Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

If given system of equations have infinitely many solutions, then
\[\begin{array}{l} \left| \begin{array}{l} 2\,\,\,\, - 1\,\,\,\,\,1\\ 1\,\,\,\, - 2\,\,\,\,\,1\\ \lambda \,\,\,\, - 1\,\,\,\,\,2 \end{array} \right| = 0\\ \Rightarrow 2\left( { - 4 + 1} \right) + 1\left( {2 - \lambda } \right) + 1\left( { - 1 + 2\lambda } \right) = 0\\ \Rightarrow - 6 + 2 - \lambda - 1 + 2\lambda = 0\\ \Rightarrow \lambda - 5 = 0\\ \Rightarrow \lambda = 5 \end{array}\]

$$\because $$ All entries of square matrix $$A$$ are integers, therefore all cofactors should also be integers.
If det $$A = \pm 1$$  then $${A^{ - 1}}$$ exists. Also all entries of $${A^{ - 1}}$$ are integers.

$$\eqalign{ & S = ABCD = A\left( {BCD} \right) = A{A^T}\,\,\,\,.....\left( 1 \right) \cr & {S^3} = \left( {ABCD} \right)\left( {ABCD} \right)\left( {ABCD} \right) \cr & = \left( {ABC} \right)\left( {DAB} \right)\left( {CDA} \right)\left( {BCD} \right) \cr & = {D^T}{C^T}{B^T}{A^T} \cr & = {\left( {BCD} \right)^T}{A^T} = A{A^T}\,\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
From (1) and (2), $$S = {S^3}$$

We know that,
$$adj\left( {adj\,A} \right) = {\left| A \right|^{n - 2}}A,\,{\text{if }}\left| A \right| \ne 0,$$       provided order of $$A$$ is $$n.$$
$$\eqalign{ & \therefore adj\left( {adj\,A} \right) = \left| A \right|A\left( {{\text{as }}n = 3} \right) \cr & \therefore \det \left( {adj\left( {adj\,A} \right)} \right) = {\left| A \right|^3}\det A = {\left| A \right|^4} \cr} $$
But \[\left| A \right| = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 1}\\ { - 1}&1&2\\ 2&{ - 1}&1 \end{array}} \right] = 14\]
$$\therefore det\left( {adj\left( {adj\,A} \right)} \right) = {\left( {14} \right)^4}$$

\[\begin{array}{l} {A^2} = \left[ {\begin{array}{*{20}{c}} { - 5}&{ - 8}&0\\ 3&5&0\\ 1&2&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 5}&{ - 8}&0\\ 3&5&0\\ 1&2&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {25 - 24 + 0}&{40 - 40 + 0}&{0 + 0 + 0}\\ { - 15 + 15 + 0}&{ - 24 + 25 + 0}&{0 + 0 + 0}\\ { - 5 + 6 - 1}&{ - 8 + 10 - 2}&{0 + 0 + 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] = I \end{array}\]
Hence, the matrix $$A$$ is involutory.

\[D = \left| \begin{array}{l} 1\,\,\,\,\,\,\,2\,\,\,\,\,\,1\\ 2\,\,\,\,\,\,3\,\,\,\,\,\,1\\ 3\,\,\,\,\,\,5\,\,\,\,\,\,2 \end{array} \right| = 0\,\,\,\,\,\,\,{D_1} = \left| \begin{array}{l} 3\,\,\,\,\,\,2\,\,\,\,\,\,1\\ 3\,\,\,\,\,\,3\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,5\,\,\,\,\,\,2 \end{array} \right| \ne 0\]
⇒ Given system, does not have any solution.
⇒ No solution

The given system of equations are :
$$\eqalign{ & {p^3}x + {\left( {p + 1} \right)^3}y = {\left( {p + 2} \right)^3}\,\,\,\,.....\left( 1 \right) \cr & px + \left( {p + 1} \right)y = \left( {p + 2} \right)\,\,\,\,\,\,.....\left( 2 \right) \cr & x + y = 1\,\,\,\,\,\,\,.....\left( 3 \right) \cr} $$
This system is consistent, if values of $$x$$ and $$y$$ from first two equation satisfy the third equation.
which \[ \Rightarrow \,\left| {\begin{array}{*{20}{c}} {{p^3}}&{{{\left( {p + 1} \right)}^3}}&{{{\left( {p + 2} \right)}^3}}\\ p&{\left( {p + 1} \right)}&{\left( {p + 2} \right)}\\ 1&1&1 \end{array}} \right| = 0\]
\[ \Rightarrow \,\left| {\begin{array}{*{20}{c}} {{p^3}}&{{{\left( {p + 1} \right)}^3} - {p^3}}&{{{\left( {p + 2} \right)}^3} - {p^3}}\\ p&1&2\\ 1&0&0 \end{array}} \right| = 0\]
$$\eqalign{ & \Rightarrow \,2{\left( {p + 1} \right)^3} - 2{p^3} - {\left( {p + 2} \right)^3} + {p^3} = 0 \cr & \Rightarrow \,2\left( {{p^3} + 1 + 3{p^2} + 3p} \right) - 2{p^3} - \left( {{p^3} + 8 + 12p + 6{p^2}} \right) + {p^3} = 0 \cr & \Rightarrow \,2{p^3} + 2 + 6{p^2} + 6p - 2{p^3} - {p^3} - 8 - 12p - 6{p^2} + {p^3} = 0 \cr & \Rightarrow \, - 6 - 6p = 0 \cr & \Rightarrow \,p = - 1 \cr} $$

$$n = 2 \times 3 \times 4 = 24.$$

Since the system has a non-trivial solution, therefore \[\left| {\begin{array}{*{20}{c}} \lambda &{\sin \alpha }&{\cos \alpha }\\ 1&{\cos \alpha }&{\sin \alpha }\\ { - 1}&{\sin \alpha }&{ - \cos \alpha } \end{array}} \right| = 0\]
$$\eqalign{ & \Rightarrow \lambda \left( { - {{\cos }^2}\alpha - {{\sin }^2}\alpha } \right) - \left( { - \sin \alpha \cos \alpha - \sin \alpha \cos \alpha } \right) - \left( {{{\sin }^2}\alpha - {{\cos }^2}\alpha } \right) = 0 \cr & \Rightarrow - \lambda + \sin 2\alpha + \cos 2\alpha = 0 \cr & \Rightarrow \lambda = \sin 2\alpha + \cos 2\alpha \cr & \Rightarrow \lambda = \sqrt 2 \cos \left( {2\alpha - \frac{\pi }{4}} \right). \cr & {\text{Since, }} - 1 \leqslant \cos \left( {2\alpha - \frac{\pi }{4}} \right) \leqslant 1\forall \in R \cr & \therefore - \sqrt 2 \leqslant \lambda \leqslant \sqrt 2 {\text{ }}\,\,{\text{i}}{\text{.e}}{\text{., }}\lambda \in \left[ { - \sqrt 2 ,\sqrt 2 } \right] \cr} $$

\[\left| {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right| = 1 \ne 0.\]     So, the rank = 3.