Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & \alpha x + y + z = \alpha - 1 \cr & x + \alpha y + z = \alpha - 1; \cr & x + y + \alpha z = \alpha - 1 \cr} $$
\[\Delta = \left| \begin{array}{l} \alpha \,\,\,\,\,\,1\,\,\,\,\,\,\,1\\ \,1\,\,\,\,\,\,\alpha \,\,\,\,\,\,1\\ \,1\,\,\,\,\,\,\,1\,\,\,\,\,\,\alpha \end{array} \right|\]
$$\eqalign{ & = \alpha \left( {{\alpha ^2} - 1} \right) - 1\left( {\alpha - 1} \right) + 1\left( {1 - \alpha } \right) \cr & = \alpha \left( {\alpha - 1} \right)\left( {\alpha + 1} \right) - 1\left( {\alpha - 1} \right) - 1\left( {\alpha - 1} \right) \cr} $$
For infinite solutions, $$\Delta = 0$$
$$\eqalign{ & \Rightarrow \,\,\left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 1 - 1} \right] = 0 \cr & \Rightarrow \,\,\left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 2} \right] = 0 \cr & \Rightarrow \,\,\alpha = - 2,1; \cr & {\text{But }}\alpha \ne 1.\,\,\,\,\,\,\therefore \,\,\alpha = - 2 \cr} $$

Let, \[\Delta = \left| {\begin{array}{*{20}{c}} {{{\cos }^2}{{54}^ \circ }}&{{{\cos }^2}{{36}^ \circ }}&{\cot {{135}^ \circ }}\\ {{{\sin }^2}{{53}^ \circ }}&{\cot {{135}^ \circ }}&{{{\sin }^2}{{37}^ \circ }}\\ {\cot {{135}^ \circ }}&{{{\cos }^2}{{25}^ \circ }}&{{{\cos }^2}{{65}^ \circ }} \end{array}} \right|\]
\[ = \left| {\begin{array}{*{20}{c}} {{{\cos }^2}{{54}^ \circ }}&{{{\sin }^2}{{54}^ \circ }}&{ - 1}\\ {{{\cos }^2}{{37}^ \circ }}&{ - 1}&{{{\sin }^2}{{37}^ \circ }}\\ { - 1}&{{{\cos }^2}{{25}^ \circ }}&{{{\sin }^2}{{25}^ \circ }} \end{array}} \right|\]
$${C_1} \to {C_1} + {C_2} + {C_3}$$
\[\begin{array}{l} = \left| {\begin{array}{*{20}{c}} {{{\cos }^2}{{54}^ \circ } + {{\sin }^2}{{54}^ \circ } - 1}&{{{\sin }^2}{{54}^ \circ }}&{ - 1}\\ {{{\cos }^2}{{37}^ \circ } - 1 + {{\sin }^2}{{37}^ \circ }}&{ - 1}&{{{\sin }^2}{{37}^ \circ }}\\ { - 1 + {{\cos }^2}{{25}^ \circ } + {{\sin }^2}{{25}^ \circ }}&{{{\cos }^2}{{25}^ \circ }}&{{{\sin }^2}{{25}^ \circ }} \end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}} 0&{{{\sin }^2}{{54}^ \circ } - 1}\\ 0&{ - 1{{\sin }^2}{{37}^ \circ }}\\ 0&{{{\cos }^2}{{25}^ \circ }{{\sin }^2}{{25}^ \circ }} \end{array}} \right| = 0 \end{array}\]

\[\left. \begin{array}{l} \,\,\,2{x_1} - 2{x_2} + {x_3} = \lambda {x_1}\\ 2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} = \lambda {x_3} \end{array} \right\}\]
$$\eqalign{ & \Rightarrow \,\,\left( {2 - \lambda } \right){x_1} - 2{x_2} + {x_3} = 0 \cr & \,\,\,\,\,\,\,2{x_1} - \left( {3 + \lambda } \right){x_2} + 2{x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} - \lambda {x_3} = 0 \cr} $$
For non - trivial solution, $$\Delta = 0$$
\[{\rm{i}}{\rm{.e}}{\rm{.,}}\,\,\left| \begin{array}{l} 2 - \lambda \,\,\,\,\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ \,\,\,\,2\,\,\,\,\,\,\,\,\,\, - \left( {3 + \lambda } \right)\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \lambda \end{array} \right| = 0\]
$$\eqalign{ & \Rightarrow \,\,\left( {2 - \lambda } \right)\left[ {\lambda \left( {3 + \lambda } \right) - 4} \right] + 2\left[ { - 2\lambda + 2} \right] + 1\left[ {4 - \left( {3 + \lambda } \right)} \right] = 0 \cr & \Rightarrow \,\,{\lambda ^3} + {\lambda ^2} - 5\lambda + 3 = 0 \cr & \Rightarrow \,\,\lambda = 1,1,3 \cr} $$
Hence, $$\lambda $$ has 2 values.

