Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} {\cos C}&{\tan A}&0\\ {\sin B}&0&{ - \tan A}\\ 0&{\sin B}&{\cos C} \end{array}} \right|\\ = \cos \,C\left( {\tan \,A\,\sin \,B} \right) - \tan \,A\left( {\sin \,B\,\cos \,C} \right)\\ = 0\\ {\rm{Hence, option 'A' is correct}}{\rm{.}} \end{array}\]

\[{\rm{Given\,\, that\,\, 10}}\,B = \left[ \begin{array}{l} \,\,\,\,4\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,2\\ - 5\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\alpha \\ \,\,\,\,1\,\,\,\, - 2\,\,\,\,\,\,\,\,\,\,3 \end{array} \right]\]
\[ \Rightarrow \,B = \frac{1}{{10}}\left[ \begin{array}{l} \,\,\,4\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,2\\ - 5\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\alpha \\ \,\,\,1\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,3 \end{array} \right]\]
$${\text{Also since, }}B = {A^{ - 1}} \Rightarrow \,\,AB = I$$
\[ \Rightarrow \,\,\frac{1}{{10}}\left[ \begin{array}{l} 1\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,1\\ 2\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\, - 3\\ 1\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right]\left[ \begin{array}{l} \,\,\,\,4\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,2\\ - 5\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\alpha \\ \,\,\,\,\,1\,\,\,\, - 2\,\,\,\,\,\,\,\,3 \end{array} \right] = \left[ \begin{array}{l} 1\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,1\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,0\,\,\,\,\,\,\,1 \end{array} \right]\]
\[ \Rightarrow \,\,\frac{1}{{10}}\left[ \begin{array}{l} 10\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5 - 2\\ \,0\,\,\,\,\,\,10\,\,\,\,\,\,\, - 5 + \alpha \\ \,0\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5 + \alpha \end{array} \right] = \left[ \begin{array}{l} 1\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,1\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,0\,\,\,\,\,\,\,1 \end{array} \right]\]
$$\eqalign{ & \Rightarrow \,\,\frac{{5 - \alpha }}{{10}} = 0 \cr & \Rightarrow \,\,\alpha = 5 \cr} $$

If the G.P. be $$a,ar,a{r^2},.....$$   then $${a_n} = a{r^{n - 1}}$$
\[D = \left| {\begin{array}{*{20}{c}} {\log a + \left( {n - 1} \right)\log r}&{\log a + n \log r}&{\log a + \left( {n + 1} \right)\log r}\\ {\log a + n\log r}&{\log a + \left( {n + 1} \right)\log r}&{\log a + \left( {n + 2} \right)\log r}\\ {\log a + \left( {n + 1} \right)\log r}&{\log a + \left( {n + 2} \right)\log r}&{\log a + \left( {n + 3} \right)\log r} \end{array}} \right|\]
$${R_3} \to {R_3} - {R_2}$$   and $${R_2} \to {R_2} - {R_1}$$   gives,
\[ = \left| {\begin{array}{*{20}{c}} {\log a + \left( {n - 1} \right)\log r}&{\log a + n \log r}&{\log a + \left( {n + 1} \right)\log r}\\ {\log r}&{\log r}&{\log r}\\ {\log r}&{\log r}&{\log r} \end{array}} \right|\]
= 0, since $$R_2$$ and $$R_3$$ are identical.

\[\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1\\ 1&2&0\\ 1&3&0 \end{array}} \right]\\ \Rightarrow A = 1\left( 0 \right) + 1\left( 0 \right) + 1\left( {3 - 2} \right)\\ \Rightarrow \left| A \right| = 1\\ {\rm{Now,\, }}\left| {{\rm{adj}}\,A} \right| = {\left| A \right|^2}\\ \Rightarrow \left| {{\rm{adj}}\,A} \right| = 1 \end{array}\]

For non - zero solution of the system of linear equations;
\[\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,\,\,k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\\ 3\,\,\,\,\,\,\,\,\,\,\,k\,\,\,\,\,\,\,\,\, - 2\\ 2\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\, - 3 \end{array} \right| = 0\]
$$ \Rightarrow \,\,k = 11$$
Now equations become
$$\eqalign{ & x + 11y + 3z = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & \,3x + 11y - 2z = 0\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr & 2x + 4y - 3z = 0\,\,\,\,\,\,\,\,\,\,.....\left( 3 \right) \cr} $$
Adding equations (1) & (3) we get
$$\eqalign{ & 3x + 15y = 0 \cr & \Rightarrow \,\,x = - 5y \cr} $$
Now put $$x = - 5y$$   in equation (1), we get
$$\eqalign{ & - 5y + 11y + 3z = 0 \cr & \Rightarrow \,\,z = - 2y \cr & \therefore \frac{{xz}}{{{y^2}}} = \frac{{\left( { - 5y} \right)\left( { - 2y} \right)}}{{{y^2}}} = 10 \cr} $$

