Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\begin{array}{l} {D^2} = \left| {\begin{array}{*{20}{c}} {{l_1}}&{{m_1}}&{{n_1}}\\ {{l_2}}&{{m_2}}&{{n_2}}\\ {{l_3}}&{{m_3}}&{{n_3}} \end{array}} \right|\,\,\left| {\begin{array}{*{20}{c}} {{l_1}}&{{m_1}}&{{n_1}}\\ {{l_2}}&{{m_2}}&{{n_2}}\\ {{l_3}}&{{m_3}}&{{n_3}} \end{array}} \right|\\ = \,\left| {\begin{array}{*{20}{c}} {{\rm{ }}l_1^2 + m_1^2 + n_1^2}&{{l_2}{l_1} + {m_2}{m_1} + {n_2}{n_1}}&{{l_1}{l_3} + {m_1}{m_3} + {n_1}{n_3}}\\ {{\rm{ }}{l_2}{l_1} + {m_2}{m_1} + {n_2}{n_1}}&{l_2^2 + m_2^2 + n_2^2}&{{l_2}{l_1} + {m_2}{m_3} + {n_2}{n_3}}\\ {{l_1}{l_3} + {m_1}{m_3} + {n_1}{n_3}}&{{\rm{ }}{l_2}{l_3} + {m_2}{m_3} + {n_2}{n_3}}&{l_3^2 + m_3^2 + n_3^2} \end{array}} \right|\\ = \,\left| {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right| = 1 \end{array}\]
$$ \Rightarrow \,D = \pm 1$$

\[\Delta = \left| \begin{array}{l} \,1\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^n }\,\,\,\,\,\,\,\,\,\,{\omega ^{2n}}\\ {\omega ^n}\,\,\,\,\,\,\,\,\,{\omega ^{2n}}\,\,\,\,\,\,\,\,\,1\\ {\omega ^{2n}}\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^n} \end{array} \right|\]
$$\eqalign{ & = 1\left( {{\omega ^{3n}} - 1} \right) - {\omega ^n}\left( {{\omega ^{2n}} - {\omega ^{2n}}} \right) + {\omega ^{2n}}\left( {{\omega ^n} - {\omega ^{4n}}} \right) \cr & = {\omega ^{3n}} - 1 - 0 + {\omega ^{3n}} - {\omega ^{6n}} \cr & = 1 - 1 + 1 - 1 = 0\,\left[ {\because \,\,{\omega ^{3n}} = 1} \right] \cr} $$

Each of the first three options contains $$m = 3.$$
When $$m = 3,$$  the last two equations become $$x + 2y + 3z = 10$$    and $$x + 2y + 3z = n.$$
Obviously, when $$n = 10$$  these equations become the same. So, we are left with only two independent equations to find the values of the three unknowns.
Consequently, there will be infinite solutions.

$${R_1} \to {R_1} - {R_2},{R_2} \to {R_2} - {R_3}$$      (reduces the determinant to :)
\[\left| {\begin{array}{*{20}{c}} {{x^2} - 2x + 1}&{x - 1}&0\\ {2x - 2}&{x - 1}&0\\ 3&3&1 \end{array}} \right|\]
$$ = {\left( {x - 1} \right)^3} - 2{\left( {x - 1} \right)^2} = {\left( {x - 1} \right)^2}\left( {x - 1 - 2} \right)$$
$$ = {\left( {x - 1} \right)^2}\left( {x - 3} \right),$$     which is clearly negative for $$x < 1$$

$$A$$ is invertible if \[\left| A \right| \ne 0,\,{\text{i}}{\text{.e}}{\text{.,}}\left| {\begin{array}{*{20}{c}} 0&{ - 4}&1 \\ 2&\lambda &{ - 3} \\ 1&2&{ - 1} \end{array}} \right| \ne 0.\]

$${\text{Let }}\frac{{{x^2}}}{{{a^2}}} = X,\frac{{{y^2}}}{{{b^2}}} = Y,\frac{{{z^2}}}{{{c^2}}} = Z$$
Then the given system of equations becomes
$$X + Y - Z = 1,X - Y + Z = 1, - X + Y + Z = 1$$
This is the new system of equations
For new system, we have
\[D = \left| \begin{array}{l} \,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\, - 1\\ \,\,\,\,\,1\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,1\\ - 1\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right|\]
$$\eqalign{ & = 1\left( { - 1 - 1} \right) - 1\left( {1 + 1} \right) - 1\left( {1 - 1} \right) \cr & = - 4 \ne 0 \cr} $$
∴ New system of equations has unique solution.
\[{D_1} = \left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\, - 1\\ 1\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right|\]
$$ = 1\left( { - 1 - 1} \right) - 1\left( {1 - 1} \right) - 1\left( {1 + 1} \right) = - 4$$
\[{D_2} = \left| \begin{array}{l} \,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\, - 1\\ \,\,\,\,\,1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ - 1\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right|\]
$$ = 1\left( {1 - 1} \right) - 1\left( {1 + 1} \right) - 1\left( {1 + 1} \right) = - 4$$
\[{D_3} = \left| \begin{array}{l} \,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1\\ \,\,\,\,\,1\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,1\\ - 1\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right|\]
$$ = 1\left( { - 1 - 1} \right) - 1\left( {1 + 1} \right) + 1\left( {1 - 1} \right) = - 4$$
$$\eqalign{ & {\text{Now, }}X = \frac{{{D_1}}}{D} = \frac{{ - 4}}{{ - 4}} = 1,Y = \frac{{{D_2}}}{D} = \frac{{ - 4}}{{ - 4}} = 1,Z = \frac{{{D_3}}}{D} = \frac{{ - 4}}{{ - 4}} = 1 \cr & \Rightarrow \,\,x = \pm \,a,\,\,y = \pm \,b,\,\,z = \pm \,c \cr} $$

