Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

Number of zeroes in a lower triangular matrix of order $$n \times n$$  is
$$1 + 2 + 3 + ..... + n = \frac{{n\left( {n + 1} \right)}}{2}$$
Number of zeros = 10
$$\eqalign{ & \Rightarrow \frac{{n\left( {n + 1} \right)}}{2} = 10 \cr & \Rightarrow {n^2} + n - 20 = 0 \cr & \Rightarrow \left( {n + 5} \right)\left( {n - 4} \right) = 0 \cr & \Rightarrow n = 4\,\,{\text{or }} - 5\left( { - 5{\text{ is meaningless}}} \right) \cr & \Rightarrow n = 4. \cr} $$
⇒ order of the matrix is $$4 \times 4$$

Let \[A{u_1} = \left( \begin{array}{l} 1\\ 0\\ 0 \end{array} \right){\rm{and }}\,\,A{u_2} = \left( \begin{array}{l} 0\\ 1\\ 0 \end{array} \right),\]
\[{\rm{Then, }}\,\,A{u_1} + A{u_2} = \left( \begin{array}{l} 1\\ 0\\ 0 \end{array} \right)\, + \left( \begin{array}{l} 0\\ 1\\ 0 \end{array} \right)\]
\[ \Rightarrow \,\,A\left( {{u_1} + {u_2}} \right) = \left( \begin{array}{l} 1\\ 1\\ 0 \end{array} \right)\,\,\,\,\,\,\,\,\,....\left( 1 \right)\]
\[{\rm{Also, }}\,\,A = \left( \begin{array}{l} 1\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\ 2\,\,\,\,\,\,\,1\,\,\,\,\,\,\,0\\ 3\,\,\,\,\,\,2\,\,\,\,\,\,\,\,1 \end{array} \right)\]
$$ \Rightarrow \,\,\left| A \right| = 1\left( 1 \right) - 0\left( 2 \right) + 0\left( {4 - 3} \right) = 1$$
We know,
$$\eqalign{ & {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj\,A \cr & \Rightarrow \,\,{A^{ - 1}} = adj\left( A \right)\,\,\,\,\,\,\left( {\because \left| A \right| = 1} \right) \cr} $$
Now, from equation (1), we have
\[{u_1} + {u_2} = {A^{ - 1}}\left( \begin{array}{l} 1\\ 1\\ 0 \end{array} \right)\]
$$ \Rightarrow \,\,{u_1} + {u_2} = {A^{ - 1}}$$
\[ = \left[ \begin{array}{l} \,\,1\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,0\\ - 2\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,0\\ \,\,1\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,1 \end{array} \right]\left( \begin{array}{l} 1\\ 1\\ 0 \end{array} \right) = \left[ \begin{array}{l} \,\,\,1\\ - 1\\ - 1 \end{array} \right]\]

We have,
\[f\left( x \right) = \left| {\begin{array}{*{20}{c}} {\cos x}&x&1\\ {2\sin x}&{{x^2}}&{2x}\\ {\tan x}&x&1 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\cos x - \tan x}&0&0\\ {2\sin x}&{{x^2}}&{2x}\\ {\tan x}&x&1 \end{array}} \right|\]
$$\eqalign{ & \left[ {{\text{Applying }}{R_1} \to {R_1} - {R_3}} \right] \cr & = \left( {\cos x - \tan x} \right)\left( {{x^2} - 2{x^2}} \right)\left[ {{\text{Expanding along }}{R_1}} \right] \cr & = - {x^2}\left( {\cos x - \tan x} \right) \cr & \therefore f'\left( x \right) = - 2x\left( {\cos x - \tan x} \right) - {x^2}\left( { - \sin x - {{\sec }^2}x} \right) \cr & \therefore \mathop {\lim }\limits_{x \to 0 } \left[ {\frac{{f'\left( x \right)}}{x}} \right] = \mathop {\lim }\limits_{x \to 0 } - 2\left( {\cos x - \tan x} \right) + x\left( {\sin x + {{\sec }^2}x} \right) \cr & = - 2 \times 1 = - 2 \cr} $$

$${x^4} - 1 = 0$$   has roots $$\alpha , \beta , \gamma , \delta .$$   Therefore, $$\alpha + \beta + \gamma + \delta = 0.$$
Use $${R_1} \to {R_1} + {R_2} + {R_3} + {R_4}.$$

\[\left| {\begin{array}{*{20}{c}} a&b&c\\ b&c&a\\ c&a&b \end{array}} \right| = - \left( {{a^3} + {b^3} + {c^3} - 3abc} \right)\]
$$\eqalign{ & = - \left( {a + b + c} \right)\left( {a + b{\omega ^2} + c\omega } \right)\left( {a + b\omega + c{\omega ^2}} \right) \cr & = - f\left( \alpha \right)f\left( \beta \right)f\left( \lambda \right)\,\,\,\,\,\left[ {\because \alpha = 1,\beta = \omega ,\lambda = {\omega ^2}} \right] \cr} $$

