Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\Delta ' = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {p{b_1}}&{q{c_1}}&{r{a_1}}\\ {p{b_2}}&{q{c_2}}&{r{a_2}}\\ {p{b_3}}&{q{c_3}}&{r{a_3}} \end{array}} \right|\]       [All other determinants vanish]
$$ = \Delta + pqr\Delta = \Delta \left( {1 + pqr} \right)$$

Using $${C_1} \to {C_1} - p \times {C_2} - {C_3},$$     we get $$\left( {x{p^2} - z} \right)\left( {xz - {y^2}} \right) = 0.$$
For all $$p,x{p^2} - z \ne 0.\,{\text{So, }}{y^2} = xz.$$

\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} {^{10}{C_4}}&{^{10}{C_5}}&{^{11}{C_m}}\\ {^{11}{C_6}}&{^{11}{C_7}}&{^{12}{C_{m + 2}}}\\ {^{12}{C_8}}&{^{12}{C_9}}&{^{13}{C_{m + 4}}} \end{array}} \right| = 0\\ \Rightarrow \left| {\begin{array}{*{20}{c}} {^{10}{C_4} + {\,^{10}}{C_5}}&{^{10}{C_5}}&{^{11}{C_m}}\\ {^{11}{C_6} + {\,^{11}}{C_7}}&{^{11}{C_7}}&{^{12}{C_{m + 2}}}\\ {^{12}{C_8} + {\,^{12}}{C_9}}&{^{12}{C_9}}&{^{13}{C_{m + 4}}} \end{array}} \right| = 0\\ \left( {{R_1} \to {R_1} + {R_2}} \right)\\ \Rightarrow \left| {\begin{array}{*{20}{c}} {^{11}{C_5}}&{^{10}{C_5}}&{^{11}{C_m}}\\ {^{12}{C_7}}&{^{11}{C_7}}&{^{12}{C_{m + 2}}}\\ {^{13}{C_9}}&{^{12}{C_9}}&{^{13}{C_{m + 4}}} \end{array}} \right| = 0 \end{array}\]
It means either two rows or two columns are identical
$$\eqalign{ & \therefore {\,^{11}}{C_5} = {\,^{11}}{C_m},{\,^{12}}{C_7} \cr & = {\,^{12}}{C_{m + 2}},\,{\,^{13}}{C_4} = {\,^{13}}{C_{m + 4}} \cr & \Rightarrow m = 5 \cr} $$

\[\det \left( {{M_r}} \right) = \left| {\begin{array}{*{20}{c}} r&{r - 1}\\ {r - 1}&r \end{array}} \right| = 2r - 1\]
$$\eqalign{ & \sum\limits_{r = 1}^{2014} {\det \left( {{M_r}} \right) = 2\sum\limits_{r = 1}^{2014} {r - 2014} } \cr & = 2 \times \frac{{2014 \times 2015}}{2} - 2014 = {\left( {2014} \right)^2} \cr} $$

$$\eqalign{ & D' = {\text{cofactor }}D \cr & \Rightarrow \left| {D'} \right| = \left| {{\text{cofactor }}D} \right| \cr & \Rightarrow \left| {D'} \right| = {\left| D \right|^{3 - 1}}. \cr & \Rightarrow \left| {D'} \right| = {\left| D \right|^2} \cr & {\text{So, }}D' = {D^2}. \cr} $$

Since the system has no solution, $$\Delta = 0$$  and any one amongst $${\Delta _x},{\Delta _y},{\Delta _z}$$   is non - zero.
\[\left| \begin{array}{l} 2\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,2\\ 1\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\lambda \end{array} \right| = 0\]
$$ \Rightarrow \,\,\lambda = 1$$
\[{\rm{Also, }}\,\,{\Delta _z} = \left| \begin{array}{l} 2\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ 1\,\,\,\,\,\, - 2\,\,\,\,\, - 4\\ 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,4 \end{array} \right| = 6 \ne 0\]

