Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

For every ‘det, with value 1’ $$\left( { \in B} \right)$$  we can find a det. with value $$- 1$$  by changing the sign of one entry of ‘1’.
Hence there are equal number of elements in $$B$$ and $$C.$$
∴ $$(B)$$ is the correct option

\[\left| \begin{array}{l} \,\,a\,\,\,\,\,\,\,\,\,a + 1\,\,\,\,\,\,\,\,\,a - 1\\ - b\,\,\,\,\,\,\,\,b + 1\,\,\,\,\,\,\,\,\,b - 1\\ \,\,c\,\,\,\,\,\,\,\,\,c - 1\,\,\,\,\,\,\,\,\,\,c + 1 \end{array} \right| + \left| \begin{array}{l} \,\,\,\,\,a + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c - 1\\ \,\,\,\,\,a - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c + 1\\ {\left( { - 1} \right)^{n + 2}}a\,\,\,\,\,\,\,\,{\left( { - 1} \right)^{n + 1}}b\,\,\,\,\,\,\,\,\,{\left( { - 1} \right)^n}c \end{array} \right| = 0\]
\[ \Rightarrow \,\,\left| \begin{array}{l} \,\,a\,\,\,\,\,\,\,\,\,a + 1\,\,\,\,\,\,\,\,\,a - 1\\ - b\,\,\,\,\,\,\,\,b + 1\,\,\,\,\,\,\,\,\,b - 1\\ \,\,c\,\,\,\,\,\,\,\,\,c - 1\,\,\,\,\,\,\,\,\,\,c + 1 \end{array} \right| + \left| \begin{array}{l} a + 1\,\,\,\,\,\,\,\,\,\,a - 1\,\,\,\,\,\,\,\,\,\,\,{\left( { - 1} \right)^{n + 2}}a\\ b + 1\,\,\,\,\,\,\,\,\,b - 1\,\,\,\,\,\,\,\,\,\,\,\,{\left( { - 1} \right)^{n + 1}}b\\ c - 1\,\,\,\,\,\,\,\,\,c + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( { - 1} \right)^n}c \end{array} \right| = 0\]
(Taking transpose of second determinant)
$${C_1} \Leftrightarrow {C_3}$$
\[ \Rightarrow \,\,\left| \begin{array}{l} \,\,a\,\,\,\,\,\,\,\,a + 1\,\,\,\,\,\,\,\,a - 1\\ - b\,\,\,\,\,\,\,b + 1\,\,\,\,\,\,\,\,\,b - 1\\ \,\,c\,\,\,\,\,\,\,\,c - 1\,\,\,\,\,\,\,\,\,\,c + 1 \end{array} \right| - \left| \begin{array}{l} \,\,\,{\left( { - 1} \right)^{n + 2}}a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,a + 1\\ {\left( { - 1} \right)^{n + 2}}\left( { - b} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,b - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,b + 1\\ \,\,\,{\left( { - 1} \right)^{n + 2}}c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c - 1 \end{array} \right| = 0\]
$${C_2} \Leftrightarrow {C_3}$$
\[ \Rightarrow \,\,\left| \begin{array}{l} \,\,a\,\,\,\,\,\,\,\,a + 1\,\,\,\,\,\,\,\,a - 1\\ - b\,\,\,\,\,\,\,b + 1\,\,\,\,\,\,\,\,\,b - 1\\ \,\,c\,\,\,\,\,\,\,\,c - 1\,\,\,\,\,\,\,\,\,\,c + 1 \end{array} \right| + {\left( { - 1} \right)^{n + 2}}\left| \begin{array}{l} \,\,a\,\,\,\,\,\,\,\,a + 1\,\,\,\,\,\,\,\,a - 1\\ - b\,\,\,\,\,\,\,b + 1\,\,\,\,\,\,\,\,\,b - 1\\ \,\,c\,\,\,\,\,\,\,\,c - 1\,\,\,\,\,\,\,\,\,\,c + 1 \end{array} \right| = 0\]
\[ \Rightarrow \,\,\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| \begin{array}{l} \,\,a\,\,\,\,\,\,\,\,a + 1\,\,\,\,\,\,\,\,a - 1\\ - b\,\,\,\,\,\,\,b + 1\,\,\,\,\,\,\,\,\,b - 1\\ \,\,c\,\,\,\,\,\,\,\,c - 1\,\,\,\,\,\,\,\,\,\,c + 1 \end{array} \right| = 0\]
$${C_2} - {C_1},{C_3} - {C_1}$$
\[ \Rightarrow \,\,\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| \begin{array}{l} \,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1\\ - b\,\,\,\,\,\,\,2b + 1\,\,\,\,\,\,\,\,\,2b - 1\\ \,\,c\,\,\,\,\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right| = 0\]
$${R_1} + {R_3}$$
\[ \Rightarrow \,\,\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| \begin{array}{l} a + c\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\ \, - b\,\,\,\,\,\,\,\,\,2b + 1\,\,\,\,\,\,\,\,\,2b - 1\\ \,\,\,\,c\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right| = 0\]
$$\eqalign{ & \Rightarrow \,\,\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left( {a + c} \right)\left( {2b + 1 + 2b - 1} \right) = 0 \cr & \Rightarrow \,\,4b\left( {a + c} \right)\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right] = 0 \cr & \Rightarrow \,\,1 + {\left( { - 1} \right)^{n + 2}} = 0\,\,{\text{as }}b\left( {a + c} \right) \ne 0 \cr} $$
⇒ $$n$$ should be an odd integer.

