Matrices and Determinants
Let $B=\left[\begin{array}{lll}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{array}\right], \alpha > 2$ be the adjoint of a matrix $A$ and $|A|=2$. Then $\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c}\alpha \\ -2 \alpha \\ \alpha\end{array}\right]$ is equal to :
The number of symmetric matrices of order 3, with all the entries from the set $\{0,1,2,3,4,5,6,7,8,9\}$ is :
Let $A=\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right]$. If $\mathrm{B}=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] A\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$, then the sum of all the elements of the matrix $\sum_\limits{n=1}^{50} B^{n}$ is equal to
If the system of linear equations
$ \begin{aligned} & 7 x+11 y+\alpha z=13 \\\\ & 5 x+4 y+7 z=\beta \\\\ & 175 x+194 y+57 z=361 \end{aligned} $
has infinitely many solutions, then $\alpha+\beta+2$ is equal to :
$\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$, then $\lambda, \frac{\lambda}{3}$ are the roots of the equation :
Let $\mathrm{A}$ be a $2 \times 2$ matrix with real entries such that $\mathrm{A}'=\alpha \mathrm{A}+\mathrm{I}$, where $\alpha \in \mathbb{R}-\{-1,1\}$. If $\operatorname{det}\left(A^{2}-A\right)=4$, then the sum of all possible values of $\alpha$ is equal to :
If $\mathrm{A}=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{ccc}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{array}\right]$, then $|\operatorname{adj}(\operatorname{adj}(2 \mathrm{~A}))|$ is equal to :
If A is a 3 $\times$ 3 matrix and $|A| = 2$, then $|3\,adj\,(|3A|{A^2})|$ is equal to :
For the system of linear equations
$2x - y + 3z = 5$
$3x + 2y - z = 7$
$4x + 5y + \alpha z = \beta $,
which of the following is NOT correct?
If $A=\left[\begin{array}{cc}1 & 5 \\ \lambda & 10\end{array}\right], \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$ and $\alpha+\beta=-2$, then $4 \alpha^{2}+\beta^{2}+\lambda^{2}$ is equal to :
Let S be the set of all values of $\theta \in[-\pi, \pi]$ for which the system of linear equations
$x+y+\sqrt{3} z=0$
$-x+(\tan \theta) y+\sqrt{7} z=0$
$x+y+(\tan \theta) z=0$
has non-trivial solution. Then $\frac{120}{\pi} \sum_\limits{\theta \in \mathrm{s}} \theta$ is equal to :
Let $A=\left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right]$. If $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))|=(16)^{n}$, then $n$ is equal to :
Let $P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $Q=P A P^{T}$. If $P^{T} Q^{2007} P=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, then $2 a+b-3 c-4 d$ equal to :
Let $P$ be a square matrix such that $P^{2}=I-P$. For $\alpha, \beta, \gamma, \delta \in \mathbb{N}$, if $P^{\alpha}+P^{\beta}=\gamma I-29 P$ and $P^{\alpha}-P^{\beta}=\delta I-13 P$, then $\alpha+\beta+\gamma-\delta$ is equal to :
For the system of equations
$x+y+z=6$
$x+2 y+\alpha z=10$
$x+3 y+5 z=\beta$, which one of the following is NOT true?
If the system of equations
$x+y+a z=b$
$2 x+5 y+2 z=6$
$x+2 y+3 z=3$
has infinitely many solutions, then $2 a+3 b$ is equal to :
Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{2 \times 2}$, where $\mathrm{a}_{\mathrm{ij}} \neq 0$ for all $\mathrm{i}, \mathrm{j}$ and $\mathrm{A}^{2}=\mathrm{I}$. Let a be the sum of all diagonal elements of $\mathrm{A}$ and $\mathrm{b}=|\mathrm{A}|$. Then $3 a^{2}+4 b^{2}$ is equal to :
For the system of linear equations $\alpha x+y+z=1,x+\alpha y+z=1,x+y+\alpha z=\beta$, which one of the following statements is NOT correct?
If $A = {1 \over 2}\left[ {\matrix{ 1 & {\sqrt 3 } \cr { - \sqrt 3 } & 1 \cr } } \right]$, then :
Let $S$ denote the set of all real values of $\lambda$ such that the system of equations
$\lambda x+y+z=1$
$x+\lambda y+z=1$
$x+y+\lambda z=1$
is inconsistent, then $\sum_\limits{\lambda \in S}\left(|\lambda|^{2}+|\lambda|\right)$ is equal to
For the system of linear equations
$x+y+z=6$
$\alpha x+\beta y+7 z=3$
$x+2 y+3 z=14$
which of the following is NOT true ?
