Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\begin{array}{l} {\rm{Let,}}\,\,\,\left[ {\begin{array}{*{20}{c}} 1&2\\ 0&3 \end{array}} \right] = B\\ {\rm{Then,}}\,\,BA = \left[ {\begin{array}{*{20}{c}} { - 1}&0\\ 6&3 \end{array}} \right]\\ \Rightarrow \,A = {B^{ - 1}}\left[ {\begin{array}{*{20}{c}} { - 1}&0\\ { - 6}&3 \end{array}} \right]\left| B \right| = 3,\\ adj\,B = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}\\ 0&1 \end{array}} \right]\\ {B^{ - 1}} = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} 3&{ - 2}\\ 0&1 \end{array}} \right]\\ \Rightarrow \,A = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} 3&{ - 2}\\ 0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&0\\ 6&3 \end{array}} \right] = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} { - 3 - 12}&{ - 6}\\ 6&3 \end{array}} \right]\\ = \,\left[ {\begin{array}{*{20}{c}} { - 5}&{ - 2}\\ 2&1 \end{array}} \right] \end{array}\]

Given, \[A = \left[ {\begin{array}{*{20}{c}} 1&2\\ 3&5 \end{array}} \right]\]
$$\eqalign{ & \Rightarrow \left| A \right| = 5 - 6 = - 1 \cr & \therefore \left| {{A^{2009}} - 5{A^{2008}}} \right| = \left| {{A^{2008}}} \right|\left| {A - 5I} \right| \cr} $$
\[ = {\left( { - 1} \right)^{2008}}\left| {\left[ {\begin{array}{*{20}{c}} 1&2\\ 3&5 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 5&0\\ 0&5 \end{array}} \right]} \right| = \left| {\begin{array}{*{20}{c}} { - 4}&2\\ 3&0 \end{array}} \right| = - 6\]

\[f\left( x \right) = \left| {\begin{array}{*{20}{c}} {1 + {{\sin }^2}x}&{{{\cos }^2}x}&{4\sin 2x}\\ {{{\sin }^2}x}&{1 + {{\cos }^2}x}&{4\sin 2x}\\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + 4\sin 2x} \end{array}} \right|\]
$${\text{Applying }}{C_1} \to {C_1} + {C_2}$$
\[ = \left| {\begin{array}{*{20}{c}} 2&{{{\cos }^2}\theta }&{4\sin 2x}\\ 2&{1 + {{\cos }^2}\theta }&{4\sin 2x}\\ 1&{{{\cos }^2}\theta }&{1 + 4\sin 2x} \end{array}} \right|\]
$$\left( {{\text{Applying }}{R_2} \to {R_2} - {R_1}{\text{ and }}{R_3} \to {R_3} - {R_1}} \right)$$
\[ = \left| {\begin{array}{*{20}{c}} 2&{{{\cos }^2}\theta }&{4\sin 2x}\\ 0&1&0\\ { - 1}&0&1 \end{array}} \right|\]
$$f\left( x \right) = 2 + 4\sin 2x$$
$$\therefore - 1 \leqslant \sin 2x \leqslant 1,$$    maximum value of $$\sin 2x = 1$$
Thus, maximum value of $$f\left( x \right) = 2 + 4 = 6$$

\[f\left( x \right) = \left| {\begin{array}{*{20}{c}} {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x}&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x}\\ {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x}&{1 + {b^2}x}&{\left( {1 + {c^2}} \right)x}\\ {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x}&{\left( {1 + {b^2}} \right)x}&{1 + {c^2}x} \end{array}} \right|\]
$${\text{Applying, }}{C_1} \to {C_1} + {C_2} + {C_3}$$
\[ = \left| {\begin{array}{*{20}{c}} 1&{\left( {1 + {b^2}} \right)x}&{\left( {1 + {c^2}} \right)x}\\ 1&{1 + {b^2}x}&{\left( {1 + {c^2}} \right)x}\\ 1&{\left( {1 + {b^2}} \right)x}&{1 + {c^2}x} \end{array}} \right|\]
$$\therefore {a^2} + {b^2} + {c^2} + 2 \ne 0$$
\[f\left( x \right) = \left| {\begin{array}{*{20}{c}} 0&{x - 1}&0\\ 0&{1 - x}&{x - 1}\\ 1&{\left( {1 + {b^2}} \right)x}&{1 + {c^2}x} \end{array}} \right|\]
$$\eqalign{ & {\text{Applying }}{R_1} \to {R_1} - {R_2},{R_2} \to {R_2} - {R_3} \cr & f\left( x \right) = {\left( {x - 1} \right)^2}\,\,\,\,{\text{Hence degree = 2}}{\text{.}} \cr} $$

\[\begin{array}{l} {\rm{Let, }}A = \left[ {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right]\\ {A^2} = \left[ {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right] = 2A\\ {A^3} = {2^2}\left[ {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right],{A^4} = {2^3}\left[ {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right] \end{array}\]
$$\eqalign{ & {A^3} = {2^2}A, \cr & {A^4} = {2^3}A \cr} $$
\[\therefore {A^n} = {2^{n - 1}}\left[ {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right]\]
$$\eqalign{ & \Rightarrow {A^{100}} = {2^{100 - 1}}A \cr & \therefore {A^{100}} = {2^{99}}A \cr} $$

\[\Delta = 0\,\, \Rightarrow \,\,\left| \begin{array}{l} 4\,\,\,\,\,\,k\,\,\,\,\,\,2\\ k\,\,\,\,\,\,4\,\,\,\,\,\,1\\ 2\,\,\,\,\,\,2\,\,\,\,\,\,1 \end{array} \right| = 0\]
$$\eqalign{ & \Rightarrow \,\,4\left( {4 - 2} \right) - k\left( {k - 2} \right) + 2\left( {2k - 8} \right) = 0 \cr & \Rightarrow \,\,8 - {k^2} + 2k + 4k - 16 = 0 \cr & \Rightarrow \,\,{k^2} - 6k + 8 = 0 \cr & \Rightarrow \,\,\left( {k - 4} \right)\left( {k - 2} \right) = 0,\,\,k = 4,2 \cr} $$

We have, $${a_{ij}} = \left( {j + i} \right)\left( {j - i} \right) = - \left( {i + j} \right)\left( {i - j} \right) = - {a_{ij}}$$
Hence, matrix is skew-symmetric.