The given points are $$P \left( { - \sin \left( {\beta - \alpha } \right), - \cos \beta } \right),\,$$     $$Q \left( {\cos \left( {\beta - \alpha } \right),\sin \beta } \right)\,\,{\text{and}}$$      $$R \left( {\cos \left( {\beta - \alpha + \theta } \right),\sin \left( {\beta - \theta } \right)} \right),$$
Where $$0 < \alpha ,\beta ,\theta < \frac{\pi }{4}$$
\[\therefore \,\,\Delta = \left| \begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ - \sin \left( {\beta - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\cos \left( {\beta - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {\beta - \alpha + \theta } \right)\\ \,\,\,\,\,\, - \cos \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \left( {\beta - \theta } \right) \end{array} \right|\]
$${\text{Operating }}{C_3} - {C_1}\sin \theta - {C_2}\cos \theta ,\,{\text{we get}}$$
\[\Delta = \left| \begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \sin \theta - \cos \theta \\ - \sin \left( {\beta - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\cos \left( {\beta - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\ \,\,\, - \cos \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \end{array} \right|\]
$$\eqalign{ & = \left( {1 - \sin \theta - \cos \theta } \right)\left[ {\cos \beta \cos \left( {\beta - \alpha } \right) - \sin \beta \sin \left( {\beta - \alpha } \right)} \right] \cr & \Rightarrow \,\,\Delta = \left[ {1 - \left( {\sin \theta + \cos \theta } \right)} \right]\cos \left( {2\beta - \alpha } \right) \cr & \because 0 < \alpha ,\beta ,\theta < \frac{\pi }{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,\sin \theta + \cos \theta \ne 1 \cr & {\text{Also 2}}\beta - \alpha < \frac{\pi }{4} \cr & \Rightarrow \,\,\cos \left( {2\beta - \alpha } \right) \ne 0 \cr & \therefore \,\,\Delta \ne 0 \cr} $$
⇒ the three points are non collinear.

\[\vartriangle = \left( {\lambda + 1 + \alpha + \beta } \right)\left| {\begin{array}{*{20}{c}} 1&1&1 \\ \alpha &{\lambda + \beta }&1 \\ \beta &1&{\lambda + \alpha } \end{array}} \right|\,\,\left( {{\text{Using }}{R_1} \to {R_1} + {R_2} + {R_3}} \right)\]
\[\vartriangle = \lambda \left| {\begin{array}{*{20}{c}} 1&0&1 \\ \alpha &{\lambda + \beta - \alpha }&{1 - \alpha } \\ \beta &{1 - \beta }&{\lambda + \alpha - \beta } \end{array}} \right| = \lambda \left| {\begin{array}{*{20}{c}} {\lambda + \beta - \alpha }&{1 - \alpha } \\ {1 - \beta }&{\lambda + \alpha - \beta } \end{array}} \right|\]
$$\vartriangle = \lambda \left\{ {{\lambda ^2} - {{\left( {\alpha - \beta } \right)}^2} - \left( {1 - \alpha } \right)\left( {1 - \beta } \right)} \right\}$$
$$\vartriangle = {\lambda ^3} - \lambda \left\{ {{\alpha ^2} + {\beta ^2} - 2\alpha \beta + 1 - \alpha - \beta + \alpha \beta } \right\} = 0\,\,{\text{because }}{\alpha ^2} = \beta ,{\beta ^2} = \alpha ,\alpha \beta = 1.$$

