Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|\]
$${\text{Operating }}{R_1} \to {R_1} - {R_2} + {R_3}$$
\[=\left| {\begin{array}{*{20}{c}} 0&0&0 \\ {1 - i}&{ - 1}&{{\omega ^2} - 1} \\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|\left( {\because \,\,1 + \omega + {\omega ^2} = 0} \right)\]
$$ = 0$$

\[{\rm{We \,\,have }}\,\,A = \left[ \begin{array}{l} \,\,2\,\,\,\,\,\,\, - 3\\ - 4\,\,\,\,\,\,\,\,\,1 \end{array} \right]\]
\[ \Rightarrow \,\,{A^2} = \left[ \begin{array}{l} \,\,16\,\,\,\,\,\,\, - 9\\ - 12\,\,\,\,\,\,\,13 \end{array} \right]\]
\[ \Rightarrow \,\,3{A^2} = \left[ \begin{array}{l} \,\,48\,\,\,\,\,\,\, - 27\\ - 36\,\,\,\,\,\,\,\,\,39 \end{array} \right]\]
\[{\rm{Also\,\, 12}}A = \left[ \begin{array}{l} \,\,24\,\,\,\,\,\, - 36\\ - 48\,\,\,\,\,\,\,\,\,12 \end{array} \right]\]
\[\therefore \,\,3{A^2} + 12A = \left[ \begin{array}{l} \,\,48\,\,\,\,\,\,\, - 27\\ - 36\,\,\,\,\,\,\,\,\,39 \end{array} \right] + \left[ \begin{array}{l} \,\,24\,\,\,\,\,\, - 36\\ - 48\,\,\,\,\,\,\,\,\,12 \end{array} \right]\]
\[\,\,\,\,\,\, = \left[ \begin{array}{l} \,\,72\,\,\,\,\,\, - 63\\ - 84\,\,\,\,\,\,\,\,51 \end{array} \right]\]
\[adj\left( {3{A^2} + 12A} \right) = \left[ \begin{array}{l} 51\,\,\,\,\,\,\,63\\ 84\,\,\,\,\,\,\,72 \end{array} \right]\]

\[\begin{array}{l} {A^2} = \left[ {\begin{array}{*{20}{c}} 1&0\\ { - 1}&7 \end{array}} \right]\,\left[ {\begin{array}{*{20}{c}} 1&0\\ { - 1}&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ { - 8}&{49} \end{array}} \right]\\ {\rm{and}}\,\,\,8A + kI = 8\left[ {\begin{array}{*{20}{c}} 1&0\\ { - 1}&7 \end{array}} \right] + k\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\\ = \,\left[ {\begin{array}{*{20}{c}} 8&0\\ { - 8}&{56} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} k&0\\ 0&k \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {8 + k}&0\\ { - 8}&{56 + k} \end{array}} \right]\,\,{\rm{Thus}},\,\,{A^2} = 8A + kI\\ \Rightarrow \,\left[ {\begin{array}{*{20}{c}} 1&0\\ { - 8}&{49} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {8 + k}&0\\ { - 8}&{56 + k} \end{array}} \right]\\ \Rightarrow \,1 = 8 + k\,\,{\rm{and}}\,\,56 + k = 49\\ \Rightarrow \,k = - 7 \end{array}\]

From the given system of equations,
$$x = \frac{{{D_1}}}{D},y = \frac{{{D_2}}}{D},z = \frac{{{D_3}}}{D}$$
where, $$D = \Delta \left( {a,b,c} \right);\,\,\,{D_1} = \Delta \left( {d,b,c} \right)$$
$${D_2} = \Delta \left( {a,d,c} \right);\,{D_1} = \Delta \left( {a,b,d} \right)$$
Now, $$x = \frac{{\Delta \left( {d,b,c} \right)}}{{\Delta \left( {a,b,c} \right)}}$$
where, \[\Delta \left( {d,b,c} \right) = \left| {\begin{array}{*{20}{c}} { - {d_1}}&{{b_1}}&{{c_1}}\\ { - {d_2}}&{{b_2}}&{{c_2}}\\ { - {d_3}}&{{b_3}}&{{c_3}} \end{array}} \right|\]
\[ = \, - \left| {\begin{array}{*{20}{c}} {{b_1}}&{ - {d_1}}&{{c_1}}\\ {{b_2}}&{ - {d_2}}&{{c_2}}\\ {{b_3}}&{ - {d_3}}&{{c_3}} \end{array}} \right|\, = + \left| {\begin{array}{*{20}{c}} {{b_1}}&{{c_1}}&{ - {d_1}}\\ {{b_2}}&{{c_2}}&{ - {d_2}}\\ {{b_3}}&{{c_3}}&{ - {d_3}} \end{array}} \right|\]
$$ = \,\Delta \left( {b,c,d} \right)\,\,\,{\text{Hence,}}\,\,x = \frac{{\Delta \left( {b,c,d} \right)}}{{\Delta \left( {a,b,c} \right)}}$$

\[\begin{array}{l} AB = \left[ {\begin{array}{*{20}{c}} 0&c&{ - b}\\ { - c}&0&a\\ b&{ - a}&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{a^2}}&{ab}&{ac}\\ {ab}&{{b^2}}&{bc}\\ {ac}&{bc}&{{c^2}} \end{array}} \right]\\ AB = \left[ {\begin{array}{*{20}{c}} {abc - abc}&{{b^2}c - {b^2}c}&{b{c^2} - b{c^2}}\\ { - {a^2}c + {a^2}c}&{ - abc + abc}&{ - ac + ac}\\ {{a^2}b - {a^2}b}&{a{b^2} - a{b^2}}&{abc - abc} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 0&0&0\\ 0&0&0\\ 0&0&0 \end{array}} \right] = O \end{array}\]

