Matrices and Determinants Exercise: 5.2 Class 11 Basic Mathematics Solution [NEB UPDATED]

XI Basic Mathematics Solution Exercise 5.2

Exercise 5.2

1. Evaluate the determinant of the following matrices.

a.$\left( {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right)$

Solution:

Let A = $\left( {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right)$

Or, |A|= $\left| {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right|$ = 1*4 – 2 * 3 = –2. 

b.$\left( {\begin{array}{*{20}{c}}1&{ - 1}\\1&{ - 1}\end{array}} \right)$

Solution:

Let A = $\left( {\begin{array}{*{20}{c}}1&{ - 1}\\1&{ - 1}\end{array}} \right)$

Or, |A|= $\left| {\begin{array}{*{20}{c}}1&{ - 1}\\1&{ - 1}\end{array}} \right|$ = –1 + 1= 0. 

c.$\left( {\begin{array}{*{20}{c}}1&2&3\\{ - 1}&0&4\\1&2&3\end{array}} \right)$

Solution:

Let A = $\left( {\begin{array}{*{20}{c}}1&2&3\\{ - 1}&0&4\\1&2&3\end{array}} \right)$, Then

|A|= $\left| {\begin{array}{*{20}{c}}1&2&3\\{ - 1}&0&4\\1&2&3\end{array}} \right|$.

Or, |A| = 1$\left| {\begin{array}{*{20}{c}}0&4\\2&3\end{array}} \right|$ – 2 $\left| {\begin{array}{*{20}{c}}{ - 1}&4\\1&3\end{array}} \right|$ + 3 $\left| {\begin{array}{*{20}{c}}{ - 1}&0\\1&2\end{array}} \right|$

= 1(0 – 8) – 2(3 – 4) + 3(–2 – 0) = 0 

d.$\left( {\begin{array}{*{20}{c}}{ - 1}&0&3\\2&1&4\\{ - 2}&{ - 3}&1\end{array}} \right)$

Solution:

Let A = $\left( {\begin{array}{*{20}{c}}{ - 1}&0&3\\2&1&4\\{ - 2}&{ - 3}&1\end{array}} \right)$, Then

Or, |A|= $\left| {\begin{array}{*{20}{c}}{ - 1}&0&3\\2&1&4\\{ - 2}&{ - 3}&1\end{array}} \right|$.

Or, |A| = –1$\left| {\begin{array}{*{20}{c}}1&4\\{ - 3}&{ - 1}\end{array}} \right|$ – 0 $\left| {\begin{array}{*{20}{c}}2&4\\{ - 2}&{ - 1}\end{array}} \right|$ + 3 $\left| {\begin{array}{*{20}{c}}2&1\\{ - 2}&{ - 3}\end{array}} \right|$

= –1(–1 + 12) – 0 + 3(–6 + 2) = –23. 

e. $\left( {\begin{array}{*{20}{c}}3&{ - 2}&2\\4&1&0\\2&3&1\end{array}} \right)$

Solution:

Let A = $\left( {\begin{array}{*{20}{c}}3&{ - 2}&2\\4&1&0\\2&3&1\end{array}} \right)$, Then

Or, |A|= $\left| {\begin{array}{*{20}{c}}3&{ - 2}&2\\4&1&0\\2&3&1\end{array}} \right|$.

Or, |A| = 3$\left| {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right|$ – (–2)$\left| {\begin{array}{*{20}{c}}4&0\\2&1\end{array}} \right|$ + 2 $\left| {\begin{array}{*{20}{c}}4&1\\2&3\end{array}} \right|$

= 3(1 – 0) + 2(4 – 0) + 2(12 – 2) = 31.

f.$\left( {\begin{array}{*{20}{c}}2&{ - 2}&0\\{ - 1}&2&4\\{ - 4}&5&6\end{array}} \right)$

Solution:

Let A = $\left( {\begin{array}{*{20}{c}}2&{ - 2}&0\\{ - 1}&2&4\\{ - 4}&5&6\end{array}} \right)$, Then

Or, |A|= $\left| {\begin{array}{*{20}{c}}2&{ - 2}&0\\{ - 1}&2&4\\{ - 4}&5&6\end{array}} \right|$.

Or, |A| = 2$\left| {\begin{array}{*{20}{c}}2&4\\5&6\end{array}} \right|$ + 2$\left| {\begin{array}{*{20}{c}}{ - 1}&4\\{ - 4}&6\end{array}} \right|$ + 0 $\left| {\begin{array}{*{20}{c}}{ - 1}&2\\{ - 4}&5\end{array}} \right|$

= 2(12 – 20) + 2(–6 + 16) + 0 = 4.

= 4.

 

2. Solve for x.

a.$\left| {\begin{array}{*{20}{c}}{\rm{x}}&2&3\\{ - 1}&0&1\\2&{ - 2}&0\end{array}} \right|$=0

Solution:

$\left| {\begin{array}{*{20}{c}}{\rm{x}}&2&3\\{ - 1}&0&1\\2&{ - 2}&0\end{array}} \right|$=0

Or, $\left| {\begin{array}{*{20}{c}}{\rm{x}}&2&3\\{ - 1}&0&1\\2&{ - 2}&0\end{array}} \right|$=0

Or, x $\left| {\begin{array}{*{20}{c}}0&1\\{ - 2}&0\end{array}} \right| - 2\left| {\begin{array}{*{20}{c}}{ - 1}&1\\2&0\end{array}} \right| + 3\left| {\begin{array}{*{20}{c}}{ - 1}&0\\2&{ - 2}\end{array}} \right|$ = 0

Or, x(0 + 2) – 2(0 – 2) + 3(2 – 0) = 0

Or, 2x + 4 + 6 = 0

Or, 2x + 10 = 0

Or, x = –5. 

b.$\left| {\begin{array}{*{20}{c}}{\rm{x}}&3&3\\3&3&{\rm{x}}\\2&2&3\end{array}} \right|$=0

Solution:

$\left| {\begin{array}{*{20}{c}}{\rm{x}}&3&3\\3&3&{\rm{x}}\\2&2&3\end{array}} \right|$=0

Or, $\left| {\begin{array}{*{20}{c}}{\rm{x}}&3&3\\3&3&{\rm{x}}\\2&2&3\end{array}} \right|$=0

Or, x $\left| {\begin{array}{*{20}{c}}3&{\rm{x}}\\3&3\end{array}} \right| - 3\left| {\begin{array}{*{20}{c}}3&{\rm{x}}\\2&3\end{array}} \right| + 3\left| {\begin{array}{*{20}{c}}3&3\\2&3\end{array}} \right|$ = 0

Or, x(9 – 3x) – 3(9 – 2x) + 3(9 – 6) = 0

Or, 9x – 3x² – 27 + 6x + 9 = 0

Or, –3x² + 15x – 18 = 0

Or, x² – 5x + 6 = 0

Or, (x – 2)(x – 3) = 0

Either, x – 2 = 0.

So, x = 2

Or, x – 3 = 0

So, x = 3

So, x = 2 or 3.

 

3. Without expanding the determinants, show that the values of each of the following determinants is zero.

(i)$\left| {\begin{array}{*{20}{c}}6&1&9\\2&4&7\\{18}&3&{27}\end{array}} \right|$

Solution:

Or, $\left| {\begin{array}{*{20}{c}}6&1&9\\2&4&7\\{18}&3&{27}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}6&1&9\\2&4&7\\{3{\rm{*}}6}&{3{\rm{*}}1}&{3{\rm{*}}9}\end{array}} \right| = 3\left| {\begin{array}{*{20}{c}}6&1&9\\2&4&7\\6&1&9\end{array}} \right|$ (taking 3 common from R₃).