\[\begin{gathered} \vartriangle = \left| {\begin{array}{*{20}{c}} {4 + 6i}&0&0 \\ {4 - 20i}&0&0 \\ {20}&3&i \end{array}} \right| = 0 = 0 + i \cdot 0. \hfill \\ \left( {{R_1} \to {R_1} + {R_2},{R_2} \to {R_2} - i{R_3}} \right) \hfill \\ \end{gathered} \]

\[{\rm{Let }}\,A = \left( \begin{array}{l} a\,\,\,\,\,\,\,b\\ c\,\,\,\,\,\,\,d \end{array} \right){\rm{where }}\,\,a,b,c,d \ne 0\]
\[{A^2} = \left( \begin{array}{l} a\,\,\,\,\,\,b\\ c\,\,\,\,\,\,d \end{array} \right)\left( \begin{array}{l} a\,\,\,\,\,\,b\\ c\,\,\,\,\,\,d \end{array} \right)\]
\[ \Rightarrow \,\,{A^2} = \left( \begin{array}{l} {a^2} + bc\,\,\,\,\,\,\,\,\,\,ab + bd\\ ac + cd\,\,\,\,\,\,\,\,\,\,bc + {d^2} \end{array} \right)\]
$$\eqalign{ & \Rightarrow \,\,{a^2} + bc = 1,bc + {d^2} = 1 \cr & ab + bd = ac + cd = 0 \cr & c \ne 0\,\,{\text{and }}b \ne 0 \cr & \Rightarrow \,\,a + d = 0 \cr & \Rightarrow \,\,{\text{Tr}}\left( A \right) = 0 \cr & \left| A \right| = ad - bc = - {a^2} - bc = - 1 \cr} $$

Let $$A$$ and $$B$$ be two matrices such that $$AB = A$$   and $$BA = B$$
Now, consider $$AB = A$$
Take Transpose on both side
$$\eqalign{ & {\left( {AB} \right)^T} = {A^T} \cr & \Rightarrow {A^T} = {B^T}.{A^T}\,\,\,\,.....\left( 1 \right) \cr & {\text{Now, }}BA = B \cr} $$
Take, Transpose on both side
$$\eqalign{ & {\left( {BA} \right)^T} = {B^T} \cr & \Rightarrow {B^T} = {A^T}.{B^T}\,\,\,\,.....\left( 2 \right) \cr} $$
Now, from equation (1) and (2). we have
$$\eqalign{ & {A^T} = \left( {{A^T}.{B^T}} \right){A^T} \cr & {A^T} = {A^T}\left( {{B^T}{A^T}} \right) \cr & = {A^T}{\left( {AB} \right)^T}\,\,\,\left( {\because {{\left( {AB} \right)}^T} = {B^T} = {B^T}{A^T}} \right) \cr & = {A^T}.{A^T} \cr & {\text{Thus, }}{A^T} = {\left( {{A^T}} \right)^2} \cr} $$

\[A = \left[ \begin{array}{l} \alpha \,\,\,\,\,\,\,2\\ 2\,\,\,\,\,\,\,\,\alpha \end{array} \right]{\rm{ and }}\left| {{A^3}} \right| = 125\]
$$\eqalign{ & \Rightarrow \,\,{\left| A \right|^3} = 125 \cr & {\text{Now, }}\left| A \right| = {\alpha ^2} - 4 \cr & \Rightarrow \,\,{\left( {{\alpha ^2} - 4} \right)^3} = 125 = {5^3} \cr & \Rightarrow \,\,{\alpha ^2} - 4 = 5 \cr & \Rightarrow \,\,\alpha = \pm 3 \cr} $$

\[\begin{array}{l} {A^n} = \left[ {\begin{array}{*{20}{c}} {\cos \,n\theta }&{\sin \,n\theta }\\ { - \sin \,n\theta }&{\cos \,n\theta } \end{array}} \right]\\ \frac{1}{n}{A^n} = \left[ {\begin{array}{*{20}{c}} {\frac{{\cos \,n\theta }}{n}}&{\frac{{\sin \,n\theta }}{n}}\\ { - \frac{{\sin \,n\theta }}{n}}&{\frac{{\cos \,n\theta }}{n}} \end{array}} \right] \end{array}\]
But $$ - 1 \leqslant \cos \,n\theta \leqslant 1{\text{ and }} - 1 \leqslant \sin \,n\theta \leqslant 1$$
$$\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \,n\theta }}{n} = 0,\mathop {\lim }\limits_{n \to \infty } \frac{{\cos \,n\theta }}{n} = 0$$
\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}{A^n} = \left[ {\begin{array}{*{20}{c}} 0&0\\ 0&0 \end{array}} \right].\]