As given : $$\left| {{A_{n \times n}}} \right| = 3{\text{ and }}\left| {adj\,A} \right| = 243$$
Dterminant of adjoint $$A$$ 13 given by : $$\left| {adj\,A} \right| = {\left| {{A_{n \times n}}} \right|^{n - 1}}$$
$$\eqalign{ & \Rightarrow 243 = {3^{n - 1}} \cr & \Rightarrow {3^5} = {3^{n - 1}} \cr & \Rightarrow n - 1 = 5 \cr & \Rightarrow n = 6 \cr} $$

\[\begin{array}{l} P = \left[ \begin{array}{l} 1\,\,\,\,\,\,\,0\,\,\,\,\,\,0\\ 4\,\,\,\,\,\,\,1\,\,\,\,\,\,0\\ 16\,\,\,\,\,4\,\,\,\,\,1 \end{array} \right] = I + \left[ \begin{array}{l} 0\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\ 4\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\ 16\,\,\,\,\,4\,\,\,\,\,\,\,0 \end{array} \right] = I + A\\ {A^2} = \left[ \begin{array}{l} \,0\,\,\,\,\,\,0\,\,\,\,\,\,0\\ \,0\,\,\,\,\,\,0\,\,\,\,\,\,0\\ 16\,\,\,\,\,0\,\,\,\,\,\,0 \end{array} \right]\,\,{\rm{and }}\,\,{A^3} = \left[ \begin{array}{l} 0\,\,\,\,\,\,\,0\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,0\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,0\,\,\,\,\,\,0 \end{array} \right] \end{array}\]
$$\eqalign{ & \therefore \,\,{A^n} = O,\forall \,\,n \geqslant 3 \cr & {\text{Now }}{P^{50}} = {\left( {I + A} \right)^{50}} = {\,^{50}}{C_0}\,{I^{50}} + {\,^{50}}{C_1}\,{I^{49}}A + {\,^{50}}{C_2}\,{I^{48}}{A^2} + O \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = I + 50A + 25 \times 49{A^2}. \cr & \therefore \,\,Q = {P^{50}} - I = 50A + 25 \times 49{A^2}. \cr & \Rightarrow \,\,{q_{21}} = 50 \times 4 = 200 \cr & \Rightarrow \,\,{q_{31}} = 50 \times 16 + 25 \times 49 \times 16 = 20400 \cr & \Rightarrow \,\,{q_{32}} = 50 \times 4 = 200 \cr & \therefore \,\,\frac{{{q_{31}} + {q_{32}}}}{{{q_{21}}}} = \frac{{20600}}{{200}} = 103 \cr} $$

\[\begin{array}{l} {A^2} = \left( {\begin{array}{*{20}{c}} p&q\\ 0&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} p&q\\ 0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{p^2}}&{pq + q}\\ 0&1 \end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}} {{p^2}}&{q\left( {p + 1} \right)}\\ 0&1 \end{array}} \right)\\ {A^3} = \left( {\begin{array}{*{20}{c}} p&q\\ 0&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{p^2}}&{pq + q}\\ 0&1 \end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}} {{p^3}}&{q\left( {{p^2} + p + 1} \right)}\\ 0&1 \end{array}} \right) \end{array}\]
Similarly, \[{A^4} = \left( {\begin{array}{*{20}{c}} {{p^4}}&{q\left( {{p^3} + {p^2} + p + 1} \right)}\\ 0&1 \end{array}} \right)\]      and so on.
\[\therefore {A^8} = \left( {\begin{array}{*{20}{c}} {{p^8}}&{q\left( {{p^7} + {p^8} + ..... + 1} \right)}\\ 0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{p^8}}&{q\left( {\frac{{{p^8} - 1}}{{p - 1}}} \right)}\\ 0&1 \end{array}} \right)\]

Given matrix is \[A = \left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&1&1\\ 1&1&1 \end{array}} \right]\]
\[\begin{array}{l} {\rm{So, }}{A^2} = \left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&1&1\\ 1&1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&1&1\\ 1&1&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 3&3&3\\ 3&3&3\\ 3&3&3 \end{array}} \right] = 3\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&1&1\\ 1&1&1 \end{array}} \right]\\ {\rm{and }}{A^3} = 3\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&1&1\\ 1&1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&1&1\\ 1&1&1 \end{array}} \right]\\ = 3\left[ {\begin{array}{*{20}{c}} 3&3&3\\ 3&3&3\\ 3&3&3 \end{array}} \right] = 9\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&1&1\\ 1&1&1 \end{array}} \right] = {3^2}A \end{array}\]
Similarly, $$A^4 = 3^3 A.$$   Hence, $$A^n = 3^{n - 1} A$$

\[{\left| A \right|_{\max }} = \left[ {\begin{array}{*{20}{c}} 0&1&1\\ 1&0&1\\ 1&1&0 \end{array}} \right] = 0 - \left( { - 1} \right) + 1\left( 1 \right) = 2\]
$$\therefore {\left| A \right|_{\min }} = - 2$$