= 3.0 = 0   (So, R₁ = R₃). 

(ii)$\left| {\begin{array}{*{20}{c}}1&{\rm{a}}&{{\rm{b}} + {\rm{c}}}\\1&{\rm{b}}&{{\rm{c}} + {\rm{a}}}\\1&{\rm{c}}&{{\rm{a}} + {\rm{b}}}\end{array}} \right|$

Soln:

Or, $\left| {\begin{array}{*{20}{c}}1&{\rm{a}}&{{\rm{b}} + {\rm{c}}}\\1&{\rm{b}}&{{\rm{c}} + {\rm{a}}}\\1&{\rm{c}}&{{\rm{a}} + {\rm{b}}}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&{\rm{a}}&{{\rm{b}} + {\rm{c}} + {\rm{a}}}\\1&{\rm{b}}&{{\rm{c}} + {\rm{a}} + {\rm{b}}}\\1&{\rm{c}}&{{\rm{a}} + {\rm{b}} + {\rm{c}}}\end{array}} \right|{\rm{\: }}$(C₃→ C₂ + C₃)

= (a + b + c) .$\left| {\begin{array}{*{20}{c}}1&{\rm{a}}&1\\1&{\rm{b}}&1\\1&{\rm{c}}&1\end{array}} \right|$  (Taking a + b + c common from C₃).

= (a + b + c). 0 = 0   (C₁ = C₃). 

(iii)$\left| {\begin{array}{*{20}{c}}{\left( {{\rm{a}} - {\rm{b}}} \right)}&{\left( {{\rm{b}} - {\rm{c}}} \right)}&{\left( {{\rm{c}} - {\rm{a}}} \right)}\\{\left( {{\rm{b}} - {\rm{c}}} \right)}&{\left( {{\rm{c}} - {\rm{a}}} \right)}&{\left( {{\rm{a}} - {\rm{b}}} \right)}\\{\left( {{\rm{c}} - {\rm{a}}} \right)}&{\left( {{\rm{a}} - {\rm{b}}} \right)}&{\left( {{\rm{b}} - {\rm{c}}} \right)}\end{array}} \right|$

Solution:

Or, $\left| {\begin{array}{*{20}{c}}{\left( {{\rm{a}} - {\rm{b}}} \right)}&{\left( {{\rm{b}} - {\rm{c}}} \right)}&{\left( {{\rm{c}} - {\rm{a}}} \right)}\\{\left( {{\rm{b}} - {\rm{c}}} \right)}&{\left( {{\rm{c}} - {\rm{a}}} \right)}&{\left( {{\rm{a}} - {\rm{b}}} \right)}\\{\left( {{\rm{c}} - {\rm{a}}} \right)}&{\left( {{\rm{a}} - {\rm{b}}} \right)}&{\left( {{\rm{b}} - {\rm{c}}} \right)}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{{\rm{a}} - {\rm{b}} + {\rm{b}} - {\rm{c}}}&{{\rm{b}} - {\rm{c}}}&{{\rm{c}} - {\rm{a}}}\\{{\rm{b}} - {\rm{c}} + {\rm{c}} - {\rm{a}}}&{{\rm{c}} - {\rm{a}}}&{{\rm{a}} - {\rm{b}}}\\{{\rm{c}} - {\rm{a}} + {\rm{a}} - {\rm{b}}}&{{\rm{a}} - {\rm{b}}}&{{\rm{b}} - {\rm{c}}}\end{array}} \right|{\rm{\: }}$(C₁→ C₁ + C₂)

=$\left| {\begin{array}{*{20}{c}}{{\rm{a}} - {\rm{c}}}&{{\rm{b}} - {\rm{c}}}&{{\rm{c}} - {\rm{a}}}\\{{\rm{b}} - {\rm{a}}}&{{\rm{c}} - {\rm{a}}}&{{\rm{a}} - {\rm{b}}}\\{{\rm{c}} - {\rm{b}}}&{{\rm{a}} - {\rm{b}}}&{{\rm{b}} - {\rm{c}}}\end{array}} \right|{\rm{\: }}$= $ - \left| {\begin{array}{*{20}{c}}{{\rm{c}} - {\rm{a}}}&{{\rm{b}} - {\rm{c}}}&{{\rm{c}} - {\rm{a}}}\\{{\rm{a}} - {\rm{b}}}&{{\rm{c}} - {\rm{a}}}&{{\rm{a}} - {\rm{b}}}\\{{\rm{b}} - {\rm{c}}}&{{\rm{a}} - {\rm{b}}}&{{\rm{b}} - {\rm{c}}}\end{array}} \right|$ (Taking –1 common from C₁).

= 0 (C₁ = C₃). 

(iv)$\left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{{\rm{a}}\left( {{\rm{b}} + {\rm{c}}} \right)}\\1&{{\rm{ca}}}&{{\rm{b}}\left( {{\rm{c}} + {\rm{a}}} \right)}\\1&{{\rm{ab}}}&{{\rm{c}}\left( {{\rm{a}} + {\rm{b}}} \right)}\end{array}} \right| $

Solution:

Or, $\left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{{\rm{a}}\left( {{\rm{b}} + {\rm{c}}} \right)}\\1&{{\rm{ca}}}&{{\rm{b}}\left( {{\rm{c}} + {\rm{a}}} \right)}\\1&{{\rm{ab}}}&{{\rm{c}}\left( {{\rm{a}} + {\rm{b}}} \right)}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{{\rm{ab}} + {\rm{ac}}}\\1&{{\rm{ca}}}&{{\rm{bc}} + {\rm{ba}}}\\1&{{\rm{ab}}}&{{\rm{ca}} + {\rm{cb}}}\end{array}} \right|{\rm{\: }}$

= $\left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{{\rm{ab}} + {\rm{ac}} + {\rm{bc}}}\\1&{{\rm{ca}}}&{{\rm{bc}} + {\rm{ab}} + {\rm{ca}}}\\1&{{\rm{ab}}}&{{\rm{ac}} + {\rm{bc}} + {\rm{ab}}}\end{array}} \right|{\rm{\: }}\left( {{\rm{C}}1{\rm{\: \: C}}1{\rm{\: }} + {\rm{\: C}}2} \right)$

 = (ab + bc + ca) $\left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&1\\1&{{\rm{ca}}}&1\\1&{{\rm{ab}}}&1\end{array}} \right|$ (Taking ab + bc common from C₃).

= (ab + bc + ca) . 0 = 0   (C₁ = C₃). 

