Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

Let $$B = I + A + {A^2} + ..... + {A^{k - 1}}$$
Now multiply both sides by $$\left( {I - A} \right),$$  we get
$$\eqalign{ & B\left( {I - A} \right) = \left( {I + A + {A^2} + ..... + {A^{k - 1}}} \right)\left( {I - A} \right) \cr & = I - A + A - {A^2} + {A^2} - {A^3} + ..... - {A^{k - 1}} + {A^{k - 1}} - {A^k} \cr & = I - {A^k} = I,\,\,{\text{since, }}{A^k} = 0 \cr & \Rightarrow B = {\left( {I - A} \right)^{ - 1}} \cr & {\text{Hence, }}{\left( {I - A} \right)^{ - 1}} = I + A + {A^2} + ..... + {A^{k - 1}} \cr & {\text{Thus, }}p = - 1. \cr} $$

\[\begin{array}{l} {\rm{Now,\, }}{P^T}P = \left[ \begin{array}{l} \frac{{\sqrt 3 }}{2}\,\,\,\,\, - \frac{1}{2}\\ \frac{1}{2}\,\,\,\,\,\,\,\,\frac{{\sqrt 3 }}{2} \end{array} \right]\left[ \begin{array}{l} - \frac{{\sqrt 3 }}{2}\,\,\,\,\,\frac{1}{2}\\ - \frac{1}{2}\,\,\,\,\,\,\,\,\frac{{\sqrt 3 }}{2} \end{array} \right]\\ \Rightarrow {P^T}P = \left[ \begin{array}{l} 1\,\,\,\,\,1\\ 0\,\,\,\,\,1 \end{array} \right]\\ \Rightarrow {P^T}P = I\\ \Rightarrow {P^T} = {P^{ - I}}\\ {\rm{Since,\, }}Q = PA{P^T}\\ \therefore \,{P^T}{Q^{2005}}P = {P^T}\left[ {\left( {PA{P^T}} \right)\left( {PA{P^T}} \right)......\,2005{\rm{ \,times}}} \right]P\\ = \frac{{\left( {PA{P^T}} \right)A\left( {PA{P^T}} \right)A\left( {PA{P^T}} \right)......\left( {PA{P^T}} \right)A\left( {PA{P^T}} \right)}}{{2005{\rm{ \,times}}}}\\ = I{A^{2005}} = {A^{2005}}\\ \therefore \,{A^1} = \left[ \begin{array}{l} 1\,\,\,\,\,1\\ 0\,\,\,\,\,1 \end{array} \right]\\ {A^2} = \left[ \begin{array}{l} 1\,\,\,\,\,1\\ 0\,\,\,\,\,1 \end{array} \right]\left[ \begin{array}{l} 1\,\,\,\,\,1\\ 0\,\,\,\,\,1 \end{array} \right] = \left[ \begin{array}{l} 1\,\,\,\,\,2\\ 0\,\,\,\,\,1 \end{array} \right]\\ {A^3} = \left[ \begin{array}{l} 1\,\,\,\,\,2\\ 0\,\,\,\,\,1 \end{array} \right]\left[ \begin{array}{l} 1\,\,\,\,\,1\\ 0\,\,\,\,\,1 \end{array} \right] = \left[ \begin{array}{l} 1\,\,\,\,\,3\\ 0\,\,\,\,\,1 \end{array} \right]\\ {A^{2005}} = \left[ \begin{array}{l} 1\,\,\,\,\,2005\\ 0\,\,\,\,\,\,\,\,1 \end{array} \right]\\ \therefore \,{P^T}{Q^{2005}}P = \left[ \begin{array}{l} 1\,\,\,\,\,2005\\ 0\,\,\,\,\,\,\,\,1 \end{array} \right]\\ {\rm{Therefore\, option\, A\, is\, the\, right\, answer}}{\rm{.}} \end{array}\]

Note that, $$P' = {P^{ - 1}}$$
$$\eqalign{ & {\text{Now, }}Q = PAP' = PA{P^{ - 1}} \cr & \Rightarrow {Q^{2007}} = P{A^{2007}}{P^{ - 1}} \cr & \therefore P'{Q^{2007}}P = {P^{ - 1}}\left( {P{A^{2007}}{P^{ - 1}}} \right)P = {A^{2007}} = {\left( {I + B} \right)^{2007}} \cr} $$
where, \[B = \left[ {\begin{array}{*{20}{c}} 0&1\\ 0&0 \end{array}} \right].\]
As, $$B^2 = 0,$$  we get $${B^r} = 0\forall r \geqslant 2.$$
Thus, by binomial theorem,
\[{A^{2007}} = I + 2007B = \left[ {\begin{array}{*{20}{c}} 1&{2007}\\ 0&1 \end{array}} \right]\]

\[\vartriangle = \frac{1}{{\log x}} \cdot \frac{1}{{\log y}} \cdot \frac{1}{{\log z}}\left| {\begin{array}{*{20}{c}} {\log x}&{\log y}&{\log z} \\ {\log x}&{\log y}&{\log z} \\ {\log x}&{\log y}&{\log z} \end{array}} \right| = 0.\]