Since the system of linear equations are
$$\eqalign{ & x + y + z = 2\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & 2x + 3y + 2z = 5\,\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr & 2x + 3y + \left( {{a^2} - 1} \right)z = a + 1\,\,\,\,\,\,\,\,.....\left( 3 \right) \cr} $$
\[{\rm{Now, }}\Delta = \left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 2\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ 2\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,{a^2} - 1 \end{array} \right|\]
\[ \Rightarrow \,\,\Delta = \left| \begin{array}{l} 1\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 2\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ 0\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,{a^2} - 3 \end{array} \right|\left( {{\rm{Applying }}{R_3} \to {R_3} - {R_2}} \right)\]
$$\eqalign{ & = {a^2} - 3 \cr & {\text{When, }}\Delta = 0 \cr & \Rightarrow \,\,{a^2} - 3 = 0 \cr & \Rightarrow \,\,\left| a \right| = \sqrt 3 \cr} $$
If $${a^2} = 3,$$   then plane represented by eqn (2) and eqn (3) are parallel.
Hence, the given system of equations is inconsistent.

\[AB = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\theta }&{\cos \theta \cdot \sin \theta } \\ {\cos \theta \sin \theta }&{{{\sin }^2}\theta } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\phi }&{\cos \phi \sin \phi } \\ {\cos \phi \sin \phi }&{{{\sin }^2}\phi } \end{array}} \right]\]
\[AB = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\theta {{\cos }^2}\phi + \cos \theta \sin \theta \cos \phi \sin \phi }&{{{\cos }^2}\theta \cos \phi \sin \phi + \cos \theta \sin \theta {{\sin }^2}\phi } \\ {\cos \theta \sin \theta {{\cos }^2}\phi + \cos \phi \sin \phi \cdot {{\sin }^2}\theta }&{\cos \theta \sin \theta \cos \phi \sin \phi + {{\sin }^2}\theta {{\sin }^2}\phi } \end{array}} \right]\]
\[AB = \left[ {\begin{array}{*{20}{c}} {\cos \theta \cos \phi \cos \left( {\theta - \phi } \right)}&{\cos \theta \sin \phi \cos \left( {\theta - \phi } \right)} \\ {\sin \theta \cos \phi \cos \left( {\theta - \phi } \right)}&{\sin \theta \sin \phi \cos \left( {\theta - \phi } \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]\]
$$ \Rightarrow \,\,\cos \left( {\theta - \phi } \right) = 0$$
$$ \Rightarrow \,\,\left| {\theta - \phi } \right| = $$   odd multiple of $$\frac{\pi }{2}.$$

\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} {{i^m}}&{{i^{m + 1}}}&{{i^{m + 2}}}\\ {{i^{m + 5}}}&{{i^{m + 4}}}&{{i^{m + 3}}}\\ {{i^{m + 6}}}&{{i^{m + 7}}}&{{i^{m + 8}}} \end{array}} \right|\\ = {i^m}{i^{m + 1}}{i^{m + 2}}\left| \begin{array}{l} 1\,\,\,\,\,\,\,1\,\,\,\,\,\,\,1\\ {i^5}\,\,\,\,{i^3}\,\,\,\,i\\ {i^6}\,\,\,\,{i^6}\,\,\,\,{i^6} \end{array} \right|\\ = {i^{3m + 3}}{i^6}\left| \begin{array}{l} 1\,\,\,\,\,1\,\,\,\,\,1\\ {i^5}\,\,\,{i^3}\,\,\,\,i\\ 1\,\,\,\,\,1\,\,\,\,\,1 \end{array} \right|\\ = 0 \end{array}\]
Hence, for all real $$m,$$ the value of given determinant is 0.

$$\eqalign{ & {\text{We have, }}BC = CB,{\text{ and }}{A^{N + 1}} = {\left( {B + C} \right)^{N + 1}} \cr & = {\,^{N + 1}}{C_0}{B^{N + 1}} + {\,^{N + 1}}{C_1}{B^N}C + {\,^{N + 1}}{C_2}{B^{N - 1}}{C^2} + ..... + {\,^{N + 1}}{C_r}{B^{N + 1}}{\,^{ - r}}{C_r} + ..... \cr & {\text{But given that }}{C^2} = 0 \cr & \Rightarrow {C^3} = {C^4} = ..... = {C^r} = 0 \cr & {\text{Hence, }}{A^{N + 1}} = {\,^{N + 1}}{C_N}{B^{N + 1}} + {\,^{N + 1}}{C_1}{B^N}C \cr & = {B^{N + 1}} + \left( {N + 1} \right){B^N}C = {B^N}\left[ {B + \left( {N + 1} \right)C} \right] \cr & {\text{Thus, }}K = N \cr} $$

Given matrix is :
\[\begin{array}{l} X = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 0&3 \end{array}} \right]\\ \therefore {X^2} = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 0&3 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2 - 6}\\ 0&9 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{ - 8}\\ 0&9 \end{array}} \right] \end{array}\]
So, the given expression is :
\[\begin{array}{l} {X^4} - 2X + 3I = \left[ {\begin{array}{*{20}{c}} 1&{ - 8}\\ 0&9 \end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 0&3 \end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 8}\\ 0&9 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { - 2}&{ + 4}\\ 0&{ - 6} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3&0\\ 0&3 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {1 - 2 + 3}&{ - 8 + 4}\\ 0&{9 - 6 + 3} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 2&{ - 4}\\ 0&6 \end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 0&3 \end{array}} \right] = 2X \end{array}\]