(v)$\left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{{\rm{bc}}\left( {{\rm{b}} + {\rm{c}}} \right)}\\1&{{\rm{ca}}}&{{\rm{ca}}\left( {{\rm{c}} + {\rm{a}}} \right)}\\1&{{\rm{ab}}}&{{\rm{ab}}\left( {{\rm{a}} + {\rm{b}}} \right)}\end{array}} \right|$

Solution:

Or, $\left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{{\rm{bc}}\left( {{\rm{b}} + {\rm{c}}} \right)}\\1&{{\rm{ca}}}&{{\rm{ca}}\left( {{\rm{c}} + {\rm{a}}} \right)}\\1&{{\rm{ab}}}&{{\rm{ab}}\left( {{\rm{a}} + {\rm{b}}} \right)}\end{array}} \right| = \frac{1}{{{\rm{abc}}}}\left| {\begin{array}{*{20}{c}}{\rm{a}}&{{\rm{abc}}}&{{\rm{abc}}\left( {{\rm{b}} + {\rm{c}}} \right)}\\{\rm{b}}&{{\rm{abc}}}&{{\rm{abc}}\left( {{\rm{c}} + {\rm{a}}} \right)}\\{\rm{c}}&{{\rm{abc}}}&{{\rm{abc}}\left( {{\rm{a}} + {\rm{b}}} \right)}\end{array}} \right|{\rm{\: }}\begin{array}{*{20}{c}}{{{\rm{R}}_1} \to {\rm{a}}{{\rm{R}}_1}}\\{{{\rm{R}}_2} \to {\rm{b}}{{\rm{R}}_2}}\\{{{\rm{R}}_3} \to {\rm{ac}}}\end{array}$

= $\frac{1}{{{\rm{abc}}}}.{\rm{abc}}.{\rm{abc}}\left| {\begin{array}{*{20}{c}}{\rm{a}}&1&{{\rm{b}} + {\rm{c}}}\\{\rm{b}}&1&{{\rm{c}} + {\rm{a}}}\\{\rm{c}}&1&{{\rm{a}} + {\rm{b}}}\end{array}} \right|$   (taking abc common from C₂ and C₃).

 = abc $\left| {\begin{array}{*{20}{c}}{\rm{a}}&1&{{\rm{b}} + {\rm{c}} + {\rm{a}}}\\{\rm{b}}&1&{{\rm{c}} + {\rm{a}} + {\rm{b}}}\\{\rm{c}}&1&{{\rm{a}} + {\rm{b}} + {\rm{c}}}\end{array}} \right|$ (C₃→C₃ + C₁)

= abc(a + b + c) $\left| {\begin{array}{*{20}{c}}{\rm{a}}&1&1\\{\rm{b}}&1&1\\{\rm{c}}&1&1\end{array}} \right|$  (taking a + b + c common from C₃).

= abc (a + b + c).0 = 0  (C₂ = C₃).

(vi)$\left| {\begin{array}{*{20}{c}}1&{\rm{w}}&{{{\rm{w}}^2}}\\{\rm{w}}&{{{\rm{w}}^2}}&1\\{{{\rm{w}}^2}}&1&{\rm{w}}\end{array}} \right|$

Solution:

Or, $\left| {\begin{array}{*{20}{c}}1&{\rm{w}}&{{{\rm{w}}^2}}\\{\rm{w}}&{{{\rm{w}}^2}}&1\\{{{\rm{w}}^2}}&1&{\rm{w}}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{1 + {\rm{w}} + {{\rm{w}}^2}}&{\rm{w}}&{{{\rm{w}}^2}}\\{{\rm{w}} + {{\rm{w}}^2} + 1}&{{{\rm{w}}^2}}&1\\{{{\rm{w}}^2} + 1 + {\rm{w}}}&1&{\rm{w}}\end{array}} \right|{\rm{\: }}$ (C₁→ C₁ + C₂ + C₃)

= $\left| {\begin{array}{*{20}{c}}0&{\rm{w}}&{{{\rm{w}}^2}}\\0&{{{\rm{w}}^2}}&1\\0&1&{\rm{w}}\end{array}} \right|$ = 0  (Each element of C₁ = 0).

 

4) Evaluate:

(i)$\left| {\begin{array}{*{20}{c}}{51}&{61}&{71}\\5&6&7\\1&1&1\end{array}} \right|$

Solution:

$\left| {\begin{array}{*{20}{c}}{51}&{61}&{71}\\5&6&7\\1&1&1\end{array}} \right|$

Or, $\left| {\begin{array}{*{20}{c}}{51}&{61}&{71}\\5&6&7\\1&1&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{51 - 1}&{61 - 1}&{71 - 1}\\5&6&7\\1&1&1\end{array}} \right|$ (R₁→R₁ – R₃)

= $\left| {\begin{array}{*{20}{c}}{50}&{60}&{70}\\5&6&7\\1&1&1\end{array}} \right|$

= 10 $\left| {\begin{array}{*{20}{c}}5&6&7\\5&6&7\\1&1&1\end{array}} \right|$ (Taking 10 common from R₁).

= 10 * 0 = 0.   (R₁ = R₂). 

(ii)$\left| {\begin{array}{*{20}{c}}{16}&{19}&{23}\\{15}&{18}&{22}\\{13}&{17}&{20}\end{array}} \right|$

Solution:

$\left| {\begin{array}{*{20}{c}}{16}&{19}&{23}\\{15}&{18}&{22}\\{13}&{17}&{20}\end{array}} \right| $

Or, $\left| {\begin{array}{*{20}{c}}{16}&{19}&{23}\\{15}&{18}&{22}\\{13}&{17}&{20}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{16 - 15}&{19 - 18}&{23 - 22}\\{15}&{18}&{22}\\{13}&{17}&{20}\end{array}} \right|$ (R₁→R₁ – R₃)

= $\left| {\begin{array}{*{20}{c}}1&1&1\\{15}&{18}&{22}\\{13}&{17}&{20}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{1 - 1}&1&{1 - 1}\\{15 - 18}&{18}&{22 - 18}\\{13 - 17}&{17}&{20 - 17}\end{array}} \right|\begin{array}{*{20}{c}}{{{\rm{C}}_1} \to {{\rm{C}}_1} - {{\rm{C}}_2}}\\{{{\rm{C}}_3} \to {{\rm{C}}_3} - {{\rm{C}}_2}}\end{array}$

= $\left| {\begin{array}{*{20}{c}}0&1&0\\{ - 3}&{18}&4\\{ - 4}&{17}&3\end{array}} \right|$

= –1 $\left| {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 4}&3\end{array}} \right|$

= –1(–9 + 16) = –7.

 

5) Without expanding the determinants, prove that

(i)$\left| {\begin{array}{*{20}{c}}1&7&8\\4&{ - 3}&4\\{ - 5}&{10}&2\end{array}} \right| $= $\left| {\begin{array}{*{20}{c}}8&4&2\\1&4&{ - 5}\\7&{ - 3}&{10}\end{array}} \right|$ 

Solution:

Or, $\left| {\begin{array}{*{20}{c}}1&7&8\\4&{ - 3}&4\\{ - 5}&{10}&2\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&4&{ - 5}\\7&{ - 3}&{10}\\8&4&2\end{array}} \right|$ (Rows and columns are interchanged).

= $ - \left| {\begin{array}{*{20}{c}}1&4&{ - 5}\\8&4&2\\7&{ - 3}&{10}\end{array}} \right|$ (Two adjacent rows are interchanged).

= $\left| {\begin{array}{*{20}{c}}8&4&2\\1&4&{ - 5}\\7&{ - 3}&{10}\end{array}} \right|$ (Two adjacent rows are interchanged).

(ii)$\left| {\begin{array}{*{20}{c}}3&5&7\\9&2&1\\7&8&5\end{array}} \right|$+$\left| {\begin{array}{*{20}{c}}9&3&7\\2&5&8\\1&7&5\end{array}} \right|$

Solution:

Or, $\left| {\begin{array}{*{20}{c}}3&5&7\\9&2&1\\7&8&5\end{array}} \right|$ and $\Delta ' = \left| {\begin{array}{*{20}{c}}9&3&7\\2&5&8\\1&7&5\end{array}} \right|$ (Rows and columns are interchanged).

Or, $\Delta $ = $\left| {\begin{array}{*{20}{c}}3&5&7\\9&2&1\\7&8&5\end{array}} \right|$ = $\left| {\begin{array}{*{20}{c}}3&9&7\\5&2&8\\7&1&5\end{array}} \right|$ (Rows and columns are interchanged).

