Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

Here, \[\left| {\begin{array}{*{20}{c}} {x - 5}&2&5 \\ 0&x&3 \\ {5 - x}&4&x \end{array}} \right| = 0\,\,\,{\text{or, }}\left( {x - 5} \right)\left| {\begin{array}{*{20}{c}} 1&2&5 \\ 0&x&3 \\ { - 1}&4&x \end{array}} \right| = 0\]
or, \[\left( {x - 5} \right)\left| {\begin{array}{*{20}{c}} 0&6&{5 + x} \\ 0&x&3 \\ { - 1}&4&x \end{array}} \right| = 0\,\,\,{\text{or, }}\left( {x - 5} \right)\left\{ {18 - x\left( {5 + x} \right)} \right\} = 0\]
∴ the non-integral roots are the roots of $${x^2} + 5x - 18 = 0.$$

Given that, $$X = AB + BA$$
$$\eqalign{ & \Rightarrow X = {X^T}{\text{ and }}Y = AB - BA \cr & \Rightarrow Y = - {Y^T}. \cr & {\text{Now, }}{\left( {XY} \right)^T} = {Y^T}{X^T} = - YX \cr} $$

\[\left| {\begin{array}{*{20}{c}} x&{\sin \theta }&{\cos \theta }\\ { - \sin \theta }&{ - x}&1\\ {\cos \theta }&1&x \end{array}} \right|\]
$$\eqalign{ & = x\left( {{x^2} - 1} \right) - \sin \theta \left( { - x\sin \theta - \cos \theta } \right) + \cos \theta \left( { - \sin \theta + x\cos \theta } \right) \cr & = \left. { - {x^3} - x + x{{\sin }^2}\theta + \sin \theta \cos \theta - \cos \theta \sin \theta + x{{\cos }^2}\theta } \right) \cr & = {x^3} - x + x = {x^3} \cr} $$

Given matrix is :
\[\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} 2&2\\ 2&2 \end{array}} \right]\\ {A^2} = \left[ {\begin{array}{*{20}{c}} 2&2\\ 2&2 \end{array}} \right]\,\left[ {\begin{array}{*{20}{c}} 2&2\\ 2&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {4 + 4}&{4 + 4}\\ {4 + 4}&{4 + 4} \end{array}} \right]\\ = \,\left[ {\begin{array}{*{20}{c}} {{2^3}}&{{2^3}}\\ {{2^3}}&{{2^3}} \end{array}} \right]\\ {A^3} = \left[ {\begin{array}{*{20}{c}} 8&8\\ 8&8 \end{array}} \right]\,\left[ {\begin{array}{*{20}{c}} 2&2\\ 2&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {16 + 16}&{16 + 16}\\ {16 + 16}&{16 + 16} \end{array}} \right]\\ = \,\left[ {\begin{array}{*{20}{c}} {32}&{32}\\ {32}&{32} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{2^5}}&{{2^5}}\\ {{2^5}}&{{2^5}} \end{array}} \right] \end{array}\]
Going this way, we get,
\[\begin{array}{l} {A^4} = \left[ {\begin{array}{*{20}{c}} {{2^7}}&{{2^7}}\\ {{2^7}}&{{2^7}} \end{array}} \right]\\ \Rightarrow \,{A^n} = \left[ {\begin{array}{*{20}{c}} {{2^{2n - 1}}}&{{2^{2n - 1}}}\\ {{2^{2n - 1}}}&{{2^{2n - 1}}} \end{array}} \right] \end{array}\]

$$\eqalign{ & tr\left( A \right) = \sum\limits_{i = j} {{a_{ij}}} = \left( {{a_{11}} + {a_{22}} + {a_{33}} + ..... + {a_{10}}} \right) \cr & = \left( {{w^2} + {w^4} + {w^6} + ..... + {w^{20}}} \right) = {w^2}\left( {1 + {w^2} + {w^4} + ..... + {w^{18}}} \right) \cr & = {w^2}\left[ {\left( {1 + w + {w^2}} \right) + ..... + \left( {1 + w + {w^2}} \right) + 1} \right] = {w^2} \times 1 = {w^2} \cr} $$

$$\eqalign{ & {\text{Given 2}}\omega {\text{ + 1}} = z; \cr & z = \sqrt 3 i \cr & \Rightarrow \,\,\omega = \frac{{\sqrt 3 i - 1}}{2} \cr} $$
⇒ $$\omega $$ is complex cube root of unity
Applying $${R_1} \to {R_1} + {R_2} + {R_3}$$
\[ = \left| {\begin{array}{*{20}{l}} {3{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,{\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,0}\\ {1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} - {\omega ^2} - 1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,{\omega ^2}}\\ {1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,{\mkern 1mu} \,\,\,\,{\omega ^2}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} {\mkern 1mu} {\omega }} \end{array}} \right|\]
$$\eqalign{ & = 3\left( { - 1 - \omega - \omega } \right) \cr & = - 3\left( {1 + 2\omega } \right) \cr & = - 3z \cr & \Rightarrow \,\,k = - z \cr} $$

