Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\left[ {\begin{array}{*{20}{c}} \alpha &0 \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} \alpha &0 \\ 1&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 5&1 \end{array}} \right]\]
\[{\text{or, }}\left[ {\begin{array}{*{20}{c}} {{\alpha ^2}}&0 \\ {\alpha + 1}&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 5&1 \end{array}} \right]\]
$$ \Rightarrow \,\,{\alpha ^2} = 1,\alpha + 1 = 5$$
$$ \Rightarrow \,\,\alpha = 1, - 1$$    satisfying $$\alpha = 4$$  (absurd).

\[{C_3} \to {C_3} + {C_2} - {C_1}\,\,{\text{gives }}\vartriangle \left( x \right) = \left| {\begin{array}{*{20}{c}} 1&{\cos x}&0 \\ {1 + \sin x}&{\cos x}&0 \\ {\sin x}&{\sin x}&1 \end{array}} \right| = \cos x - \cos x\left( {1 + \sin x} \right) = - \sin x \cdot \cos x\]
$$\therefore \,\,\int\limits_0^{\frac{\pi }{2}} {\vartriangle \left( x \right)dx = - \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2x\,dx} } $$
\[\int\limits_0^{\frac{\pi }{2}} {\vartriangle \left( x \right)dx = - \frac{1}{2}\left[ { - \frac{{\cos 2x}}{2}} \right]_0^{\frac{\pi }{2}} = \frac{1}{4}\left( {\cos \pi - \cos 0} \right) = - \frac{1}{2}.} \]

\[{\rm{Given }}\,\,A = \left[ \begin{array}{l} 0\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,1\\ 0\,\,\,\,\, - 2\,\,\,\,\,\,\,\,4 \end{array} \right]\]
∴ Characteristic eqn of above matrix $$A$$ is given by
\[\left| {A - \lambda I} \right| = 0 \Rightarrow \,\left[ \begin{array}{l} 1 - \lambda \,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\ \,\,\,\,0\,\,\,\,\,\,\,\,\,\,1 - \lambda \,\,\,\,\,\,\,\,\,\,\,\,1\\ \,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,4 - \lambda \end{array} \right] = 0\]
$$\eqalign{ & \Rightarrow \,\,\left( {1 - \lambda } \right)\left( {4 - 5\lambda + {\lambda ^2} + 2} \right) = 0 \cr & \Rightarrow \,\,{\lambda ^3} - 6{\lambda ^2} + 11\lambda - 6 = 0 \cr} $$
Also by Cayley Hamilton thm (every square matrix satisfies its characteristic equation) we obtain
$${A^3} - 6{A^2} + 11A - 6I = 0$$
Multiplying by $${A^{ - 1}},$$ we get
$$\eqalign{ & {A^2} - 6A + 11I - 6{A^{ - 1}} = 0 \cr & \Rightarrow \,\,{A^{ - 1}} = \frac{1}{6}\left( {{A^2} - 6A + 11I} \right) \cr} $$
Comparing it with given relation,
$${A^{ - 1}} = \frac{1}{6}\left( {{A^2} - cA + dI} \right)$$
$${\text{we get }}c = - 6\,\,{\text{and}}\,{\text{ }}d = 11$$

Let, $$A^2 = I$$
⇒ $$A^2 A^{- 1} = IA^{- 1}$$
⇒ $$A = A^{- 1}$$

For homogeneous system of equations to have non zero solution, $$\Delta = 0$$
\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} 1&{2a}&a\\ 1&{3b}&b\\ 1&{4c}&c \end{array}} \right| = 0\left[ {\therefore {C_2} \to {C_2} - 2{C_3}} \right]\\ \left| {\begin{array}{*{20}{c}} 1&0&a\\ 1&b&b\\ 1&{2c}&c \end{array}} \right| = 0\left[ {{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}} \right] \end{array}\]
On simplification, $$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$
∴ $$a, b, c$$  are in Harmonic Progression.