= $ - \left| {\begin{array}{*{20}{c}}9&3&7\\2&5&8\\1&7&5\end{array}} \right|$ (Two adjacent columns are interchanged.)

= $ - \Delta {\rm{'}}$So, $\Delta  + \Delta {\rm{'}}$ = 0. 

(iii)$\left| {\begin{array}{*{20}{c}}1&{\rm{x}}&{{{\rm{x}}^2}}\\1&{\rm{y}}&{{{\rm{y}}^2}}\\1&{\rm{z}}&{{{\rm{z}}^2}}\end{array}} \right| $=$\left| {\begin{array}{*{20}{c}}1&{\rm{x}}&{{\rm{yz}}}\\1&{\rm{y}}&{{\rm{zx}}}\\1&{\rm{z}}&{{\rm{xy}}}\end{array}} \right|$

Solution:

Or, $\left| {\begin{array}{*{20}{c}}1&{\rm{x}}&{{{\rm{x}}^2}}\\1&{\rm{y}}&{{{\rm{y}}^2}}\\1&{\rm{z}}&{{{\rm{z}}^2}}\end{array}} \right| = \frac{1}{{{\rm{xyz}}}}\left| {\begin{array}{*{20}{c}}{{\rm{xyz}}}&{\rm{x}}&{{{\rm{x}}^2}}\\{{\rm{xyz}}}&{\rm{y}}&{{{\rm{y}}^2}}\\{{\rm{xyz}}}&{\rm{z}}&{{{\rm{z}}^2}}\end{array}} \right|$

= $\frac{1}{{{\rm{xyz}}}}.{\rm{xyz}}\left| {\begin{array}{*{20}{c}}{{\rm{yz}}}&1&{\rm{x}}\\{{\rm{zx}}}&1&{\rm{y}}\\{{\rm{xy}}}&1&{\rm{z}}\end{array}} \right|$  (x,y,z are taken common from R₁,R₂ and R₃).

= $\left| {\begin{array}{*{20}{c}}1&{{\rm{yz}}}&{\rm{x}}\\1&{{\rm{zx}}}&{\rm{y}}\\1&{{\rm{xy}}}&{\rm{z}}\end{array}} \right|$ (Two adjacent column are interchanged).

= $\left| {\begin{array}{*{20}{c}}1&{\rm{x}}&{{\rm{yz}}}\\1&{\rm{y}}&{{\rm{zx}}}\\1&{\rm{z}}&{{\rm{xy}}}\end{array}} \right|$ (Two adjacent columns are interchanged). 

(iv)$\left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{{\rm{b}} + {\rm{c}}}\\1&{{\rm{ca}}}&{{\rm{c}} + {\rm{a}}}\\1&{{\rm{ab}}}&{{\rm{a}} + {\rm{b}}}\end{array}} \right| $=$\left| {\begin{array}{*{20}{c}}1&{\rm{a}}&{{{\rm{a}}^2}}\\1&{\rm{b}}&{{{\rm{b}}^2}}\\1&{\rm{c}}&{{{\rm{c}}^2}}\end{array}} \right|$

Solution:

L.H.S. =$\left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{{\rm{b}} + {\rm{c}}}\\1&{{\rm{ca}}}&{{\rm{c}} + {\rm{a}}}\\1&{{\rm{ab}}}&{{\rm{a}} + {\rm{b}}}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{\left( {{\rm{a}} + {\rm{b}} + {\rm{c}}} \right) - {\rm{a}}}\\1&{{\rm{ca}}}&{\left( {{\rm{b}} + {\rm{c}} + {\rm{a}}} \right) - {\rm{b}}}\\1&{{\rm{ab}}}&{\left( {{\rm{a}} + {\rm{b}} + {\rm{c}}} \right) - {\rm{c}}}\end{array}} \right|$

= $\left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{{\rm{a}} + {\rm{b}} + {\rm{c}}}\\1&{{\rm{ca}}}&{{\rm{b}} + {\rm{c}} + {\rm{a}}}\\1&{{\rm{ab}}}&{{\rm{a}} + {\rm{b}} + {\rm{c}}}\end{array}} \right| - \left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&{\rm{a}}\\1&{{\rm{ca}}}&{\rm{b}}\\1&{{\rm{ab}}}&{\rm{c}}\end{array}} \right|$

= (a + b + c)$\left| {\begin{array}{*{20}{c}}1&{{\rm{bc}}}&1\\1&{{\rm{ca}}}&1\\1&{{\rm{ab}}}&1\end{array}} \right| - \frac{1}{{{\rm{abc}}}}\left| {\begin{array}{*{20}{c}}{\rm{a}}&{{\rm{abc}}}&{{{\rm{a}}^2}}\\{\rm{b}}&{{\rm{abc}}}&{{{\rm{b}}^2}}\\{\rm{c}}&{{\rm{abc}}}&{{{\rm{c}}^2}}\end{array}} \right|$

= (a + b + c). 0 – $\frac{1}{{{\rm{abc}}}}$.abc $\left| {\begin{array}{*{20}{c}}{\rm{a}}&1&{{{\rm{a}}^2}}\\{\rm{b}}&1&{{{\rm{b}}^2}}\\{\rm{c}}&1&{{{\rm{c}}^2}}\end{array}} \right|$ = $\left| {\begin{array}{*{20}{c}}1&{\rm{a}}&{{{\rm{a}}^2}}\\1&{\rm{b}}&{{{\rm{b}}^2}}\\1&{\rm{c}}&{{{\rm{c}}^2}}\end{array}} \right|$ = R.H.S. 

(v) $\left| {\begin{array}{*{20}{c}}{{\rm{bc}}}&{\rm{a}}&{{{\rm{a}}^2}}\\{{\rm{ca}}}&{\rm{b}}&{{{\rm{b}}^2}}\\{{\rm{ab}}}&{\rm{c}}&{{{\rm{c}}^2}}\end{array}} \right|$= $\left| {\begin{array}{*{20}{c}}1&{{{\rm{a}}^2}}&{{{\rm{a}}^3}}\\1&{{{\rm{b}}^2}}&{{{\rm{b}}^3}}\\1&{{{\rm{c}}^2}}&{{{\rm{c}}^3}}\end{array}} \right|$

Solution:

L.H.S. = $\left| {\begin{array}{*{20}{c}}{{\rm{bc}}}&{\rm{a}}&{{{\rm{a}}^2}}\\{{\rm{ca}}}&{\rm{b}}&{{{\rm{b}}^2}}\\{{\rm{ab}}}&{\rm{c}}&{{{\rm{c}}^2}}\end{array}} \right| = \frac{1}{{{\rm{abc}}}}\left| {\begin{array}{*{20}{c}}{{\rm{abc}}}&{{{\rm{a}}^2}}&{{{\rm{a}}^3}}\\{{\rm{abc}}}&{{{\rm{b}}^2}}&{{{\rm{b}}^3}}\\{{\rm{abc}}}&{{{\rm{c}}^2}}&{{{\rm{c}}^3}}\end{array}} \right|$

= $\frac{1}{{{\rm{abc}}}}.{\rm{abc}}$$\left| {\begin{array}{*{20}{c}}1&{{{\rm{a}}^2}}&{{{\rm{a}}^3}}\\1&{{{\rm{b}}^2}}&{{{\rm{b}}^3}}\\1&{{{\rm{c}}^2}}&{{{\rm{c}}^3}}\end{array}} \right|$

= $\left| {\begin{array}{*{20}{c}}1&{{{\rm{a}}^2}}&{{{\rm{a}}^3}}\\1&{{{\rm{b}}^2}}&{{{\rm{b}}^3}}\\1&{{{\rm{c}}^2}}&{{{\rm{c}}^3}}\end{array}} \right|$  = R.H.S.