\[A = \left[ \begin{array}{l} 1\,\,\,\,\,\,2\\ 3\,\,\,\,\,4 \end{array} \right]\,\,\,\,\,{\rm{ }}B = \left[ \begin{array}{l} a\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,b \end{array} \right]\]
\[AB = \left[ \begin{array}{l} \,a\,\,\,\,\,\,\,2b\\ 3a\,\,\,\,\,\,4b \end{array} \right]\]
\[BA = \left[ \begin{array}{l} a\,\,\,\,\,0\\ 0\,\,\,\,\,b \end{array} \right]\left[ \begin{array}{l} 1\,\,\,\,\,\,\,2\\ 3\,\,\,\,\,\,4 \end{array} \right] = \left[ \begin{array}{l} \,a\,\,\,\,\,\,\,2b\\ 3a\,\,\,\,\,4b \end{array} \right]\]
Hence, $$AB = BA$$   only when $$a = b$$
∴ There can be infinitely many $$B’s$$  for which $$AB = BA$$

We have
\[\frac{1}{2}\left| {\left| \begin{array}{l} \,\,k\,\,\,\,\, - 3k\,\,\,\,\,\,\,1\\ \,\,5\,\,\,\,\,\,\,\,\,\,\,k\,\,\,\,\,\,\,\,1\\ - k\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,1 \end{array} \right|} \right| = 28\]
$$\eqalign{ & \Rightarrow \,\,5{k^2} + 13k - 46 = 0 \cr & {\text{or }}5{k^2} + 13k + 66 = 0 \cr & {\text{Now,}}\,\,5{k^2} + 13k - 46 = 0 \cr} $$

$$\eqalign{ & k = \frac{{ - 13 \pm \sqrt {1089} }}{{10}} \cr & \therefore \,\,k = \frac{{ - 23}}{5};\,k = 2 \cr} $$
since $$k$$ is an integer, $$\therefore \,\,k = 2$$
Also $$5{k^2} + 13k + 66 = 0$$
$$ \Rightarrow \,\,k = \frac{{ - 13 \pm \sqrt { - 1151} }}{{10}}$$
So no real solution exist
For orthocentre
$$BH \bot AC$$
$$\eqalign{ & \therefore \,\,\left( {\frac{{\beta - 2}}{{\alpha - 5}}} \right)\left( {\frac{8}{{ - 4}}} \right) = - 1 \cr & \Rightarrow \,\,\alpha - 2\beta = 1\,\,\,\,.....\left( 1 \right) \cr} $$
Also $$CH \bot AB$$
$$\eqalign{ & \therefore \,\,\left( {\frac{{\beta - 2}}{{\alpha + 2}}} \right)\left( {\frac{8}{3}} \right) = - 1 \cr & \Rightarrow \,\,3\alpha + 8\beta = 10\,\,\,\,\,.....\left( 2 \right) \cr} $$
Solving (1) and (2), we get
$$\alpha = 2,\beta = \frac{1}{2}$$
orthocentre is $$\left( {2,\frac{1}{2}} \right)$$

\[\left[ {\begin{array}{*{20}{c}} 1&0 \\ { - 1}&7 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0 \\ { - 1}&7 \end{array}} \right] = 8\left[ {\begin{array}{*{20}{c}} 1&0 \\ { - 1}&7 \end{array}} \right] + k\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]\]
\[{\text{or, }}\left[ {\begin{array}{*{20}{c}} {1 + 0}&{0 + 0} \\ { - 1 - 7}&{0 + 49} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 8&0 \\ { - 8}&{56} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} k&0 \\ 0&k \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {8 + k}&0 \\ { - 8}&{56 + k} \end{array}} \right]\]
$$\eqalign{ & \Rightarrow \,\,1 = 8 + k,49 = 56 + k \cr & \Rightarrow \,\,k = - 7. \cr} $$