Let $$A$$ be the $$1^{st}$$ term and $$R$$ the common ratio of G.P., then ;
$$\eqalign{ & a = {T_p} = A{R^{p - 1}} \cr & \therefore \log a = \log A + \left( {p - 1} \right)\log R \cr} $$
Similarly, $$\log b = \log A + \left( {q - 1} \right)\log R$$
and $$\log c = \log A + \left( {r - 1} \right)\log R$$
\[\therefore \Delta = \left| {\begin{array}{*{20}{c}} {\log A + \left( {p - 1} \right)\log R}&p&1\\ {\log A + \left( {q - 1} \right)\log R}&q&1\\ {\log A + \left( {r - 1} \right)\log R}&r&1 \end{array}} \right|\]
Split into two determinants and in the first take $$\log A$$  common and in the second take $$\log R$$  common
\[\Delta = \log A\left| {\begin{array}{*{20}{c}} 1&p&1\\ 1&q&1\\ 1&r&1 \end{array}} \right| + \log R\left| {\begin{array}{*{20}{c}} {p - 1}&p&1\\ {q - 1}&q&1\\ {r - 1}&r&1 \end{array}} \right|\]
Apply $${C_1} \to {C_1} - {C_2} + {C_3}$$    in the second
\[\Delta = 0 + \log R\left| {\begin{array}{*{20}{c}} 0&p&1\\ 0&q&1\\ 0&r&1 \end{array}} \right| = 0\]

Given, \[\left| {\begin{array}{*{20}{c}} {2a}&{3r}&x\\ {4b}&{6s}&{2y}\\ { - 2c}&{ - 3t}&{ - z} \end{array}} \right| = \lambda \left| {\begin{array}{*{20}{c}} a&r&x\\ b&s&y\\ c&t&z \end{array}} \right|\]
Taking 2 common from $$C_1$$ and 3 from $$C_2$$ in L.H.S.
\[\therefore 2 \times 3\left| {\begin{array}{*{20}{c}} a&r&x\\ {2b}&{2s}&{2y}\\ { - c}&{ - t}&{ - z} \end{array}} \right| = \lambda \left| {\begin{array}{*{20}{c}} a&r&x\\ b&s&y\\ c&t&z \end{array}} \right|\]
Taking 2 common from $$R_2$$ and $$- 1$$ from $$R_3$$ in L.H.S.
\[\therefore \, - 12\left| {\begin{array}{*{20}{c}} a&r&x\\ b&s&y\\ c&t&z \end{array}} \right| = \lambda \left| {\begin{array}{*{20}{c}} a&r&x\\ b&s&y\\ c&t&z \end{array}} \right|\]
$$ \Rightarrow \,\lambda = - 12$$

Since, $$ - 1 < x < 0$$
$$\eqalign{ & \therefore \left[ x \right] = - 1 \cr & 0 < y < 1 \cr & \therefore \left[ y \right] = 0, \cr & 1 < z < 2 \cr & \therefore \left[ z \right] = 1 \cr} $$
∴ Given determinant \[ = \left| {\begin{array}{*{20}{c}} 0&0&1\\ { - 1}&1&1\\ { - 1}&0&2 \end{array}} \right| = 1 = \left[ z \right]\]

\[\begin{array}{l} {\rm{Given,\, }}{\Delta _1} = \left| {\begin{array}{*{20}{c}} {10}&4&3\\ {17}&7&4\\ 4&{ - 5}&7 \end{array}} \right|\\ {C_1} \leftrightarrow {C_2}\\ {\Delta _1} = - \left| {\begin{array}{*{20}{c}} 4&{10}&3\\ 7&{17}&4\\ { - 5}&4&7 \end{array}} \right|\\ {\rm{Also\, given\, }}{\Delta _1} + {\Delta _2} = 0\\ \Rightarrow - \left| {\begin{array}{*{20}{c}} 4&{10}&3\\ 7&{17}&4\\ { - 5}&4&7 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} 4&{x + 5}&3\\ 7&{x + 12}&4\\ { - 5}&{x - 1}&7 \end{array}} \right| = 0\\ \Rightarrow \left| {\begin{array}{*{20}{c}} 4&{x - 5}&3\\ 7&{x - 5}&4\\ { - 5}&{x - 5}&7 \end{array}} \right| = 0\\ \Rightarrow \left( {x - 5} \right)\left| {\begin{array}{*{20}{c}} 4&1&3\\ 7&1&4\\ { - 5}&1&7 \end{array}} \right| = 0\\ \Rightarrow \left( {x - 5} \right)\left( { - 21} \right) = 0\\ \Rightarrow x - 5 = 0\\ \Rightarrow x = 5 \end{array}\]

$${C_1} \to {C_1} - {C_2}$$
\[ \Rightarrow \left| {\begin{array}{*{20}{c}} 4&{{{\left( {{e^{i\alpha }} - {e^{ - i\alpha }}} \right)}^2}}&4\\ 4&{{{\left( {{e^{i\beta }} - {e^{ - i\beta }}} \right)}^2}}&4\\ 4&{{{\left( {{e^{i\gamma }} - {e^{ - i\gamma }}} \right)}^2}}&4 \end{array}} \right| = 0\]