 

6) Show that:

(i)$\left| {\begin{array}{*{20}{c}}1&1&1\\{\rm{a}}&{\rm{b}}&{\rm{c}}\\{{\rm{bc}}}&{{\rm{ca}}}&{{\rm{ab}}}\end{array}} \right|$=(a – b)(c – a)(b – c) 
Solution:

L.H.S. = $\left| {\begin{array}{*{20}{c}}1&1&1\\{\rm{a}}&{\rm{b}}&{\rm{c}}\\{{\rm{bc}}}&{{\rm{ca}}}&{{\rm{ab}}}\end{array}} \right|$

= $\left| {\begin{array}{*{20}{c}}1&0&0\\{\rm{a}}&{{\rm{b}} - {\rm{a}}}&{{\rm{c}} - {\rm{a}}}\\{{\rm{bc}}}&{{\rm{ca}} - {\rm{bc}}}&{{\rm{ab}} - {\rm{bc}}}\end{array}} \right|\left( {\begin{array}{*{20}{c}}{{{\rm{C}}_2} \to {{\rm{C}}_2} - {{\rm{C}}_1}}\\{{{\rm{C}}_3} \to {{\rm{C}}_3} - {{\rm{C}}_1}}\end{array}} \right)$

= $\left| {\begin{array}{*{20}{c}}{{\rm{b}} - {\rm{a}}}&{{\rm{c}} - {\rm{a}}}\\{{\rm{c}}\left( {{\rm{a}} - {\rm{b}}} \right)}&{{\rm{b}}\left( {{\rm{a}} - {\rm{c}}} \right)}\end{array}} \right|$ 

= $\left| {\begin{array}{*{20}{c}}{ - \left( {{\rm{a}} - {\rm{b}}} \right)}&{\left( {{\rm{c}} - {\rm{a}}} \right)}\\{{\rm{c}}\left( {{\rm{a}} - {\rm{b}}} \right)}&{ - {\rm{b}}\left( {{\rm{c}} - {\rm{a}}} \right)}\end{array}} \right|$

= (a – b)(c – a) $\left| {\begin{array}{*{20}{c}}{ - 1}&1\\{\rm{c}}&{ - {\rm{b}}}\end{array}} \right|$ = (a – b)(c – a)(b – c) 

= R.H.S. 

(ii)$\left| {\begin{array}{*{20}{c}}{\rm{a}}&{\rm{b}}&{\rm{c}}\\{{{\rm{a}}^2}}&{{{\rm{b}}^2}}&{{{\rm{c}}^2}}\\{{\rm{b}} + {\rm{c}}}&{{\rm{c}} + {\rm{a}}}&{{\rm{a}} + {\rm{b}}}\end{array}} \right|\left( {\begin{array}{*{20}{c}}{{{\rm{C}}_2} \to {{\rm{C}}_2} - {{\rm{C}}_1}}\\{{{\rm{C}}_3} \to {{\rm{C}}_3} - {{\rm{C}}_1}}\end{array}} \right)$=(a – b)(b – c)(c – a)(a + b + c)

Solution:

L.H.S. = $\left| {\begin{array}{*{20}{c}}{\rm{a}}&{\rm{b}}&{\rm{c}}\\{{{\rm{a}}^2}}&{{{\rm{b}}^2}}&{{{\rm{c}}^2}}\\{{\rm{b}} + {\rm{c}}}&{{\rm{c}} + {\rm{a}}}&{{\rm{a}} + {\rm{b}}}\end{array}} \right|\left( {\begin{array}{*{20}{c}}{{{\rm{C}}_2} \to {{\rm{C}}_2} - {{\rm{C}}_1}}\\{{{\rm{C}}_3} \to {{\rm{C}}_3} - {{\rm{C}}_1}}\end{array}} \right)$

= $\left| {\begin{array}{*{20}{c}}{\rm{a}}&{ - \left( {{\rm{a}} - {\rm{b}}} \right)}&{\left( {{\rm{c}} - {\rm{a}}} \right)}\\{{{\rm{a}}^2}}&{ - \left( {{\rm{a}} - {\rm{b}}} \right)\left( {{\rm{a}} + {\rm{b}}} \right)}&{\left( {{\rm{c}} - {\rm{a}}} \right)\left( {{\rm{c}} + {\rm{a}}} \right)}\\{{\rm{b}} + {\rm{c}}}&{\left( {{\rm{a}} - {\rm{b}}} \right)}&{ - \left( {{\rm{c}} - {\rm{a}}} \right)}\end{array}} \right|$

= (a – b)(c – a) $\left| {\begin{array}{*{20}{c}}{\rm{a}}&{ - 1}&1\\{{{\rm{a}}^2}}&{ - \left( {{\rm{a}} + {\rm{b}}} \right)}&{\left( {{\rm{c}} + {\rm{a}}} \right)}\\{{\rm{b}} + {\rm{c}}}&1&{ - 1}\end{array}} \right|$

= (a – b)(c – a) $\left| {\begin{array}{*{20}{c}}{\rm{a}}&{ - 1}&0\\{{{\rm{a}}^2}}&{ - \left( {{\rm{a}} + {\rm{b}}} \right)}&{{\rm{c}} - {\rm{b}}}\\{{\rm{b}} + {\rm{c}}}&1&0\end{array}} \right|$ (C­₃àC₃ + C₂).

= –(a – b)(c – a)(c – b) $\left| {\begin{array}{*{20}{c}}{\rm{a}}&{ - 1}\\{{\rm{b}} + {\rm{c}}}&1\end{array}} \right|$

= (a – b)(b – c)(c – a)(a + b + c) = R.H.S. 

(iii)$\left| {\begin{array}{*{20}{c}}1&1&1\\{\rm{a}}&{\rm{b}}&{\rm{c}}\\{{{\rm{a}}^3}}&{{{\rm{b}}^3}}&{{{\rm{c}}^3}}\end{array}} \right|$=(a – b)(b – c)(c – a)(a + b + c)

Solution:

L.H.S. = $\left| {\begin{array}{*{20}{c}}1&1&1\\{\rm{a}}&{\rm{b}}&{\rm{c}}\\{{{\rm{a}}^3}}&{{{\rm{b}}^3}}&{{{\rm{c}}^3}}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{1 - 1}&1&{1 - 1}\\{{\rm{a}} - {\rm{b}}}&{\rm{b}}&{{\rm{c}} - {\rm{b}}}\\{{{\rm{a}}^3} - {{\rm{b}}^3}}&{{{\rm{b}}^3}}&{{{\rm{c}}^3} - {{\rm{b}}^3}}\end{array}} \right|\left( {\begin{array}{*{20}{c}}{{{\rm{C}}_1} \to {{\rm{C}}_1} - {{\rm{C}}_2}}\\{{{\rm{C}}_3} \to {{\rm{C}}_3} - {{\rm{C}}_1}}\end{array}} \right)$

= $\left| {\begin{array}{*{20}{c}}0&1&0\\{{\rm{a}} - {\rm{b}}}&{\rm{b}}&{{\rm{c}} - {\rm{b}}}\\{{{\rm{a}}^3} - {{\rm{b}}^3}}&{{{\rm{b}}^3}}&{{{\rm{c}}^3} - {{\rm{b}}^3}}\end{array}} \right|$

= $ - 1\left| {\begin{array}{*{20}{c}}{{\rm{a}} - {\rm{b}}}&{{\rm{c}} - {\rm{b}}}\\{{{\rm{a}}^3} - {{\rm{b}}^3}}&{{{\rm{c}}^3} - {{\rm{b}}^3}}\end{array}} \right|$

= –(a – b)(c – b) $\left| {\begin{array}{*{20}{c}}1&1\\{{{\rm{a}}^2} + {\rm{ab}} + {{\rm{b}}^2}}&{{{\rm{c}}^2} + {\rm{cb}} + {{\rm{b}}^2}}\end{array}} \right|$

= (a – b)(b – c)(c² + cb + b² – a² – ab – b²)

= (a – b)(b – c){(c + a)(c – a) + b(c – a)}

= (a – b)(b – c)(c – a)(a + b + c) = R.H.S. 

