Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\left[ \begin{array}{l} \,1\,\,\,\,\,\,\,...\,\,\,\,\,\,...\\ ...\,\,\,\,\,\,\,1\,\,\,\,\,\,\,...\\ ...\,\,\,\,\,\,...\,\,\,\,\,\,\,1 \end{array} \right]\]   are 6 non-singular matrices because 6 blanks will be filled by 5 zeros and 1 one.
Similarly, \[\left[ \begin{array}{l} ...\,\,\,\,\,\,\,...\,\,\,\,\,\,\,1\\ ...\,\,\,\,\,\,\,1\,\,\,\,\,\,\,...\\ \,1\,\,\,\,\,\,...\,\,\,\,\,\,... \end{array} \right]\]   are 6 non-singular matrices.
So, required cases are more than 7, non-singular $$3 \times 3$$  matrices.

\[\vartriangle = abc\left| {\begin{array}{*{20}{c}} {\frac{1}{a}}&{\frac{1}{b}}&{\frac{1}{c}} \\ p&q&r \\ 1&1&1 \end{array}} \right| = abc\left| {\begin{array}{*{20}{c}} {x + \left( {p - 1} \right)d}&{x + \left( {q - 1} \right)d}&{x + \left( {r - 1} \right)d} \\ {p - 1}&{q - 1}&{r - 1} \\ 1&1&1 \end{array}} \right|\]
\[\vartriangle = abc\left| {\begin{array}{*{20}{c}} x&x&x \\ {p - 1}&{q - 1}&{r - 1} \\ 1&1&1 \end{array}} \right| = abcx\left| {\begin{array}{*{20}{c}} 1&1&1 \\ {p - 1}&{q - 1}&{r - 1} \\ 1&1&1 \end{array}} \right| = 0.\]
$$\left( {{R_1} \to {R_1} - d \times {R_2}} \right)$$

$$\eqalign{ & \left( {A - 2I} \right)\left( {A + I} \right) = 0 \cr & \Rightarrow AA - A - 2I = 0 \cr & \Rightarrow A\left( {\frac{{A - I}}{2}} \right) = I \cr & \therefore \frac{{A - I}}{2} = {A^{ - 1}} \cr} $$

Here, \[\left[ {\begin{array}{*{20}{c}} 2&4&{2\mu + 9} \\ 1&{5 + 2\lambda }&{\mu - 3\lambda } \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&4&1 \\ 1&{ - 1}&{13} \end{array}} \right]\]
$$ \Rightarrow \,\,2\mu + 9 = 1,5 + 2\lambda = - 1,\mu - 3\lambda = 13$$
$$\therefore \,\,\mu = - 4,\lambda = - 3,$$    which does not satisfy $$\mu - 3\lambda = 13.$$

\[A + B = \left[ {\begin{array}{*{20}{c}} {1 + \omega }&{\omega + {\omega ^2}}&{{\omega ^2} + 1} \\ {\omega + {\omega ^2}}&{{\omega ^2} + 1}&{1 + \omega } \\ {{\omega ^2} + \omega }&{1 + {\omega ^2}}&{\omega + 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - {\omega ^2}}&{ - 1}&{ - \omega } \\ { - 1}&{ - \omega }&{ - {\omega ^2}} \\ { - 1}&{ - \omega }&{ - {\omega ^2}} \end{array}} \right]\]
\[\therefore \,\,\left( {A + B} \right)C = \left[ {\begin{array}{*{20}{c}} { - {\omega ^2}}&{ - 1}&{ - \omega } \\ { - 1}&{ - \omega }&{ - {\omega ^2}} \\ { - 1}&{ - \omega }&{ - {\omega ^2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ \omega \\ {{\omega ^2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - {\omega ^2}}&{ - \omega }&{ - {\omega ^3}} \\ { - 1}&{ - {\omega ^2}}&{ - {\omega ^4}} \\ { - 1}&{ - {\omega ^2}}&{ - {\omega ^4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ 0 \\ 0 \end{array}} \right],\,{\text{using }}{\omega ^3} = 1,1 + \omega + {\omega ^2} = 0.\]

\[{\text{Let }}A = \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right].{\text{Then }}\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ 1&4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 6&0 \\ 0&6 \end{array}} \right]\]
\[ \Rightarrow \,\,\left[ {\begin{array}{*{20}{c}} {a + b}&{ - 2a + 4b} \\ {c + d}&{ - 2c + 4d} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 6&0 \\ 0&6 \end{array}} \right]\]
$$\eqalign{ & \Rightarrow \,\,a + b = 6, - 2a + 4b = 0\,\,\,\,\,\, \Rightarrow \,\,a = 4,b = 2 \cr & \Rightarrow \,\,c + d = 0, - 2c + 4d = 6\,\,\,\,\,\, \Rightarrow \,\,c = - 1,d = 1. \cr} $$

\[{\text{Here,}}\,\left| A \right| = \left[ {\begin{array}{*{20}{c}} {\sin \alpha }&{ - \cos \alpha }&0 \\ {\cos \alpha }&{\sin \alpha }&0 \\ 0&0&1 \end{array}} \right] = 1\]
∴ by definition, $${A^{ - 1}} = \frac{1}{{\left| A \right|}}{\text{adj}}\,A.$$    We get $${A^{ - 1}} = {\text{adj}}\,A.$$

Let, $$f\left( x \right) = {a_0}{x^2} + {a_1}x + {a_2}$$
$$\eqalign{ & g\left( x \right) = {b_0}{x^2} + {b_1}x + {b_2} \cr & h\left( x \right) = {c_0}{x^2} + {c_1}x + {c_2} \cr} $$
Then, \[\Delta \left( x \right) = \left| {\begin{array}{*{20}{c}} {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\ {2{a_0}x + {a_1}}&{2{b_0}x + {b_1}}&{2{c_0}x + {c_1}}\\ {2{a_0}}&{2{b_0}}&{2{c_0}} \end{array}} \right|\]
\[ = x\left| {\begin{array}{*{20}{c}} {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\ {2{a_0}}&{2{b_0}}&{2{c_0}}\\ {2{a_0}}&{2{b_0}}&{2{c_0}} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {2{a_0}}&{2{b_0}}&{2{c_0}} \end{array}} \right|\]
\[ = 0 + 2\left| {\begin{array}{*{20}{c}} {f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_0}}&{{b_0}}&{{c_0}} \end{array}} \right|\]
$$ = 2\left[ {\left( {{b_1}{c_0} - {b_0}{c_1}} \right]f\left( x \right) - \left( {{a_1}{c_0} - {a_0}{c_1}} \right)g\left( x \right) + \left( {{a_1}{b_0} - {a_0}{b_1}} \right)h\left( x \right)} \right]$$
Hence degree of $$\Delta \left( x \right) \leqslant 2.$$

The elements in the leading diagonal are 1, 2, 3, 4, 5. On one side of the leading diagonal all the elements are zero.
∴ the value of the determinant = the product of the elements in the leading diagonal.

$$\eqalign{ & \left| P \right| = 1\left( {12 - 12} \right) - \alpha \left( {4 - 6} \right) + 3\left( {4 - 6} \right) = 2\alpha - 6 \cr & {\text{Now, }}adj\,A = P \cr & \Rightarrow \,\,\left| {adj\,A} \right| = \left| P \right| \cr & \Rightarrow \,\,{\left| A \right|^2} = \left| P \right| \cr & \Rightarrow \,\,\left| P \right| = 16 \cr & \Rightarrow \,\,2\alpha - 6 = 16 \cr & \Rightarrow \,\,\alpha = 11 \cr} $$