Let $A = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 4 & { - 1} \cr 0 & {12} & { - 3} \cr } } \right)$. Then the sum of the diagonal elements of the matrix ${(A + I)^{11}}$ is equal to :
$ \begin{aligned} & x-y+z=5 \\ & 2 x+2 y+\alpha z=8 \\ & 3 x-y+4 z=\beta \end{aligned} $
has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of :
Let the system of linear equations
$x+y+kz=2$
$2x+3y-z=1$
$3x+4y+2z=k$
have infinitely many solutions. Then the system
$(k+1)x+(2k-1)y=7$
$(2k+1)x+(k+5)y=10$
has :
Let $A=\left(\begin{array}{cc}\mathrm{m} & \mathrm{n} \\ \mathrm{p} & \mathrm{q}\end{array}\right), \mathrm{d}=|\mathrm{A}| \neq 0$ and $\mathrm{|A-d(A d j A)|=0}$. Then
The set of all values of $\mathrm{t\in \mathbb{R}}$, for which the matrix
$\left[ {\matrix{
{{e^t}} & {{e^{ - t}}(\sin t - 2\cos t)} & {{e^{ - t}}( - 2\sin t - \cos t)} \cr
{{e^t}} & {{e^{ - t}}(2\sin t + \cos t)} & {{e^{ - t}}(\sin t - 2\cos t)} \cr
{{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr
} } \right]$ is invertible, is :
Let $\alpha$ and $\beta$ be real numbers. Consider a 3 $\times$ 3 matrix A such that $A^2=3A+\alpha I$. If $A^4=21A+\beta I$, then
Consider the following system of equations
$\alpha x+2y+z=1$
$2\alpha x+3y+z=1$
$3x+\alpha y+2z=\beta$
for some $\alpha,\beta\in \mathbb{R}$. Then which of the following is NOT correct.
Let A, B, C be 3 $\times$ 3 matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements
(S1) A$^{13}$ B$^{26}$ $-$ B$^{26}$ A$^{13}$ is symmetric
(S2) A$^{26}$ C$^{13}$ $-$ C$^{13}$ A$^{26}$ is symmetric
Then,
Let $A = \left[ {\matrix{ {{1 \over {\sqrt {10} }}} & {{3 \over {\sqrt {10} }}} \cr {{{ - 3} \over {\sqrt {10} }}} & {{1 \over {\sqrt {10} }}} \cr } } \right]$ and $B = \left[ {\matrix{ 1 & { - i} \cr 0 & 1 \cr } } \right]$, where $i = \sqrt { - 1} $. If $\mathrm{M=A^T B A}$, then the inverse of the matrix $\mathrm{AM^{2023}A^T}$ is
Let $x,y,z > 1$ and $A = \left[ {\matrix{ 1 & {{{\log }_x}y} & {{{\log }_x}z} \cr {{{\log }_y}x} & 2 & {{{\log }_y}z} \cr {{{\log }_z}x} & {{{\log }_z}y} & 3 \cr } } \right]$. Then $\mathrm{|adj~(adj~A^2)|}$ is equal to
Let S$_1$ and S$_2$ be respectively the sets of all $a \in \mathbb{R} - \{ 0\} $ for which the system of linear equations
$ax + 2ay - 3az = 1$
$(2a + 1)x + (2a + 3)y + (a + 1)z = 2$
$(3a + 5)x + (a + 5)y + (a + 2)z = 3$
has unique solution and infinitely many solutions. Then
Let A be a 3 $\times$ 3 matrix such that $\mathrm{|adj(adj(adj~A))|=12^4}$. Then $\mathrm{|A^{-1}~adj~A|}$ is equal to
If the system of equations
$x+2y+3z=3$
$4x+3y-4z=4$
$8x+4y-\lambda z=9+\mu$
has infinitely many solutions, then the ordered pair ($\lambda,\mu$) is equal to :
If A and B are two non-zero n $\times$ n matrices such that $\mathrm{A^2+B=A^2B}$, then :
Let $\alpha$ be a root of the equation $(a - c){x^2} + (b - a)x + (c - b) = 0$ where a, b, c are distinct real numbers such that the matrix $\left[ {\matrix{ {{\alpha ^2}} & \alpha & 1 \cr 1 & 1 & 1 \cr a & b & c \cr } } \right]$ is singular. Then, the value of ${{{{(a - c)}^2}} \over {(b - a)(c - b)}} + {{{{(b - a)}^2}} \over {(a - c)(c - b)}} + {{{{(c - b)}^2}} \over {(a - c)(b - a)}}$ is
Let $X=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ and $A=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]$. For $\mathrm{k} \in N$, if $X^{\prime} A^{k} X=33$, then $\mathrm{k}$ is equal to _______.