(iv)$\left| {\begin{array}{*{20}{c}}{\rm{x}}&{\rm{y}}&{\rm{z}}\\{{{\rm{x}}^2}}&{{{\rm{y}}^2}}&{{{\rm{z}}^2}}\\{{\rm{yz}}}&{{\rm{zx}}}&{{\rm{xy}}}\end{array}} \right|$=(b-c)(c-a)(a-b)(a+b+c)

Solution:

L.H.S. = $\left| {\begin{array}{*{20}{c}}{\rm{x}}&{\rm{y}}&{\rm{z}}\\{{{\rm{x}}^2}}&{{{\rm{y}}^2}}&{{{\rm{z}}^2}}\\{{\rm{yz}}}&{{\rm{zx}}}&{{\rm{xy}}}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{{\rm{x}} - {\rm{y}}}&{\rm{y}}&{{\rm{z}} - {\rm{y}}}\\{{{\rm{x}}^2} - {{\rm{y}}^2}}&{{{\rm{y}}^2}}&{{{\rm{z}}^2} - {{\rm{y}}^2}}\\{{\rm{yz}} - {\rm{zx}}}&{{\rm{zx}}}&{{\rm{xy}} - {\rm{zx}}}\end{array}} \right|\left( {\begin{array}{*{20}{c}}{{{\rm{C}}_1} \to {{\rm{C}}_1} - {{\rm{C}}_2}}\\{{{\rm{C}}_3} \to {{\rm{C}}_3} - {{\rm{C}}_2}}\end{array}} \right)$

= (x – y)(z – y) $\left| {\begin{array}{*{20}{c}}1&{\rm{y}}&1\\{{\rm{x}} + {\rm{y}}}&{{{\rm{y}}^2}}&{{\rm{z}} + {\rm{y}}}\\{ - {\rm{z}}}&{{\rm{zx}}}&{ - {\rm{x}}}\end{array}} \right|$

= (x – y)(z – y)${\rm{\: }}\left| {\begin{array}{*{20}{c}}0&{\rm{y}}&1\\{{\rm{x}} - {\rm{z}}}&{{{\rm{y}}^2}}&{{\rm{z}} + {\rm{y}}}\\1&{{\rm{zx}}}&{ - {\rm{x}}}\end{array}} \right|$ (C₁àC₁ – C₃)

= (x – y)(z – y)(z – x) $\left| {\begin{array}{*{20}{c}}0&{\rm{y}}&1\\1&{{{\rm{y}}^2}}&{{\rm{z}} + {\rm{y}}}\\1&{{\rm{zx}}}&{ - {\rm{x}}}\end{array}} \right|$  (Taking x – z common from C₁).

= (x – y)(z – y)(z – x) $\left| {\begin{array}{*{20}{c}}0&{\rm{y}}&1\\0&{{{\rm{y}}^2} - {\rm{zx}}}&{{\rm{z}} + {\rm{y}}}\\1&{{\rm{zx}}}&{ - {\rm{x}}}\end{array} + {\rm{x}}} \right|$  (R₂àR₂ – R₃)

= (x – y)(z – y)(z – x).1. $\left| {\begin{array}{*{20}{c}}{\rm{y}}&1\\{{{\rm{y}}^2} - {\rm{zx}}}&{{\rm{z}} + {\rm{y}} + {\rm{x}}}\end{array}} \right|$

= (x – y)(z – y)(z – x){yz + y² + xy – y² + zx}

= (x – y)(z – y)(z – x)(yz + zx + xy) = R.H.S. 

(v) $\left| {\begin{array}{*{20}{c}}{{\rm{b}} + {\rm{c}}}&{\rm{a}}&{\rm{b}}\\{{\rm{c}} + {\rm{a}}}&{\rm{c}}&{\rm{a}}\\{{\rm{a}} + {\rm{b}}}&{\rm{b}}&{\rm{c}}\end{array}} \right|$=(a + b + c)(a – c)²

Solution:

L.H.S. = $\left| {\begin{array}{*{20}{c}}{{\rm{b}} + {\rm{c}}}&{\rm{a}}&{\rm{b}}\\{{\rm{c}} + {\rm{a}}}&{\rm{c}}&{\rm{a}}\\{{\rm{a}} + {\rm{b}}}&{\rm{b}}&{\rm{c}}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{2\left( {{\rm{a}} + {\rm{b}} + {\rm{c}}} \right)}&{{\rm{a}} + {\rm{b}} + {\rm{c}}}&{{\rm{a}} + {\rm{b}} + {\rm{c}}}\\{{\rm{c}} + {\rm{a}}}&{\rm{c}}&{\rm{a}}\\{{\rm{a}} + {\rm{b}}}&{\rm{b}}&{\rm{c}}\end{array}} \right|$ (C₁àC₁ + C₂ + C₃).

= (a + b + c)$\left| {\begin{array}{*{20}{c}}2&1&1\\{{\rm{c}} + {\rm{a}}}&{\rm{c}}&{\rm{a}}\\{{\rm{a}} + {\rm{b}}}&{\rm{b}}&{\rm{c}}\end{array}} \right|$(Taking a + b + c common from R₁).

= (a + b + c)$\left| {\begin{array}{*{20}{c}}0&1&1\\0&{\rm{c}}&{\rm{a}}\\{{\rm{a}} - {\rm{c}}}&{\rm{b}}&{\rm{c}}\end{array}} \right|$ (C₁àC₁ – C₂ – C₃)

= (a + b + c)(a – c) $\left| {\begin{array}{*{20}{c}}1&1\\{\rm{c}}&{\rm{a}}\end{array}} \right|$ = (a + b + c)(a – c)(a – c)

= (a + b + c)(a – c)²= R.H.S. 

(vi)$\left| {\begin{array}{*{20}{c}}{{\rm{a}} - {\rm{b}} - {\rm{c}}}&{2{\rm{a}}}&{2{\rm{a}}}\\{2{\rm{b}}}&{{\rm{b}} - {\rm{c}} - {\rm{a}}}&{2{\rm{b}}}\\{2{\rm{c}}}&{2{\rm{c}}}&{{\rm{c}} - {\rm{a}} - {\rm{b}}}\end{array}} \right|$= (a + b + c)³ 

Solution:

L.H.S. = $\left| {\begin{array}{*{20}{c}}{{\rm{a}} - {\rm{b}} - {\rm{c}}}&{2{\rm{a}}}&{2{\rm{a}}}\\{2{\rm{b}}}&{{\rm{b}} - {\rm{c}} - {\rm{a}}}&{2{\rm{b}}}\\{2{\rm{c}}}&{2{\rm{c}}}&{{\rm{c}} - {\rm{a}} - {\rm{b}}}\end{array}} \right|$

$ = \left| {\begin{array}{*{20}{c}}{{\rm{a}} + {\rm{b}} + {\rm{c}}}&{{\rm{a}} + {\rm{b}} + {\rm{c}}}&{{\rm{a}} + {\rm{b}} + {\rm{c}}}\\{2{\rm{b}}}&{{\rm{b}} - {\rm{c}} - {\rm{a}}}&{2{\rm{b}}}\\{2{\rm{c}}}&{2{\rm{c}}}&{{\rm{c}} - {\rm{a}} - {\rm{b}}}\end{array}} \right|$ (R₁àR₁ + R₂ + R₃).