We have, \[\left[ {\begin{array}{*{20}{c}} 2&0&7\\ 0&1&0\\ 1&{ - 2}&1 \end{array}} \right]\]
\[\left[ {\begin{array}{*{20}{c}} { - x}&{14x}&{7x}\\ 0&1&0\\ x&{ - 4x}&{ - 2x} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {5x}&0&0\\ 0&1&0\\ 0&{10x - 2}&{5x} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\]
$$ \Rightarrow x = \frac{1}{5}$$

\[M = \left[ \begin{array}{l} \,\,{\sin ^4}\theta {\rm{ }} - 1\sin \theta \\ {\rm{1 + co}}{{\rm{s}}^2}\theta \,\,\,\,\,\,\,\,\,\,{\cos ^4}\theta \end{array} \right]\]
$$\eqalign{ & \left| M \right| = {\sin ^4}\theta {\cos ^4}\theta + 1 + {\sin ^2}\theta + {\cos ^2}\theta + {\sin ^2}\theta {\cos ^2}\theta \cr & = 2 + {\sin ^2}\theta {\cos ^2}\theta + {\sin ^4}\theta {\cos ^4}\theta \cr} $$
\[{M^{ - 1}} = \frac{1}{{\left| M \right|}}\left[ \begin{array}{l} \,\,\,{\cos ^4}\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + {\sin ^2}\theta \\ - 1 - {\cos ^2}\theta \,\,\,\,\,\,\,\,\,\,\,{\sin ^4}\theta \end{array} \right]\]
$${\text{Given}}\,{\text{that}}\,M = \alpha I + \beta {M^{ - 1}}$$
\[ \Rightarrow \,\left[ \begin{array}{l} \,\,\,{\sin ^4}\theta \,\,\,\,\,\,\,\,\, - 1 - {\sin ^2}\theta \\ 1 + {\cos ^2}\theta \,\,\,\,\,\,\,\,{\cos ^4}\theta \end{array} \right] = \left[ \begin{array}{l} \alpha\,\,\,\,\,0\\ 0\,\,\,\,\,a \end{array} \right] + \frac{\beta }{{\left| M \right|}}\left[ \begin{array}{l} \,\,\,{\cos ^4}\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + {\sin ^2}\theta \\ - 1 - {\cos ^2}\theta \,\,\,\,\,\,\,\,\,\,\,\,\,{\sin ^4}\theta \end{array} \right]\]
$$\eqalign{ & \Rightarrow \,\frac{\beta }{{\left| M \right|}} = - 1\,{\text{and}}\,\alpha \, + \frac{\beta }{{\left| M \right|}}{\cos ^4}\theta = {\sin ^4}\theta \cr & \Rightarrow \,\alpha = {\sin ^4}\theta + {\cos ^4}\theta \cr & \Rightarrow \,\beta = - \left[ {2 + {{\sin }^2}\theta {{\cos }^2}\theta + {{\sin }^4}\theta {{\cos }^4}\theta } \right] \cr & {\text{Now,}}\, \alpha \, = {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^2} - 2{\sin ^2}\theta {\cos ^2}\theta \cr & = 1 - 2{\sin ^2}\theta {\cos ^2}\theta = 1 - \frac{1}{2}{\sin ^2}2\theta \cr} $$
For $$\alpha $$ to be minimum $${\sin ^2}2\theta $$  is maximum i.e., 1.
$$\eqalign{ & \therefore \,{\alpha ^ * } = 1 - \frac{1}{2} = \frac{1}{2} \cr & {\text{Also}},\beta = - \left[ {2 + \frac{1}{4}{{\sin }^2}2\theta + \frac{1}{{16}}{{\sin }^4}2\theta } \right] \cr} $$
For $$\beta $$ to be minimum, $${\sin ^2}2\theta $$  is maximum i.e.,
$$\eqalign{ & \therefore \,\,{\beta ^ * } = - \left[ {2 + \frac{1}{4} + \frac{1}{{16}}} \right] = - \frac{{32 + 4 + 1}}{{16}} = \frac{{ - 37}}{{16}} \cr & \therefore \,\,{\alpha ^ * } + {\beta ^ * } = \frac{1}{2} - \frac{{37}}{{16}} = \frac{{ - 29}}{{16}} \cr} $$