Explanation:
$A^{2}=\left[\begin{array}{lll}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad A^{4}=\left[\begin{array}{ccc}1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow A^{k}=\left[\begin{array}{ccc}1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\therefore \quad X^{\prime} A^{k} X=[111]\left[\begin{array}{ccc}1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=[3 k+3]$
$\Rightarrow[3 k+3]=33$ (here it shall be [33] as matrix can't be equal to a scalar)
i.e. $[3 k+3]=33$
$3 k+3=[33] \quad \Rightarrow k=10$
If $k$ is odd and apply above process, we don't get odd value of $k$
$\therefore k=10$
Let p and p + 2 be prime numbers and let
$ \Delta=\left|\begin{array}{ccc} \mathrm{p} ! & (\mathrm{p}+1) ! & (\mathrm{p}+2) ! \\ (\mathrm{p}+1) ! & (\mathrm{p}+2) ! & (\mathrm{p}+3) ! \\ (\mathrm{p}+2) ! & (\mathrm{p}+3) ! & (\mathrm{p}+4) ! \end{array}\right| $
Then the sum of the maximum values of $\alpha$ and $\beta$, such that $\mathrm{p}^{\alpha}$ and $(\mathrm{p}+2)^{\beta}$ divide $\Delta$, is __________.
Explanation:
$\Delta = \left| {\matrix{ {p!} & {(p + 1)!} & {(p + 2)!} \cr {(p + 1)!} & {(p + 2)!} & {(p + 3)!} \cr {(p + 2)!} & {(p + 3)!} & {(p + 4)!} \cr } } \right|$
$ = p!\,.\,(p + 1)!\,.\,(p + 2)!\left| {\matrix{ 1 & {(p + 1)} & {(p + 1)(p + 2)} \cr 1 & {(p + 2)} & {(p + 2)(p + 3)} \cr 1 & {(p + 3)} & {(p + 3)(p + 4)} \cr } } \right|$
$ = p!\,.\,(p + 1)!\,.\,(p + 2)!\left| {\matrix{ 1 & {p + 1} & {{p^2} + 3p + 2} \cr 0 & 1 & {2p + 4} \cr 0 & 1 & {2p + 6} \cr } } \right|$
$ = 2(p!)\,.\,\left( {(p + 1)!} \right)\,.\,\left( {(p + 2)!} \right)$
$ = 2(p + 1)\,.\,{(p!)^2}\,.\,\left( {(p + 2)!} \right)$
$ = 2{(p + 1)^2}\,.\,{(p!)^3}\,.\,\left( {(p + 2)!} \right)$
$\therefore$ Maximum value of $\alpha$ is 3 and $\beta$ is 1.
$\therefore$ $\alpha + \beta = 4$
Let $A=\left[\begin{array}{cc}1 & -1 \\ 2 & \alpha\end{array}\right]$ and $B=\left[\begin{array}{cc}\beta & 1 \\ 1 & 0\end{array}\right], \alpha, \beta \in \mathbf{R}$. Let $\alpha_{1}$ be the value of $\alpha$ which satisfies $(\mathrm{A}+\mathrm{B})^{2}=\mathrm{A}^{2}+\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$ and $\alpha_{2}$ be the value of $\alpha$ which satisfies $(\mathrm{A}+\mathrm{B})^{2}=\mathrm{B}^{2}$. Then $\left|\alpha_{1}-\alpha_{2}\right|$ is equal to ___________.
