Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & B = {A_1} + 3A_3^3 + .....\left( {2n - 1} \right){\left( {{A_{2n - 1}}} \right)^{2n - 1}} \cr & {B^T} = - \left[ {{A_1} + 3A_3^3 + .....\left( {2n - 1} \right){{\left( {{A_{2r - 1}}} \right)}^{2r - 1}}} \right] \cr & = - B,{\text{ so skew - symmetric}} \cr} $$

For trivial solution,
\[\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\lambda \,\,\,\,\,\, - 1\\ \lambda \,\,\,\, - 1\,\,\,\,\,\, - 1\\ 1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\, - \lambda \end{array} \right| = 0\]
$$\eqalign{ & \Rightarrow \, - \lambda \left( {\lambda + 1} \right)\left( {\lambda - 1} \right) = 0 \cr & \Rightarrow \,\,\lambda = 0, + 1, - 1 \cr} $$

Taking $$x^5$$ common from last row, we get
\[{x^5}\left| {\begin{array}{*{20}{c}} {{x^n}}&{{x^{n + 2}}}&{{x^{2n}}}\\ 1&{{x^a}}&a\\ {{x^n}}&{{x^{a + 1}}}&{{x^{2n}}} \end{array}} \right| = 0\,\,\forall \,\,x \in R,\]
⇒ $$a + 1 = n + 2$$
⇒ $$a = n + 1$$
(as it will make first and third row identical)

\[f\left( x \right) = x \cdot x \cdot \left( {x - 1} \right)\left| {\begin{array}{*{20}{c}} 1&x&{x + 1} \\ 2&{x - 1}&{x + 1} \\ 3&{x - 2}&{x + 1} \end{array}} \right|\]
\[f\left( x \right) = {x^2}\left( {x - 1} \right)\left| {\begin{array}{*{20}{c}} 1&x&1 \\ 2&{x - 1}&2 \\ 3&{x - 2}&3 \end{array}} \right| = 0\]
$$\eqalign{ & \left( {{C_3} \to {C_3} - {C_2}} \right) \cr & \therefore \,\,\,f\left( {100} \right) = 0. \cr} $$

$$\eqalign{ & {A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right) \cr & {A^2} - {B^2} = {A^2} + AB - BA - {B^2} \cr & \Rightarrow AB = BA \cr} $$

Check \[\left| {\begin{array}{*{20}{c}} 0&c&b \\ c&0&a \\ b&a&0 \end{array}} \right| \times \left| {\begin{array}{*{20}{c}} 0&c&b \\ c&0&a \\ b&a&0 \end{array}} \right| = \]      the given determinant.

Let $$r$$ be the common ratio, then
\[\left| \begin{array}{l} \,\,\log {a_n}\,\,\,\,\,\,\,\,\log {a_{n + 1}}\,\,\,\,\,\,\,\,\,\,\log {a_{n + 2}}\\ \log {a_{n + 3}}\,\,\,\,\,\,\,\log {a_{n + 4}}\,\,\,\,\,\,\,\,\,\log {a_{n + 5}}\\ \log {a_{n + 6}}\,\,\,\,\,\,\,\log {a_{n + 7}}\,\,\,\,\,\,\,\,\,\log {a_{n + 8}} \end{array} \right|\]
\[ = \left| \begin{array}{l} \log {a_1}{r^{n - 1}}\,\,\,\,\,\,\log {a_1}{r^n}\,\,\,\,\,\,\,\,\,\log {a_1}{r^{n + 1}}\\ \log {a_1}{r^{n + 2}}\,\,\,\,\,\,\log {a_1}{r^{n + 3}}\,\,\,\,\,\,\log {a_1}{r^{n + 4}}\\ \log {a_1}{r^{n + 5}}\,\,\,\,\,\,\log {a_1}{r^{n + 6}}\,\,\,\,\,\,\log {a_1}{r^{n + 7}} \end{array} \right|\]
\[ = \left| \begin{array}{l} \log \,{a_1} + \left( {n - 1} \right)\log r\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\log {a_1} + n\log r\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\log {a_1} + \left( {n + 1} \right)\log r\\ \log {a_1} + \left( {n + 2} \right)\log r\,\,\,\,\,\,\,\,\,\,\,\log {a_1} + \left( {n + 3} \right)\log r\,\,\,\,\,\,\,\,\,\,\,\log {a_1} + \left( {n + 4} \right)\log r\\ \log {a_1} + \left( {n + 5} \right)\log r\,\,\,\,\,\,\,\,\,\,\,\log {a_1} + \left( {n + 6} \right)\log r\,\,\,\,\,\,\,\,\,\,\,\log {a_1} + \left( {n + 7} \right)\log r \end{array} \right|\]
$$ = 0\left[ {{\text{Apply }}{c_2} \to {c_2} - \frac{1}{2}{c_1} - \frac{1}{2}{c_3}} \right]$$

We have, \[{A^2} = \left[ {\begin{array}{*{20}{c}} \alpha &0\\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} \alpha &0\\ 1&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{\alpha ^2}}&0\\ {\alpha + 1}&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 9&a\\ b&c \end{array}} \right]\]
⇒ we get, $${\alpha ^2} = 9$$
$$ \Rightarrow \alpha = \pm 3$$   and $$a = 0, c = 1, b = \alpha + 1 = 3 + 1 = 4$$        or $$b = - 3 + 1 = - 2$$
So, $$a + b + c = \left( {0 + 4 + 1} \right) = 5\,\,{\text{or }}\left( {0 - 2 + 1} \right) = - 1$$

\[A = \left[ {\begin{array}{*{20}{c}} {x + y}&y\\ {2x}&{x - y} \end{array}} \right],B = \left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right]{\rm{ and }}\,\,C = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right]\]
Here, $$AB = C$$
\[\begin{array}{l} \therefore \left[ {\begin{array}{*{20}{c}} {x + y}&y\\ {2x}&{x - y} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right]\\ \Rightarrow \left[ {\begin{array}{*{20}{c}} {2\left( {x + y} \right)}&{ - y}\\ {4x}&{ - x + y} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right] \end{array}\]
$$\eqalign{ & 2x + y = 3\,\,\,.....\left( {\text{i}} \right) \cr & 3x + y = 2\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
From equations (i) and (ii), we get
$$x = - 1$$  and $$y = 5$$
\[\begin{array}{l} \therefore A = \left[ {\begin{array}{*{20}{c}} 4&5\\ { - 2}&{ - 6} \end{array}} \right]\\ {\rm{Now, }}\,\,{A^2} = \left[ {\begin{array}{*{20}{c}} 4&5\\ { - 2}&{ - 6} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4&5\\ { - 2}&{ - 6} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {16 - 10}&{20 - 30}\\ { - 8 + 12}&{ - 10 + 36} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 6&{ - 10}\\ 4&{26} \end{array}} \right] \end{array}\]

$$\eqalign{ & {A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right) \cr & {A^2} - {B^2} = {A^2} + AB - BA - {B^2} \cr & \Rightarrow \,\,AB = BA \cr} $$