We know that $$\det \left( A \right) = \frac{1}{{\det \left( {{A^{ - 1}}} \right)}}$$
\[\begin{array}{l} {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ { - 2}&2 \end{array}} \right]\\ \Rightarrow \left| {{A^{ - 1}}} \right| = \left| {\begin{array}{*{20}{c}} 1&{ - 2}\\ { - 2}&2 \end{array}} \right| = 2 - 4 = - 2 \end{array}\]
$$ \Rightarrow det\left( A \right) = - \frac{1}{2}$$

\[\vartriangle = \left| {\begin{array}{*{20}{c}} x&{{x^2}}&1 \\ y&{{y^2}}&1 \\ z&{{z^2}}&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} x&{ - yz}&1 \\ y&{ - zx}&1 \\ z&{ - xy}&1 \end{array}} \right|\]
\[\vartriangle = \left| {\begin{array}{*{20}{c}} x&{{x^2}}&1 \\ y&{{y^2}}&1 \\ z&{{z^2}}&1 \end{array}} \right| - \frac{1}{{xyz}}\left| {\begin{array}{*{20}{c}} {{x^2}}&{xyz}&x \\ {{y^2}}&{zxy}&y \\ {{z^2}}&{xyz}&z \end{array}} \right|,\,\,{\text{etc}}{\text{.}}\]

$$\eqalign{ & {\text{Given,}}\,{A^2} = 2A - I \cr & {\text{Now,}}\,{A^3} = A\left( {{A^2}} \right) = A\left( {2A = I} \right) \cr & = \,2{A^2} - A = 2\left( {2A - I} \right) - A = 3A - 2I \cr & {A^4} = A\left( {{A^3}} \right) = A\left( {3A - 2I} \right) \cr & = \,3{A^2} - 2A = 3\left( {2A - I} \right) - 2A = 4A - 3I \cr} $$
Following this, we can say $${A^n} = nA - \left( {n - 1} \right)I$$

$$\eqalign{ & {\text{Given, }}x = {a_3}y + {a_2}z\,\,\,\,.....\left( {\text{i}} \right) \cr & y = {a_1}z + {a_3}x\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & z = {a_2}x + {a_1}y\,\,\,\,\,.....\left( {{\text{iii}}} \right) \cr} $$
Since, $$x, y, z$$  are not all zero, therefore given system of equations has non-trivial solution.
\[\therefore \left| {\begin{array}{*{20}{c}} 1&{ - {a_3}}&{ - {a_2}}\\ {{a_3}}&{ - 1}&{{a_1}}\\ {{a_2}}&{{a_1}}&{ - 1} \end{array}} \right| = 0\]
$$ \Rightarrow {a_1}^2 + {a_2}^2 + {a_3}^2 + 2{a_1}{a_2}{a_3} = 1\,\,\,\,\,.....\left( {{\text{iv}}} \right)$$
Since, $${a_1} = m - \left[ m \right]$$   and $$m$$ is not an integer.
$$\eqalign{ & \therefore 0 < {a_1} < 1 \cr & \Rightarrow 0 < 1 - {a_1}^2 < 1\,\,\,\,.....\left( {\text{v}} \right) \cr & {\text{From Eq}}{\text{.}}\left( {{\text{iv}}} \right),1 - {a_2}^2 - {a_3}^2 = {a_1}^2 + 2{a_1}{a_2}{a_3} \cr & \Rightarrow 1 - {a_2}^2 - {a_3}^2 + {a_2}^2{a_3}^2 = {a_1}^2 + 2{a_1}{a_2}{a_3} + {a_2}^2{a_3}^2 \cr & \Rightarrow \left( {1 - {a_2}^2} \right)\left( {1 - {a_3}^2} \right) = {\left( {{a_1} + {a_2}{a_3}} \right)^2}\,\,\,\,\,.....\left( {{\text{vi}}} \right) \cr & {\text{Similarly, }}\left( {1 - {a_1}^2} \right)\left( {1 - {a_3}^2} \right) = {\left( {{a_2} + {a_1}{a_3}} \right)^2}\,\,\,\,.....\left( {{\text{vii}}} \right) \cr & \left( {1 - {a_1}^2} \right)\left( {1 - {a_2}^2} \right) = {\left( {{a_3} + {a_1}{a_2}} \right)^2}\,\,\,\,.....\left( {{\text{viii}}} \right) \cr & {\text{From Eq}}{\text{.}}\left( {{\text{vii}}} \right),1 - {a_2}^2 = \,\, > 0 \cdot \frac{{{{\left( {{a_3} + {a_1}{a_2}} \right)}^2}}}{{1 - {a_1}^2}} \cr & {\text{From Eq}}{\text{.}}\left( {{\text{viii}}} \right),1 - {a_3}^2 > 0 \cr & \Rightarrow 3 - \left( {{a_1}^2 + {a_2}^2 + {a_3}^2} \right) > 0 \cr & \Rightarrow {a_1}^2 + {a_2}^2 + {a_3}^2 < 3 \cr & \Rightarrow 1 - 2{a_1}{a_2}{a_3} < 3\left[ {{\text{From Eq}}{\text{.}}\left( {{\text{iv}}} \right)} \right] \cr & \Rightarrow {a_1}{a_2}{a_3} > - 1. \cr} $$