Explanation:
${(A + B)^2} = {A^2} + {B^2} + AB + BA$
$ = {A^2} + \left[ {\matrix{ 2 & 2 \cr 2 & 2 \cr } } \right]$
$\therefore$ ${B^2} + AB + BA = \left[ {\matrix{ 2 & 2 \cr 2 & 2 \cr } } \right]$ ..... (1)
$AB = \left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right]\left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right] = \left[ {\matrix{ {\beta - 1} & 1 \cr {\alpha + 2\beta } & 2 \cr } } \right]$
$BA = \left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right] = \left[ {\matrix{ {\beta + 2} & {\alpha - \beta } \cr 1 & { - 1} \cr } } \right]$
${B^2} = \left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right]\left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right] = \left[ {\matrix{ {{\beta ^2} + 1} & \beta \cr \beta & 1 \cr } } \right]$
By (1) we get
$\left[ {\matrix{ {{\beta ^2} + 2\beta } + 2 & {\alpha + 1} \cr {\alpha + 3\beta + 1} & 2 \cr } } \right] = \left[ {\matrix{ 2 & 2 \cr 2 & 2 \cr } } \right]$
$\therefore$ $\alpha = 1\,\,\beta = 0\,\, \Rightarrow {\alpha _1} = 1$
Similarly if ${A^2} + AB + BA = 0$ then
$\left( {{A^2} = \left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right]\left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right] = \left[ {\matrix{ { - 1} & { - 1 - \alpha } \cr {2 + 2\alpha } & {{\alpha ^2} - 2} \cr } } \right]} \right)$
$\left[ {\matrix{ {2\beta } & {\alpha - \beta + 1 - 1 - \alpha } \cr {\alpha + 2\beta + 1 + 2 + 2\alpha } & {{\alpha ^2} - 2 + 1} \cr } } \right] = \left[ {\matrix{ 0 & 0 \cr 0 & 0 \cr } } \right]$
$ \Rightarrow \beta = 0$ and $\alpha = - 1\,\, \Rightarrow {\alpha _2} = - 1$
$\therefore$ $|{\alpha _1} - {\alpha _2}| = |2| = 2.$
Consider a matrix $A=\left[\begin{array}{ccc}\alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta\end{array}\right]$, where $\alpha, \beta, \gamma$ are three distinct natural numbers.
If $\frac{\operatorname{det}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))))}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}=2^{32} \times 3^{16}$, then the number of such 3 - tuples $(\alpha, \beta, \gamma)$ is ____________.
Explanation:
$\det (A) = \left| {\matrix{ \alpha & \beta & \gamma \cr {{\alpha ^2}} & {{\beta ^2}} & {{\gamma ^2}} \cr {\beta + \gamma } & {\gamma + \alpha } & {\alpha + \beta } \cr } } \right|$
${R_3} \to {R_3} + {R_1}$
$ \Rightarrow (\alpha + \beta + \gamma )\left| {\matrix{ \alpha & \beta & \gamma \cr {{\alpha ^2}} & {{\beta ^2}} & {{\gamma ^2}} \cr 1 & 1 & 1 \cr } } \right|$
$\therefore$ $\det (A) = (\alpha + \beta + \gamma )(\alpha - \beta )(\beta - \gamma )(\gamma - \alpha )$
Also, $\det (adj\,(adj\,(adj\,(adj\,(A)))))$
$ = {(\det (A))^{{2^4}}} = (\det {(A)^{16}}$
$\therefore$ ${{{{(\alpha + \beta + \gamma )}^{16}}{{(\alpha - \beta )}^{16}}{{(\beta - \gamma )}^{16}}{{(\gamma - \alpha )}^{16}}} \over {{{(\alpha - \beta )}^{16}}{{(\beta - \gamma )}^{16}}{{(\gamma - \alpha )}^{16}}}} = {(4.13)^{16}}$
$ \Rightarrow \alpha + \beta + \gamma = 12$
$ \Rightarrow (\alpha ,\beta ,\gamma )$ distinct natural triplets
$ = {}^{11}{C_2} - 1 - {}^3{C_2}(4) = 55 - 1 - 12$
$ = 42$
Let $S$ be the set containing all $3 \times 3$ matrices with entries from $\{-1,0,1\}$. The total number of matrices $A \in S$ such that the sum of all the diagonal elements of $A^{\mathrm{T}} A$ is 6 is ____________.
Explanation:
Sum of all diagonal elements is equal to sum of square of each element of the matrix.
i.e., $A = \left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{b_1}} & {{b_2}} & {{b_3}} \cr {{c_1}} & {{c_2}} & {{c_3}} \cr } } \right]$
then ${t_r}\,(A\,.\,{A^T})$
$ = a_1^2 + a_2^2 + a_3^2 + b_1^2 + b_2^2 + b_3^2 + c_1^2 + c_2^2 + c_3^2$
$\because$ ${a_i},{b_i},{c_i} \in \{ - 1,0,1\} $ for $i = 1,2,3$
$\therefore$ Exactly three of them are zero and rest are 1 or $-$1.
