) of a determinant be interchanged,the sign of determinant is changed,the numerical value remaining unaltered

Example:

$\left| {\begin{array}{c} {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right|$ $ = - \left| {\begin{array}{c} {{a_3}}&{{b_3}}&{{c_3}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_1}}&{{b_1}}&{{c_1}} \end{array}} \right|$ ${R_1} \leftrightarrow {R_3}$

  • If two rows or columns of a determinant are identical,the value of the determinant is "zero".

Example:
$\left| {\begin{array}{c} {{a_1}}&{{b_1}}&{{a_1}} \\ {{a_2}}&{{b_2}}&{{a_2}} \\ {{a_3}}&{{b_3}}&{{a_3}} \end{array}} \right| = 0$ $(\because {C_1} = {C_3})$

  • If all the elements of a row (column ) is multiplied by a non zero number $k$ , then the value of the new determinant is $k$ times the vaue of the original determinant
Example:

$k\left| {\begin{array}{c} {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = \left| {\begin{array}{c} {k{a_1}}&{k{b_1}}&{k{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right|$



  • If each element of any row (or column ) consists of two terms ,then the determinant can be expressed as the sum of two other determinants
Example:
$\left| {\begin{array}{c} {{a_1} + x}&{{b_1}}&{{c_1}} \\ {{a_2} + y}&{{b_2}}&{{c_2}} \\ {{a_3} + z}&{{b_3}}&{{c_3}} \end{array}} \right| = \left| {\begin{array}{c} {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| + \left| {\begin{array}{c} x&{{b_1}}&{{c_1}} \\ y&{{b_2}}&{{c_2}} \\ z&{{b_3}}&{{c_3}} \end{array}} \right|$

  • If all elements below leading diagonal or above leading diagonal or except leading diagonal elements are zero, then the value of the determinant is equal to product of all leading diagonal elements
Example:

$\left| {\begin{array}{c} {{a_1}}&{{b_1}}&{{c_1}} \\ 0&{{b_2}}&{{c_2}} \\ 0&0&{{c_3}} \end{array}} \right| = \left| {\begin{array}{c} {{a_1}}&0&0 \\ {{a_2}}&{{b_2}}&0 \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = \left| {\begin{array}{c} {{a_1}}&0&0 \\ 0&{{b_2}}&0 \\ 0&0&{{c_3}} \end{array}} \right| = {a_1}{b_2}{c_3}$

  • If a determinant $D$ becomes zero on putting $x$=$\alpha $ ,then we say that ($x$-$\alpha $) is factor of $\Delta $

Example:

$D = \left| {\begin{array}{c} x&5&2 \\ {{x^2}}&9&4 \\ {{x^3}}&{16}&8 \end{array}} \right|$

At $x$=2 , $D$=0 (because ${C_1}$ and ${C_2}$ are identical at $x$=2)

Hence ($x$-2) is a factor of $D$.


  • The value of a determinant is unaltered when any row (or column) is multilied by a number of or any expression and then added or subtracted from any other row (or column)
Example:

$\left| {\begin{array}{c} {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = \left| {\begin{array}{c} {{a_1} + m{b_1} - n{c_1}}&{{b_1}}&{{c_1}} \\ {{a_2} + m{b_2} - n{c_2}}&{{b_2}}&{{c_2}} \\ {{a_3} + m{b_3} - n{c_3}}&{{b_3}}&{{c_3}} \end{array}} \right|$ ${C_1} \to {C_1} + m{C_2} - n{C_3}$

Note:
  • It should be noted that while applying operations on determinant then atleast one row (or column) must remain unchanged.
  • It should be noted that if the row (or column) which is changed by multiplying a non-zero number, then the determinant will be divided by that number


Question 9.