If $$A$$ and $$B$$ are square matrices of same degree then matrices $$A$$ and $$B$$ can be added or subtracted or multiplied. By algebra of matrices the only correct option is $$A + B = B + A$$

Since $$C_1$$ has variable terms and $$C_2$$ and $$C_3$$ are constant, summation runs on $$C_1 .$$ Therefore,
\[\begin{array}{l} \sum\limits_{r = 1}^n {{\Delta _r}} = \left| {\begin{array}{*{20}{c}} {\sum\limits_1^n {\left( {r - 1} \right)} }&n&6\\ {\sum\limits_1^n {{{\left( {r - 1} \right)}^2}} }&{2{n^2}}&{4n - 2}\\ {\sum\limits_1^n {{{\left( {r - 1} \right)}^3}} }&{3{n^3}}&{3{n^2} - 3n} \end{array}} \right|\\ = \,\left| {\begin{array}{*{20}{c}} {\frac{1}{2}\left( {n - 1} \right)n}&n&6\\ {\frac{1}{6}\left( {n - 1} \right)n\left( {2n - 1} \right)}&{2{n^2}}&{4n - 2}\\ {\frac{1}{4}{{\left( {n - 1} \right)}^2}{n^2}}&{3{n^3}}&{3{n^2} - 3n} \end{array}} \right| \end{array}\]
Taking $$\frac{1}{{12}}n\left( {n - 1} \right)$$   common from $$C_1$$ and $$n$$ common from $$C_2 ,$$ we get
\[\sum {{\Delta _r} = \frac{1}{{12}}{n^2}\left( {n - 1} \right) \times \left| {\begin{array}{*{20}{c}} 6&1&6\\ {2\left( {2n - 1} \right)}&{2n}&{2\left( {2n - 1} \right)}\\ {3n\left( {n - 1} \right)}&{3{n^3}}&{3n\left( {n - 1} \right)} \end{array}} \right|} \]
$$ = \,0\,\,\,\left[ {\because \,{C_1}\,{\text{and}}\,{C_3}\,{\text{are}}\,{\text{identical}}} \right]$$

The determinent of a orthogonal matrix is always $$ \pm 1$$
$$\left| A \right| = \pm 1$$
\[B = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ { - 3}&0&2\\ 2&5&0 \end{array}} \right]\]
$$\eqalign{ & \left| B \right| = - 10 - 2\left( { - 4} \right) + 3\left( { - 15} \right) \cr & = - 47 \cr & \left| {AB} \right| = \left| A \right|\left| B \right| \cr & = \left( { \pm 1} \right)\left( { - 47} \right) \cr & = \pm 47 \cr} $$
Taking $$A$$ as identity matrix we can prove $$AB = BA$$

\[A = \left[ {\begin{array}{*{20}{c}} 1&3&2\\ 1&{x - 1}&1\\ 2&7&{x - 3} \end{array}} \right]\]
$$\eqalign{ & \left| A \right| = 1\left[ {\left( {x - 1} \right)\left( {x - 3} \right) - 7} \right] - 3\left[ {\left( {x - 3} \right) - 2} \right] + 2\left[ {7 - 2\left( {x - 1} \right)} \right] \cr & = {x^2} - 11x + 29 \cr} $$
If inverse will not exist then $$\left| A \right| = 0$$
$$\eqalign{ & {x^2} - 11x + 29 = 0 \cr & x = \frac{{11 \pm \sqrt 5 }}{2} \cr} $$

Consider,
\[{A^2} = \left[ {\begin{array}{*{20}{c}} { - 5}&{ - 8}&0\\ 3&5&0\\ 1&2&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 5}&{ - 8}&0\\ 3&5&0\\ 1&2&{ - 1} \end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] = I\,\,{\rm{ So, }}\,\,{A^3} = \left[ {\begin{array}{*{20}{c}} { - 5}&{ - 8}&0\\ 3&5&0\\ 1&2&{ - 1} \end{array}} \right]\]         and so on $$tr\left( A \right) + tr\left( {{A^2}} \right) + tr\left( {{A^3}} \right) + ..... + tr\left( {{A^{100}}} \right)$$
$$ = \left( { - 1} \right) + \left( 3 \right) + \left( { - 1} \right) + \left( 3 \right) + ..... + \left( { - 1} \right) + \left( 3 \right) = 200$$

\[\begin{array}{l} AB = \left[ {\begin{array}{*{20}{c}} 1&2\\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} a&0\\ 0&b \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a&{2b}\\ {3a}&{4b} \end{array}} \right]\\ {\rm{and, }}\,\,BA = \left[ {\begin{array}{*{20}{c}} a&0\\ 0&b \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&2\\ 3&4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a&{2a}\\ {3b}&{4b} \end{array}} \right]\\ {\rm{If, }}\,\,AB = BA\\ \Rightarrow \left[ {\begin{array}{*{20}{c}} a&{2b}\\ {3a}&{4b} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a&{2a}\\ {3b}&{4b} \end{array}} \right]\\ \Rightarrow a = b \end{array}\]
From the above it is clear that there exist infinitely many $$B's$$  such that $$AB = BA.$$