Let, \[D = \left| {\begin{array}{*{20}{c}} {a + b + c}&{a + b}&a\\ {4a + 3b + 2c}&{3a + 2b}&{2a}\\ {10a + 6b + 3c}&{6a + 3b}&{3a} \end{array}} \right|\]
\[ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {a + b + c}&{a + b}&a\\ {4a + 3b + 2c}&{3a + 2b}&{2a}\\ {10a + 6b + 3c}&{6a + 3b}&{3a} \end{array}} \right|\]
By $${R_2} \to {R_2} - 2{R_1}$$    and $${R_3} \to {R_3} - 3{R_1},$$    we get:
\[ \Rightarrow \left| {\begin{array}{*{20}{c}} {a + b + c}&{a + b}&a\\ {2a + b}&a&0\\ {7a + 3b}&{3a}&0 \end{array}} \right|\]
By $${C_1} \to {C_1} - {C_2}$$    gives:
\[ \Rightarrow \left| {\begin{array}{*{20}{c}} c&{a + b}&a\\ {a + b}&a&0\\ {4a + 3b}&{3a}&0 \end{array}} \right|\]
Again by, $${R_3} \to {R_3} - 3{R_1},$$    we get:
\[D = \left| {\begin{array}{*{20}{c}} {a + b + c}&{a + b}&a\\ {a + b}&a&0\\ a&0&0 \end{array}} \right|\]
$$ = a\left\{ {0.\left( {a + b} \right) - a.a} \right\}$$
$$ = - {a^3}$$ which is independent of $$b$$ and $$c.$$

For the given matrix to be non - singular
\[\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,b\\ \omega \,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,c\\ {\omega ^2}\,\,\,\,\omega \,\,\,\,\,\,\,\,1 \end{array} \right| \ne 0\]
$$\eqalign{ & \Rightarrow \,\,1 - \left( {a + c} \right)\omega + a\,c\,{\omega ^2} \ne 0 \cr & \Rightarrow \,\,\left( {1 - a\,\omega } \right)\left( {1 - c\,\omega } \right) \ne 0 \cr} $$
$$ \Rightarrow \,\,a \ne {\omega ^2}\,\,{\text{and }}c \ne {\omega ^2}$$     where $$\omega $$ is complex cube root of unity.
As $$a, b$$  and $$c$$ are complex cube roots of unity
∴ $$a$$ and $$c$$ can take only one value i.e. $$\omega $$  while $$b$$ can take 2 values i.e. $$\omega $$  and $${\omega ^2}.$$
∴ Total number of distinct matrices $$ = 1 \times 1 \times 2 = 2$$

$$\eqalign{ & {\text{As, }}{A^2} = 0,{A^k} = 0\,\,\forall \,\,k \geqslant 2. \cr & {\text{Thus, }}{\left( {A + I} \right)^{50}} = I + 50A \cr & \Rightarrow {\left( {A + I} \right)^{50}} - 50A = I \cr & \therefore a = 1,b = 0,c = 0,d = 1 \cr & abc + abd + bcd + acd = 0 \cr} $$

We have, $$I = AA^{-1}$$
\[\begin{array}{l} = \,\frac{1}{2}\left[ {\begin{array}{*{20}{c}} 0&1&2\\ 1&2&3\\ 3&a&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1\\ { - 8}&6&{2\,c}\\ 5&{ - 3}&1 \end{array}} \right]\\ = \,\left[ {\begin{array}{*{20}{c}} 1&0&{c + 1}\\ 0&1&{2\left( {c + 1} \right)}\\ {4\left( {1 - a} \right)}&{3\left( {a - 1} \right)}&{2 + a\,c} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] \end{array}\]
Comparing the elements we get $$c + 1 = 0$$
⇒ $$c = - 1$$  and $$a - 1 = 0$$
⇒ $$a = 1$$