= (a + b + c) $\left| {\begin{array}{*{20}{c}}1&1&1\\{2{\rm{b}}}&{{\rm{b}} - {\rm{c}} - {\rm{a}}}&{2{\rm{b}}}\\{2{\rm{c}}}&{2{\rm{c}}}&{{\rm{c}} - {\rm{a}} - {\rm{b}}}\end{array}} \right|$

= (a + b + c)$\left| {\begin{array}{*{20}{c}}0&1&1\\0&{{\rm{b}} - {\rm{c}} - {\rm{a}}}&{2{\rm{b}}}\\{{\rm{c}} + {\rm{a}} + {\rm{b}}}&{2{\rm{c}}}&{{\rm{c}} - {\rm{a}} - {\rm{b}}}\end{array}} \right|$ (C₁àC₁ – C₃)

= (a + b + c)²$\left| {\begin{array}{*{20}{c}}1&1\\{{\rm{b}} - {\rm{c}} - {\rm{a}}}&{2{\rm{b}}}\end{array}} \right|$

= (a + b + c)²(2b – b + c + a) = (a + b + c)³ = R.H.S.

 

(vii) $\left| {\begin{array}{*{20}{c}}{1 + {{\rm{a}}_1}}&{{{\rm{a}}_2}}&{{{\rm{a}}_3}}\\{{{\rm{a}}_1}}&{1 + {{\rm{a}}_2}}&{{{\rm{a}}_3}}\\{{{\rm{a}}_1}}&{{{\rm{a}}_2}}&{1 + {{\rm{a}}_3}}\end{array}} \right|$= 1 + a₁ + a₂ + a₃

Solution:

L.H.S. = $\left| {\begin{array}{*{20}{c}}{1 + {{\rm{a}}_1}}&{{{\rm{a}}_2}}&{{{\rm{a}}_3}}\\{{{\rm{a}}_1}}&{1 + {{\rm{a}}_2}}&{{{\rm{a}}_3}}\\{{{\rm{a}}_1}}&{{{\rm{a}}_2}}&{1 + {{\rm{a}}_3}}\end{array}} \right|$$ = \left| {\begin{array}{*{20}{c}}{1 + {{\rm{a}}_1} + {{\rm{a}}_2} + {{\rm{a}}_3}}&{{{\rm{a}}_2}}&{{{\rm{a}}_3}}\\{1 + {{\rm{a}}_1} + {{\rm{a}}_2} + {{\rm{a}}_3}}&{1 + {{\rm{a}}_2}}&{{{\rm{a}}_3}}\\{1 + {{\rm{a}}_1} + {{\rm{a}}_2} + {{\rm{a}}_3}}&{{{\rm{a}}_2}}&{{\rm{a}} + {{\rm{a}}_3}}\end{array}} \right|$ (C₁àC₁ + C₂ + C₃).

= (1 + a₁ + a₂ + a₃) $\left| {\begin{array}{*{20}{c}}1&{{{\rm{a}}_2}}&{{{\rm{a}}_3}}\\1&{1 + {{\rm{a}}_2}}&{{{\rm{a}}_3}}\\1&{{{\rm{a}}_2}}&{1 + {{\rm{a}}_3}}\end{array}} \right|$

= (1 + a₁ + a₂ + a₃)$\left| {\begin{array}{*{20}{c}}0&0&{ - 1}\\1&{1 + {{\rm{a}}_2}}&{{{\rm{a}}_3}}\\1&{{{\rm{a}}_2}}&{1 + {{\rm{a}}_3}}\end{array}} \right|$ (R₁àR₁– R₃)

= –(1 + a₁ + a₂ + a₃)$\left| {\begin{array}{*{20}{c}}1&{1 + {{\rm{a}}_2}}\\1&{{{\rm{a}}_2}}\end{array}} \right|$

= –(1 + a₁ + a₂ + a₃)(a₂ – 1 – a₂) = 1 + a₁ + a₂ + a₃ = R.H.S.

 

(viii) $\left| {\begin{array}{*{20}{c}}{1 + {\rm{x}}}&1&1\\1&{1 + {\rm{y}}}&1\\1&1&{1 + {\rm{z}}}\end{array}} \right|$= xyz$\left( {\frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} + \frac{1}{{\rm{z}}} + 1} \right)$

Solution:

L.H.S. = $\left| {\begin{array}{*{20}{c}}{1 + {\rm{x}}}&1&1\\1&{1 + {\rm{y}}}&1\\1&1&{1 + {\rm{z}}}\end{array}} \right|$

$ = {\rm{xyz}}.\left| {\begin{array}{*{20}{c}}{\frac{{1 + {\rm{x}}}}{{\rm{x}}}}&{\frac{1}{{\rm{y}}}}&{\frac{1}{{\rm{z}}}}\\{\frac{1}{{\rm{x}}}}&{\frac{{1 + {\rm{y}}}}{{\rm{y}}}}&{\frac{1}{{\rm{z}}}}\\{\frac{1}{{\rm{x}}}}&{\frac{1}{{\rm{y}}}}&{\frac{{1 + {\rm{z}}}}{{\rm{z}}}}\end{array}} \right|$

= ${\rm{xyz}}.\left| {\begin{array}{*{20}{c}}{\frac{1}{{\rm{x}}} + 1}&{\frac{1}{{\rm{y}}}}&{\frac{1}{{\rm{z}}}}\\{\frac{1}{{\rm{x}}}}&{\frac{1}{{\rm{y}}} + 1}&{\frac{1}{{\rm{z}}}}\\{\frac{1}{{\rm{x}}}}&{\frac{1}{{\rm{y}}}}&{\frac{1}{{\rm{z}}} + 1}\end{array}} \right|$

= ${\rm{xyz}}.\left| {\begin{array}{*{20}{c}}{\frac{1}{{\rm{x}}} + 1 + \frac{1}{{\rm{y}}} + \frac{1}{{\rm{z}}}}&{\frac{1}{{\rm{y}}}}&{\frac{1}{{\rm{z}}}}\\{\frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} + 1 + \frac{1}{{\rm{z}}}}&{\frac{1}{{\rm{y}}} + 1}&{\frac{1}{{\rm{z}}}}\\{\frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} + \frac{1}{{\rm{z}}} + 1}&{\frac{1}{{\rm{y}}}}&{\frac{1}{{\rm{z}}} + 1}\end{array}} \right|$ (C₁àC₁ + C₂ + C₃)

= xyz.$\left( {\frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} + \frac{1}{{\rm{z}}} + 1} \right)\left| {\begin{array}{*{20}{c}}1&{\frac{1}{{\rm{y}}}}&{\frac{1}{{\rm{z}}}}\\1&{\frac{1}{{\rm{y}}} + 1}&{\frac{1}{{\rm{z}}}}\\1&{\frac{1}{{\rm{y}}}}&{\frac{1}{{\rm{z}}} + 1}\end{array}} \right|$

= xyz.${\rm{\: }}\left( {\frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} + \frac{1}{{\rm{z}}} + 1} \right)\left| {\begin{array}{*{20}{c}}0&0&{ - 1}\\1&{\frac{1}{{\rm{y}}} + 1}&{\frac{1}{{\rm{z}}}}\\1&{\frac{1}{{\rm{y}}}}&{\frac{1}{{\rm{z}}} + 1}\end{array}} \right|$  (R₁àR₁ – R₃)

= –xyz $\left( {\frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} + \frac{1}{{\rm{z}}} + 1} \right)$$\left| {\begin{array}{*{20}{c}}1&{\frac{1}{{\rm{y}}} + 1}\\1&{\frac{1}{{\rm{y}}}}\end{array}} \right|$

= –xyz $\left( {\frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} + \frac{1}{{\rm{z}}} + 1} \right)$$\left\{ {\frac{1}{{\rm{y}}} - \frac{1}{{\rm{y}}} - 1} \right\}$

= xyz$\left( {\frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} + \frac{1}{{\rm{z}}} + 1} \right)$ = R.H.S.