Given, \[\left| {\begin{array}{*{20}{c}} a&{\cot \frac{A}{2}}&\lambda \\ b&{\cot \frac{B}{2}}&\mu \\ c&{\cot \frac{C}{2}}&\gamma \end{array}} \right| = 0\]
\[\begin{array}{l} \Rightarrow \left| {\begin{array}{*{20}{c}} a&{\frac{{s\left( {s - a} \right)}}{\Delta }}&\lambda \\ b&{\frac{{s\left( {s - b} \right)}}{\Delta }}&\mu \\ c&{\frac{{s\left( {s - c} \right)}}{\Delta }}&\gamma \end{array}} \right| = 0\\ \Rightarrow \frac{1}{r}\left| {\begin{array}{*{20}{c}} a&{s - a}&\lambda \\ b&{s - b}&\mu \\ c&{s - c}&\gamma \end{array}} \right| = 0 \end{array}\]
Apply, $${C_2} \to {C_2} + {C_1}$$
\[\begin{array}{l} \Rightarrow \frac{1}{r}\left| {\begin{array}{*{20}{c}} a&s&\lambda \\ b&s&\mu \\ c&s&\gamma \end{array}} \right| = 0,\,\,{\rm{where}}\,\,r = \frac{\Delta }{s}\\ \Rightarrow \frac{\Delta }{{{r^2}}}\left| {\begin{array}{*{20}{c}} a&1&\lambda \\ b&1&\mu \\ c&1&\gamma \end{array}} \right| = 0 \end{array}\]
Apply, $${R_1} \to {R_1} - {R_2},{R_2} \to {R_2} - {R_3}$$
\[ \Rightarrow \frac{\Delta }{{{r^2}}}\left| {\begin{array}{*{20}{c}} {a - b}&0&{\lambda - \mu }\\ {b - c}&0&{\mu - \gamma }\\ c&1&\gamma \end{array}} \right| = 0\]
$$\eqalign{ & \Rightarrow \frac{\Delta }{{{r^2}}}\left[ {\left( {b - c} \right)\left( {\lambda - \mu } \right) - \left( {\mu - \gamma } \right)\left( {a - b} \right)} \right] = 0 \cr & \Rightarrow b\left( {\lambda - \mu } \right) - c\left( {\lambda - \mu } \right) - a\left( {\mu - \gamma } \right) + b\left( {\mu - \gamma } \right) = 0 \cr & \Rightarrow - a\left( {\mu - \gamma } \right) + b\left( {\lambda - \mu + \mu - \gamma } \right) - c\left( {\lambda - \mu } \right) = 0 \cr & \Rightarrow - a\left( {\mu - \gamma } \right) + b\left( {\lambda - \gamma } \right) - c\left( {\lambda - \mu } \right) = 0 \cr & \Rightarrow a\left( {\mu - \gamma } \right) + b\left( {\gamma - \lambda } \right) + c\left( {\lambda - \mu } \right) = 0 \cr} $$

Here,
$$ \Rightarrow $$ \[B = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} 4&2&2\\ { - 5}&0&\alpha \\ 1&{ - 2}&3 \end{array}} \right]\]
Also since, $$B = A^{- 1}$$
⇒ $$AB = I$$
\[\begin{array}{l} \Rightarrow \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1\\ 2&1&{ - 3}\\ 1&1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4&2&2\\ { - 5}&0&\alpha \\ 1&{ - 2}&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\\ \Rightarrow \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {10}&0&{5 - 2}\\ 0&{10}&{ - 5 + \alpha }\\ 0&0&{5 + \alpha } \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] \end{array}\]
$$\eqalign{ & \Rightarrow \frac{{5 - \alpha }}{{10}} = 0 \cr & \Rightarrow \alpha = 5 \cr} $$

$${a_r} = {\left( {\cos 2r\pi + i\sin 2r\pi } \right)^{\frac{1}{9}}} = {a_r} = {\left( {{e^{i2r\pi }}} \right)^{\frac{1}{9}}} = {e^{\frac{{i2r\pi }}{9}}}$$
\[\begin{array}{l} \Rightarrow \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{a_4}}&{{a_5}}&{{a_6}}\\ {{a_7}}&{{a_8}}&{{a_9}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{e^{\frac{{i2\pi }}{9}}}}&{{e^{\frac{{i4\pi }}{9}}}}&{{e^{\frac{{i6\pi }}{9}}}}\\ {{e^{\frac{{i2\pi }}{9}}}}&{{e^{\frac{{i10\pi }}{9}}}}&{{e^{\frac{{i12\pi }}{9}}}}\\ {{e^{\frac{{i14\pi }}{9}}}}&{{e^{\frac{{i16\pi }}{9}}}}&{{e^{\frac{{i18\pi }}{9}}}} \end{array}} \right|\\ = {e^{\frac{{i8\pi }}{9}}}\left| {\begin{array}{*{20}{c}} {{e^{\frac{{i2\pi }}{9}}}}&{{e^{\frac{{i4\pi }}{9}}}}&{{e^{\frac{{i6\pi }}{9}}}}\\ {{e^{\frac{{i2\pi }}{9}}}}&{{e^{\frac{{i4\pi }}{9}}}}&{{e^{\frac{{i6\pi }}{9}}}}\\ {{e^{\frac{{i14\pi }}{9}}}}&{{e^{\frac{{i16\pi }}{9}}}}&{{e^{\frac{{i18\pi }}{9}}}} \end{array}} \right| \end{array}\]
$$=$$ (taking $${{e^{\frac{{i8\pi }}{9}}}}$$ common from second row)
$$= 0 $$

The constant term = value of the determinant when $$x = 0.$$