Consider
\[\left| \begin{array}{l} \,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + f\left( 1 \right)\,\,\,\,\,\,\,\,\,\,1 + f\left( 2 \right)\\ 1 + f\left( 1 \right)\,\,\,\,\,\,\,\,\,1 + f\left( 2 \right)\,\,\,\,\,\,\,\,\,\,1 + f\left( 3 \right)\\ 1 + f\left( 2 \right)\,\,\,\,\,\,\,\,1 + f\left( 3 \right)\,\,\,\,\,\,\,\,\,\,1 + f\left( 4 \right) \end{array} \right|\]
\[ = \left| \begin{array}{l} \,\,\,1 + 1 + 1\,\,\,\,\,\,\,\,\,\,\,\,\,1 + \alpha + \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + {\alpha ^2} + {\beta ^2}\\ \,\,1 + \alpha + \beta \,\,\,\,\,\,\,\,\,\,\,\,1 + {\alpha ^2} + {\beta ^2}\,\,\,\,\,\,\,\,\,1 + {\alpha ^3} + {\beta ^3}\\ 1 + {\alpha ^2} + {\beta ^2}\,\,\,\,\,\,1 + {\alpha ^3} + {\beta ^3}\,\,\,\,\,\,\,\,\,\,1 + {\alpha ^4} + {\beta ^4} \end{array} \right|\]
\[ = \left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\beta \\ 1\,\,\,\,\,\,\,{\alpha ^2}\,\,\,\,\,\,\,{\beta ^2} \end{array} \right| \times \,\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\beta \\ 1\,\,\,\,\,\,\,{\alpha ^2}\,\,\,\,\,\,\,{\beta ^2} \end{array} \right|\]
\[ = {\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\beta \\ 1\,\,\,\,\,\,\,{\alpha ^2}\,\,\,\,\,\,\,{\beta ^2} \end{array} \right|^2}\]
$$\eqalign{ & = {\left[ {\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {\alpha - \beta } \right)} \right]^2} \cr & {\text{So, }}\boxed{k = 1} \cr} $$

Given matrix is :
\[\left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2x}&{ - 2{\omega ^2}}\\ 2&\omega &{ - \omega }\\ 0&\omega &1 \end{array}} \right| = 0\]
By $${C_2} \to {C_2} + {C_3},$$   we get
\[ \Rightarrow \left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2x - 2{\omega ^2}}&{ - 2{\omega ^2}}\\ 2&0&{ - \omega }\\ 0&{1 + \omega }&1 \end{array}} \right| = 0\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2x - 2{\omega ^2}}&{ - 2{\omega ^2}} \\ 2&0&{ - \omega } \\ 0&{ - {\omega ^2}}&1 \end{array}} \right| = 0\left[ {\because 1 + \omega = - {\omega ^2}} \right]\]
\[ \Rightarrow {\omega ^2}\left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2{\omega ^2}}\\ 2&{ - \omega } \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2x - 2{\omega ^2}}\\ 2&{ - 0} \end{array}} \right| = 0\]
$$\eqalign{ & \Rightarrow {\omega ^2}\left( { - \omega {x^2} + 4{\omega ^2}} \right) - \left( { - 4x - 4{\omega ^2}} \right) = 0 \cr & \Rightarrow - {x^2} + 4\omega + 4x + 4{\omega ^2} = 0 \cr & \Rightarrow - {x^2} + 4\omega - 4x - 4 - 4\omega = 0 \cr & \Rightarrow - {x^2} - 4x - 4 = 0 \cr & \Rightarrow {\left( {x + 2} \right)^2} = 0 \cr & \Rightarrow x = - 2. \cr} $$

Given,
$$\eqalign{ & \left( {{a^2} + {b^2} + {c^2}} \right){x^2} - 2\left( {ab + bc + cd} \right)x + {b^2} + {c^2} + {d^2} \leqslant 0 \cr & \Rightarrow {\left( {ax - b} \right)^2} + {\left( {bx - c} \right)^2} + {\left( {cx - d} \right)^2} \leqslant 0 \cr & \Rightarrow {\left( {ax - b} \right)^2} + {\left( {bx - c} \right)^2} + {\left( {cx - d} \right)^2} = 0 \cr & \Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c} = x \cr & \Rightarrow {b^2} = ac{\text{ or }}2\log b = \log a + \log c \cr} $$
Now, \[\Delta = \left| {\begin{array}{*{20}{c}} {33}&{14}&{\log a}\\ {65}&{27}&{\log b}\\ {97}&{40}&{\log c} \end{array}} \right|\]
Apply $${R_1} \to {R_1} + {R_3};$$
\[\Delta = \left| {\begin{array}{*{20}{c}} {130}&{54}&{\log a + \log c}\\ {65}&{27}&{\log b}\\ {97}&{40}&{\log c} \end{array}} \right| = 0\]
Now, $${R_1} \to {R_1} - 2{R_2};$$
\[\Delta = \left| {\begin{array}{*{20}{c}} 0&0&0\\ {65}&{27}&{\log b}\\ {97}&{40}&{\log c} \end{array}} \right| = 0\]