 

(ix)$\left| {\begin{array}{*{20}{c}}{{{\rm{a}}^2}}&{{\rm{bc}}}&{{{\rm{c}}^2} + {\rm{ac}}}\\{{{\rm{a}}^2} + {\rm{ab}}}&{{{\rm{b}}^2}}&{{\rm{ac}}}\\{{\rm{ab}}}&{{{\rm{b}}^2} + {\rm{bc}}}&{{{\rm{c}}^2}}\end{array}} \right|$= 4a²b²c²

Solution:

L.H.S. = $\left| {\begin{array}{*{20}{c}}{{{\rm{a}}^2}}&{{\rm{bc}}}&{{{\rm{c}}^2} + {\rm{ac}}}\\{{{\rm{a}}^2} + {\rm{ab}}}&{{{\rm{b}}^2}}&{{\rm{ac}}}\\{{\rm{ab}}}&{{{\rm{b}}^2} + {\rm{bc}}}&{{{\rm{c}}^2}}\end{array}} \right|$

$ = {\rm{abc}}.\left| {\begin{array}{*{20}{c}}{\rm{a}}&{\rm{c}}&{{\rm{c}} + {\rm{a}}}\\{{\rm{a}} + {\rm{b}}}&{\rm{b}}&{\rm{a}}\\{\rm{b}}&{{\rm{b}} + {\rm{c}}}&{\rm{c}}\end{array}} \right|$

= ${\rm{abc}}.\left| {\begin{array}{*{20}{c}}{{\rm{a}} - {\rm{c}} - {\rm{c}} - {\rm{a}}}&{\rm{c}}&{{\rm{c}} + {\rm{a}}}\\{{\rm{a}} + {\rm{b}} - {\rm{b}} - {\rm{a}}}&{\rm{b}}&{\rm{a}}\\{{\rm{b}} - {\rm{b}} - {\rm{c}} - {\rm{c}}}&{{\rm{b}} + {\rm{c}}}&{\rm{c}}\end{array}} \right|$

=${\rm{abc}}.\left| {\begin{array}{*{20}{c}}{ - 2{\rm{c}}}&{\rm{c}}&{{\rm{c}} + {\rm{a}}}\\0&{\rm{b}}&{\rm{a}}\\{ - 2{\rm{c}}}&{{\rm{b}} + {\rm{c}}}&{\rm{c}}\end{array}} \right|$

=  ${\rm{abc}}.\left| {\begin{array}{*{20}{c}}{ - 2{\rm{c}}}&{\rm{c}}&{{\rm{c}} + {\rm{a}}}\\0&{\rm{b}}&{\rm{a}}\\0&{\rm{b}}&{ - {\rm{a}}}\end{array}} \right|$ (C₃ àC₃ – C₁)

= abc(–2c) $\left| {\begin{array}{*{20}{c}}{\rm{b}}&{\rm{a}}\\{\rm{b}}&{ - {\rm{a}}}\end{array}} \right|$ = –2abc²(–ab – ab)

= – 2abc²(–2ab) = 4a²b²c² = R.H.S. 

(x)$\frac{{{\rm{xyz}}}}{{{\rm{xyz}}}}\left| {\begin{array}{*{20}{c}}{{{\rm{x}}^2} + 1}&{{\rm{xy}}}&{{\rm{xz}}}\\{{\rm{xy}}}&{{{\rm{y}}^2} + 1}&{{\rm{yz}}}\\{{\rm{xz}}}&{{\rm{yz}}}&{{{\rm{z}}^2} + 1}\end{array}} \right|$= (1 + x² + y² + z²)

Solution:

L.H.S. = $\frac{{{\rm{xyz}}}}{{{\rm{xyz}}}}\left| {\begin{array}{*{20}{c}}{{{\rm{x}}^2} + 1}&{{\rm{xy}}}&{{\rm{xz}}}\\{{\rm{xy}}}&{{{\rm{y}}^2} + 1}&{{\rm{yz}}}\\{{\rm{xz}}}&{{\rm{yz}}}&{{{\rm{z}}^2} + 1}\end{array}} \right|$

$ = \frac{1}{{{\rm{xyz}}}}.\left| {\begin{array}{*{20}{c}}{{\rm{x}}\left( {{{\rm{x}}^2} + 1} \right)}&{{{\rm{x}}^2}{\rm{y}}}&{{{\rm{x}}^2}{\rm{z}}}\\{{\rm{x}}{{\rm{y}}^2}}&{{\rm{y}}\left( {{{\rm{y}}^2} + 1} \right)}&{{{\rm{y}}^2}{\rm{z}}}\\{{\rm{x}}{{\rm{z}}^2}}&{{\rm{y}}{{\rm{z}}^2}}&{{\rm{z}}\left( {{{\rm{z}}^2} + 1} \right)}\end{array}} \right|$

= $\frac{1}{{{\rm{xyz}}}}.{\rm{xyz}}\left| {\begin{array}{*{20}{c}}{{{\rm{x}}^2} + 1}&{{{\rm{x}}^2}}&{{{\rm{x}}^2}}\\{{{\rm{y}}^2}}&{{{\rm{y}}^2} + 1}&{{{\rm{y}}^2}}\\{{{\rm{z}}^2}}&{{{\rm{z}}^2}}&{{{\rm{z}}^2} + 1}\end{array}} \right|$

=$\left| {\begin{array}{*{20}{c}}{1 + {{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2}}&{1 + {{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2}}&{1 + {{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2}}\\{{{\rm{y}}^2}}&{{{\rm{y}}^2} + 1}&{{{\rm{y}}^2}}\\{{{\rm{z}}^2}}&{{{\rm{z}}^2}}&{{{\rm{z}}^2} + 1}\end{array}} \right|$

=  $\left( {1 + {{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2}} \right)\left| {\begin{array}{*{20}{c}}1&1&1\\{{{\rm{y}}^2}}&{{{\rm{y}}^2} + 1}&{{{\rm{y}}^2}}\\{{{\rm{z}}^2}}&{{{\rm{z}}^2}}&{{{\rm{z}}^2} + 1}\end{array}} \right|$=  $\left( {1 + {{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2}} \right)\left| {\begin{array}{*{20}{c}}1&0&0\\{{{\rm{y}}^2}}&1&0\\{{{\rm{z}}^2}}&0&1\end{array}} \right|\left( {\begin{array}{*{20}{c}}{{{\rm{C}}_2} \to {{\rm{C}}_2} - {{\rm{C}}_1}}\\{{{\rm{C}}_3} \to {{\rm{C}}_3} - {{\rm{C}}_1}}\end{array}} \right)$

= $\left( {1 + {{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2}} \right)\left| {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right|$

=$\left( {1 + {{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2}} \right)$(1 – 0) = (1 + x² + y² + z²) = R.H.S.