Prove that $\left| {\begin{array}{c} {{a^2} + 1}&{ab}&{ac} \\ {ab}&{{b^2} + 1}&{bc} \\ {ac}&{bc}&{{c^2} + 1} \end{array}} \right| = 1 + {a^2} + {b^2} + {c^2}$

Solution:

L.H.S.= $\left| {\begin{array}{c} {{a^2} + 1}&{ab}&{ac} \\ {ab}&{{b^2} + 1}&{bc} \\ {ac}&{bc}&{{c^2} + 1} \end{array}} \right|$

Multiplying ${C_1},{C_2},{C_3}$ by $a,b,c$ respectively

= $\frac{1}{{abc}}\left| {\begin{array}{c} {a({a^2} + 1)}&{a{b^2}}&{a{c^2}} \\ {{a^2}b}&{b({b^2} + 1)}&{b{c^2}} \\ {{a^2}c}&{{b^2}c}&{c({c^2} + 1)} \end{array}} \right|$

Now taking common $a,b,c$ from ${R_1},{R_2},{R_3}$ respectively

$\frac{{abc}}{{abc}}\left| {\begin{array}{c} {({a^2} + 1)}&{{b^2}}&{{c^2}} \\ {{a^2}}&{({b^2} + 1)}&{{c^2}} \\ {{a^2}}&{{b^2}}&{({c^2} + 1)} \end{array}} \right|$

= $\left| {\begin{array}{c} {1 + {a^2} + {b^2} + {c^2}}&{{b^2}}&{{c^2}} \\ {1 + {a^2} + {b^2} + {c^2}}&{({b^2} + 1)}&{{c^2}} \\ {1 + {a^2} + {b^2} + {c^2}}&{{b^2}}&{({c^2} + 1)} \end{array}} \right|$ $\left[ {{C_1} \to {C_1} + {C_2} + {C_3}} \right]$

= $(1 + {a^2} + {b^2} + {c^2})\left| {\begin{array}{c} 1&{{b^2}}&{{c^2}} \\ 1&{({b^2} + 1)}&{{c^2}} \\ 1&{{b^2}}&{({c^2} + 1)} \end{array}} \right|$

= $(1 + {a^2} + {b^2} + {c^2})\left| {\begin{array}{c} 1&{{b^2}}&{{c^2}} \\ 0&1&0 \\ 0&0&1 \end{array}} \right|$ $\left[ {{R_2} \to {R_2} - {R_1}} \right],\left[ {{R_3} \to {R_3} - {R_1}} \right]$

= $(1 + {a^2} + {b^2} + {c^2})(1.1.1)$

= $(1 + {a^2} + {b^2} + {c^2})$=R.H.S



Question 10.

Prove that $\left| {\begin{array}{c} 1&1&1 \\ x&y&z \\ {{x^3}}&{{y^3}}&{{z^3}} \end{array}} \right| = \left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\left( {x + y + z} \right)$
Solution:

Let $D = \left| {\begin{array}{c} 1&1&1 \\ x&y&z \\ {{x^3}}&{{y^3}}&{{z^3}} \end{array}} \right|$

$ = \left| {\begin{array}{c} 1&0&0 \\ x&{y - x}&{z - x} \\ {{x^3}}&{{y^3} - {x^3}}&{{z^3} - {x^3}} \end{array}} \right|\left[ {{C_2} \to {C_2} - {C_1},{C_3} \to {C_3} - {C_1}} \right]$

$ = \left| {\begin{array}{c} {y - x}&{z - x} \\ {{y^3} - {x^3}}&{{z^3} - {x^3}} \end{array}} \right|$

$ = \left( {y - x} \right)\left( {z - x} \right)\left| {\begin{array}{c} 1&1 \\ {{y^2} + {x^2} + yx}&{{z^2} + {x^2} + yx} \end{array}} \right|$

$ = \left( {y - x} \right)\left( {z - x} \right)\left( {{z^2} + {x^2} + yx - {y^2} - {x^2} - yx} \right)$

$ = \left( {y - x} \right)\left( {z - x} \right)\left\{ {\left( {z + y} \right)\left( {z - y} \right) + x\left( {z - y} \right)} \right\}$

$ = \left( {y - x} \right)\left( {z - x} \right)\left( {z - y} \right)\left( {z + y + x} \right)$

$\therefore D = \left( {y - x} \right)\left( {z - x} \right)\left( {z - y} \right)\left( {z + y + x} \right)$
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