\[A = \left[ \begin{array}{l} \cos \theta \,\,\,\,\,\,\, - \sin \theta \\ \sin \theta \,\,\,\,\,\,\,\,\,\,\cos \theta \end{array} \right]\]
$$ \Rightarrow \,\,\left| A \right| = 1$$
\[adj\left( A \right) = {\left[ {\begin{array}{*{20}{c}} { + \cos \theta }&{ - \sin \theta }\\ { + \sin \theta }&{ + \cos \theta } \end{array}} \right]^T}\]
\[\,\,\,\,\, = \left[ \begin{array}{l} \,\,\cos \theta \,\,\,\,\,\,\,\,\,\sin \theta \\ - \sin \theta \,\,\,\,\,\,\,\,\cos \theta \end{array} \right]\]
\[ \Rightarrow \,\,{A^{ - 1}} = \left[ \begin{array}{l} \,\cos \theta \,\,\,\,\,\,\,\,\,\,\sin \theta \\ - \sin \theta \,\,\,\,\,\,\,\,\cos \theta \end{array} \right] = B\]
\[{B^2} = \left[ \begin{array}{l} \,\,\cos \theta \,\,\,\,\,\,\,\,\,\sin \theta \\ - \sin \theta \,\,\,\,\,\,\,\,\cos \theta \end{array} \right]\left[ \begin{array}{l} \,\,\cos \theta \,\,\,\,\,\,\,\,\,\sin \theta \\ - \sin \theta \,\,\,\,\,\,\,\,\cos \theta \end{array} \right]\]
\[ = \left[ \begin{array}{l} \,\,\cos 2\theta \,\,\,\,\,\,\,\,\,\sin 2\theta \\ - \sin 2\theta \,\,\,\,\,\,\,\,\cos 2\theta \end{array} \right]\]
\[ \Rightarrow \,\,{B^3} = \left[ \begin{array}{l} \,\,\cos 3\theta \,\,\,\,\,\,\,\,\,\sin 3\theta \\ - \sin 3\theta \,\,\,\,\,\,\,\,\cos 3\theta \end{array} \right]\]
\[ \Rightarrow \,\,{A^{ - 50}} = {B^{50}} = \left[ \begin{array}{l} \,\,\,\,\cos \left( {50\theta } \right)\,\,\,\,\,\,\,\sin \left( {50\theta } \right)\\ \, - \sin \left( {50\theta } \right) \,\,\,\,\,\,\,\,\,\cos \left( {50\theta } \right) \end{array} \right]\]
\[{\left( {{A^{ - 50}}} \right)_{\theta = \frac{\pi }{12}}} = \left[ \begin{array}{l} \,\frac{{\sqrt 3 }}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{2}\\ - \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\frac{{\sqrt 3 }}{2} \end{array} \right]\]
$$\left[ {\because \,\,\cos \left( {\frac{{50\pi }}{{12}}} \right) = \cos \left( {4\pi + \frac{\pi }{6}} \right) = \cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2}} \right]$$

$$\eqalign{ & BB' = B\left( {{A^{ - 1}}A'} \right)' = B\left( {A'} \right)'\left( {{A^{ - 1}}} \right)' = BA\left( {{A^{ - 1}}} \right)' \cr & = \left( {{A^{ - 1}}A'} \right)\left( {A\left( {{A^{ - 1}}} \right)'} \right) \cr & = {A^{ - 1}}A.A'.\left( {{A^{ - 1}}} \right)'\,\,\,\,\,\left\{ {{\text{as }}AA' = A'A} \right\} \cr & = I\left( {{A^{ - 1}}A} \right)' \cr & = I.I \cr & = {I^2} \cr & = I \cr} $$

$$a\left( {x + y + z} \right) = \left( {1 + a} \right)x,b\left( {x + y + z} \right) = \left( {1 + b} \right)y,c\left( {x + y + z} \right) = \left( {1 + c} \right)z.$$
In case of non-trivial solution, $$x + y + z \ne 0.$$
$$\therefore \,\,\frac{a}{{1 + a}} = \frac{x}{{x + y + z}},\,{\text{e}}{\text{.t}}{\text{.c}}{\text{.}}$$
Adding, $$\frac{a}{{1 + a}} + \frac{b}{{1 + b}} + \frac{c}{{1 + c}} = 1.$$
$$\therefore \,\,\left( {1 - \frac{1}{{1 + a}}} \right) + \left( {1 - \frac{1}{{1 + b}}} \right) + \left( {1 - \frac{1}{{1 + c}}} \right) = 1.$$