\[{\rm{Let }}\,\Delta = \left| \begin{array}{l} \,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a^2}\\ \cos \left( {p - d} \right)x\,\,\,\,\,\,\,\,\,\,\cos px\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {p + d} \right)x\\ \sin \left( {p - d} \right)x\,\,\,\,\,\,\,\,\,\,\,\sin px\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \left( {p + d} \right)x \end{array} \right|\]
$${\text{Applying }}{C_1} \to {C_1} + {C_3}$$
\[ \Rightarrow \,\,\Delta = \left| \begin{array}{l} \,\,\,\,\,\,\,\,\,1 + {a^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a^2}\\ 2\cos px\cos dx\,\,\,\,\,\,\,\,\,\,\,\,\cos px\,\,\,\,\,\,\,\,\,\,\,\cos \left( {p + d} \right)x\\ 2\sin px\cos dx\,\,\,\,\,\,\,\,\,\,\,\,\,\sin px\,\,\,\,\,\,\,\,\,\,\,\sin \left( {p + d} \right)x \end{array} \right|\]
$${C_1} \to {C_1} - \left( {2\cos dx} \right){C_2}$$
\[\Delta = \left| \begin{array}{l} 1 + {a^2} - 2a\cos dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a^2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos px\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {p + d} \right)x\\ \,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin px\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \left( {p + d} \right)x \end{array} \right|\]
Expanding along $${C_1},$$ we get
$$\eqalign{ & \Delta = \left( {1 + {a^2} - 2a\cos dx} \right)\left[ {\sin \left( {p + d} \right)x\cos px - \sin px\cos \left( {p + d} \right)x} \right] \cr & \Rightarrow \,\,\Delta = \left( {1 + {a^2} - 2a\cos dx} \right)\left[ {\sin \left\{ {\left( {p + d} \right)x - px} \right\}} \right] \cr & \Rightarrow \,\,\Delta = \left( {1 + {a^2} - 2a\cos dx} \right)\left[ {\sin dx} \right] \cr} $$
which is independent of $$p.$$

We have,
\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} {\cos \left( {A - P} \right)}&{\cos \left( {A - Q} \right)}&{\cos \left( {A - R} \right)}\\ {\cos \left( {B - P} \right)}&{\cos \left( {B - Q} \right)}&{\cos \left( {B - R} \right)}\\ {\cos \left( {C - P} \right)}&{\cos \left( {C - Q} \right)}&{\cos \left( {C - R} \right)} \end{array}} \right|\\ \left| {\begin{array}{*{20}{c}} {\cos A}&{\sin A}&0\\ {\cos B}&{\sin B}&0\\ {\cos C}&{\sin C}&0 \end{array}} \right| \times \left| {\begin{array}{*{20}{c}} {\cos P}&{\sin P}&0\\ {\cos Q}&{\sin Q}&0\\ {\cos R}&{\sin R}&0 \end{array}} \right| = 0 \cdot 0 = 0 \end{array}\]

\[{\rm{Here,}}\,\,A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 2&3 \end{array}} \right]\,\,{\rm{and}}\,\,B = \left[ {\begin{array}{*{20}{c}} 2&3\\ { - 1}&{ - 2} \end{array}} \right]\]
$$\left| A \right| = 3 - \left( { - 2} \right) = 5\,\,{\text{and}}\,\,\left| B \right| = - 4 - \left( { - 3} \right) = - 1$$
\[\begin{array}{l} \Rightarrow \,{A^{ - 1}} = \frac{1}{5}\left[ {\begin{array}{*{20}{c}} 3&1\\ { - 2}&1 \end{array}} \right]\,\,{\rm{and}}\,\,{B^{ - 1}} = - 1\left[ {\begin{array}{*{20}{c}} { - 2}&{ - 3}\\ 1&2 \end{array}} \right]\\ AB = \left[ {\begin{array}{*{20}{c}} 3&5\\ 1&0 \end{array}} \right]\,\,{\rm{and}}\,\,{A^{ - 1}}{B^{ - 1}} = \frac{1}{5}\left[ {\begin{array}{*{20}{c}} 5&7\\ { - 5}&8 \end{array}} \right]\\ \Rightarrow \,AB\left( {{A^{ - 1}}{B^{ - 1}}} \right) = \frac{1}{5}\left[ {\begin{array}{*{20}{c}} { - 10}&{ - 12}\\ 5&7 \end{array}} \right] \ne 1. \end{array}\]
$$\left| {AB} \right| = 0 - 5 = - 5$$
\[\therefore \,{\left( {AB} \right)^{ - 1}} = \frac{{ - 1}}{5}\left[ {\begin{array}{*{20}{c}} 0&{ - 5}\\ { - 1}&3 \end{array}} \right] \ne \,{A^{ - 1}}{B^{ - 1}}\]

Here, $$X$$ is $$n \times 1,C$$   is $$n \times n$$  and $$X'$$ is $$1 \times n$$
Hence, $$X'CX$$  is $$1 \times 1$$  matrix. Let $$X'CX = k .$$
Then, $$\left( {X'CX} \right)' = X'C'X' = X'C'X' = X'\left( { - C} \right)X = - X'CX = - k$$
$$\eqalign{ & \Rightarrow k = - k \cr & \Rightarrow k = 0 \cr & \Rightarrow X'CX{\text{ is null matrix}}{\text{.}} \cr} $$

Put $$\lambda = 0$$  on both sides. Then
\[\left| {\begin{array}{*{20}{c}} 0&{ - 1}&3 \\ 1&0&{ - 4} \\ { - 3}&4&0 \end{array}} \right| = t.\]