Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\begin{array}{l} \Delta = \left| {\begin{array}{*{20}{c}} {{S_0}}&{{S_1}}&{{S_2}}\\ {{S_1}}&{{S_2}}&{{S_3}}\\ {{S_2}}&{{S_3}}&{{S_4}} \end{array}} \right|\\ = \,\left| {\begin{array}{*{20}{c}} {1 + 1 + 1}&{\alpha + \beta + \gamma }&{{\alpha ^2} + {\beta ^2} + {\gamma ^2}}\\ {\alpha + \beta + \gamma }&{{\alpha ^2} + {\beta ^2} + {\gamma ^2}}&{{\alpha ^3} + {\beta ^3} + {\gamma ^3}}\\ {{\alpha ^2} + {\beta ^2} + {\gamma ^2}}&{{\alpha ^3} + {\beta ^3} + {\gamma ^3}}&{{\alpha ^4} + {\beta ^4} + {\gamma ^4}} \end{array}} \right| \end{array}\]
The above determinant can be expressed as product of two determinants. Thus,
\[\Delta = \left| {\begin{array}{*{20}{c}} 1&1&1\\ \alpha &\beta &\gamma \\ {{\alpha ^2}}&{{\beta ^2}}&{{\gamma ^2}} \end{array}} \right|\,\left| {\begin{array}{*{20}{c}} 1&1&1\\ \alpha &\beta &\gamma \\ {{\alpha ^2}}&{{\beta ^2}}&{{\gamma ^2}} \end{array}} \right| = {\left[ {\left( {\beta - \alpha } \right)\left( {\gamma - \alpha } \right)\left( {\gamma - \beta } \right)} \right]^2}\]

For infinitely many solutions the two equations become identical
$$\eqalign{ & \Rightarrow \frac{{k + 1}}{k} = \frac{8}{{k + 3}} = \frac{{4k}}{{3k - 1}} \cr & \Rightarrow k = 1. \cr} $$

For existence of a solution of the first system,
\[\left| {\begin{array}{*{20}{c}} a&b&c \\ b&c&a \\ c&a&b \end{array}} \right| = 0.\]
The second system will have a nontrivial solution if we can prove that
\[\left| {\begin{array}{*{20}{c}} {b + c}&{c + a}&{a + b} \\ {c + a}&{a + b}&{b + c} \\ {a + b}&{b + c}&{c + a} \end{array}} \right| = 0.\]
Establish \[\left| {\begin{array}{*{20}{c}} {b + c}&{c + a}&{a + b} \\ {c + a}&{a + b}&{b + c} \\ {a + b}&{b + c}&{c + a} \end{array}} \right| = 2\left| {\begin{array}{*{20}{c}} a&b&c \\ b&c&a \\ c&a&b \end{array}} \right| = 0.\]
Remember that the existence of one nontrivial solution implies existence of infinite number of non-trivial solutions

Given that,
\[\Delta = \left[ {\begin{array}{*{20}{c}} C&1&0\\ 1&C&1\\ 6&1&C \end{array}} \right] = C\left( {{C^2} - 1} \right) - 1\left( {C - 6} \right)\]
$$\eqalign{ & \Rightarrow \Delta = 2\cos \theta \left( {4\,{{\cos }^2}\theta - 1} \right) - \left( {2\cos \theta - 6} \right)\left( {\because C = 2\cos \theta {\text{ given}}} \right) \cr & = 8\,{\cos ^3}\theta - 4\cos \theta + 6 \cr} $$

\[{\rm{Given }}\,\,D = \left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,1 + x\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,1 + y \end{array} \right|\]
$${\text{Apply }}{R_2} \to {R_2} - {R_1}\,{\text{and }}R \to {R_3} - {R_1}$$
\[\therefore \,\,D = \left| \begin{array}{l} 1\,\,\,\,\,\,\,1\,\,\,\,\,\,\,1\\ 0\,\,\,\,\,\,x\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,0\,\,\,\,\,\,y \end{array} \right| = xy\]
Hence, $$D$$ is divisible by both $$x$$ and $$y$$

We have,
\[\begin{array}{l} \frac{{{d^n}}}{{d{x^n}}}\left[ {f\left( x \right)} \right] = \left| {\begin{array}{*{20}{c}} 0&{\cos \left( {x + \frac{{n\pi }}{2}} \right)}&{\frac{{{{\left( { - 1} \right)}^n}n!}}{{{{\left( {x + 3} \right)}^{n + 1}}}}}\\ 0&{\cos \frac{{n\pi }}{2}}&{\frac{{{{\left( { - 1} \right)}^n}n!}}{{{3^{n + 1}}}}}\\ \alpha &{{\alpha ^3}}&{{\alpha ^5}} \end{array}} \right|\\ \therefore \frac{{{d^n}}}{{d{x^n}}}{\left[ {f\left( x \right)} \right]_{x = 0}} = \left| {\begin{array}{*{20}{c}} 0&{\cos \frac{{n\pi }}{2}}&{\frac{{{{\left( { - 1} \right)}^n}n!}}{{{{3}^{n + 1}}}}}\\ 0&{\cos \frac{{n\pi }}{2}}&{\frac{{{{\left( { - 1} \right)}^n}n!}}{{{3^{n + 1}}}}}\\ \alpha &{{\alpha ^3}}&{{\alpha ^5}} \end{array}} \right| = 0 \end{array}\]
$$\left( {\because {R_1}{\text{ and }}{R_2}{\text{ are identical}}} \right)$$

Use $${R_1} \to {R_1} + {R_2} - {R_3}.$$

Let, $$A = {\left[ {{a_{ij}}} \right]_{n \times m}}.$$
Since $$A$$ is skew-symmetric $$a_{ii} = 0$$
$$\left( {i = 1,2,.....,n} \right){\text{ and }}{a_{ji}} = - {a_{ji}}\left( {i \ne j} \right)$$
Also, $$A$$ is symmetric so $${a_{ji}} = {a_{ji}}\forall i{\text{ and }}j$$
$$\therefore {a_{ji}} = 0\,\,\forall \,\,i \ne j$$
Hence, $${a_{ji}} = 0\,\,\forall \,\,i{\text{ and }}j$$
⇒ $$A$$ is a null zero matrix.

Given, $$adj\,B = A,\left| P \right| = \left| Q \right| = 1$$
Consider, $$adj\,\left( {{Q^{ - 1}}B{P^{ - 1}}} \right)$$
$$\eqalign{ & = \left( {adj\,{P^{ - 1}}} \right)\left( {adj\,B} \right)\left( {adj\,{Q^{ - 1}}} \right) \cr & = {\left( {adj\,P} \right)^{ - 1}}A.{\left( {adj\,Q} \right)^{ - 1}} = {\left( {{P^{ - 1}}} \right)^{ - 1}}A{\left( {{Q^{ - 1}}} \right)^{ - 1}} \cr & = PAQ. \cr} $$

$$\eqalign{ & tr\,\left( A \right) = \sum\limits_{i = 1}^{^\lambda } {{a_{{\text{ii}}}} \geqslant \lambda \,\,\,\left( {{\text{As}}\,\,{a_{ij}} \geqslant {\text{1}}\forall \,i,j} \right)} \cr & {\text{and}}\,\,tr\,\left( {M{N^2}} \right) = \left\{ {tr\left( M \right)} \right\}{\left\{ {tr\left( N \right)} \right\}^2} \cr} $$