Total number of possible matrices ${}^9{C_3} \times {2^6}$
$ = {{9 \times 8 \times 7} \over 6} \times 64$
$ = 5376$
The number of matrices $A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$, where $a, b, c, d \in\{-1,0,1,2,3, \ldots \ldots, 10\}$, such that $A=A^{-1}$, is ___________.
Explanation:
$\because$ $A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$ then ${A^2} = \left[ {\matrix{ {{a^2} + bc} & {b(a + d)} \cr {c(a + d)} & {bc + {d^2}} \cr } } \right]$
For A$-$1 must exist $ad - bc \ne 0$ ...... (i)
and $A = {A^{ - 1}} \Rightarrow {A^2} = I$
$\therefore$ ${a^2} + bc = {d^2} + bc = 1$ ...... (ii)
and $b(a + d) = c(a + d) = 0$ ...... (iii)
Case I : When a = d = 0, then possible values of (b, c) are (1, 1), ($-$1, 1) and (1, $-$1) and ($-$1, 1).
Total four matrices are possible.
Case II : When a = $-$d then (a, d) be (1, $-$1) or ($-$1, 1).
Then total possible values of (b, c) are $(12 + 11) \times 2 = 46$.
$\therefore$ Total possible matrices $= 46 + 4 = 50$.
Let $A=\left[\begin{array}{lll}
1 & a & a \\
0 & 1 & b \\
0 & 0 & 1
\end{array}\right], a, b \in \mathbb{R}$. If for some
$n \in \mathbb{N}, A^{n}=\left[\begin{array}{ccc}
1 & 48 & 2160 \\
0 & 1 & 96 \\
0 & 0 & 1
\end{array}\right]
$ then $n+a+b$ is equal to ____________.
Explanation:
$A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] + \left[ {\matrix{ 0 & a & a \cr 0 & 0 & b \cr 0 & 0 & 0 \cr } } \right] = I + B$
${B^2} = \left[ {\matrix{ 0 & a & a \cr 0 & 0 & b \cr 0 & 0 & 0 \cr } } \right] + \left[ {\matrix{ 0 & a & a \cr 0 & 0 & b \cr 0 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 0 & {ab} \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$
${B^3} = 0$
$\therefore$ ${A^n} = {(1 + B)^n} = {}^n{C_0}I + {}^n{C_1}B + {}^n{C_2}{B^2} + {}^n{C_3}{B^3} + \,\,....$
$ = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] + \left[ {\matrix{ 0 & {na} & {na} \cr 0 & 0 & {nb} \cr 0 & 0 & 0 \cr } } \right] + \left[ {\matrix{ 0 & 0 & {{{n(n - 1)ab} \over 2}} \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$
$ = \left[ {\matrix{ 1 & {na} & {na + {{n(n - 1)} \over 2}ab} \cr 0 & 1 & {nb} \cr 0 & 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & {48} & {2160} \cr 0 & 1 & {48} \cr 0 & 0 & 1 \cr } } \right]$
On comparing we get $na = 48$, $nb = 96$ and
$na + {{n(n - 1)} \over 2}ab = 2160$
$ \Rightarrow a = 4,n = 12$ and $b = 8$
$n + a + b = 24$
Let $A=\left(\begin{array}{rrr}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{array}\right)$ and $B=A-I$. If $\omega=\frac{\sqrt{3} i-1}{2}$, then the number of elements in the $\operatorname{set}\left\{n \in\{1,2, \ldots, 100\}: A^{n}+(\omega B)^{n}=A+B\right\}$ is equal to ____________.
Explanation:
We get $A^{2}=A$ and similarly for
$ B=A-I=\left[\begin{array}{lll} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{array}\right] $
We get $B^{2}=-B \Rightarrow B^{3}=B$
$ \therefore A^{n}+(\omega B)^{n}=A+(\omega B)^{n} \quad \text { for } n \in \mathrm{N} $
For $\omega^{n}$ to be unity $n$ shall be multiple of 3 and for $B^{n}$ to be $B . n$ shell be $3,5,7, \ldots 99$
$\therefore n=\{3,9,15, \ldots . .99\}$
Number of elements $=17$
Let $M = \left[ {\matrix{ 0 & { - \alpha } \cr \alpha & 0 \cr } } \right]$, where $\alpha$ is a non-zero real number an $N = \sum\limits_{k = 1}^{49} {{M^{2k}}} $. If $(I - {M^2})N = - 2I$, then the positive integral value of $\alpha$ is ____________.
Explanation:
$N=M^{2}+M^{4}+\ldots+M^{98}$
$=\left[-\alpha^{2}+\alpha^{4}-\alpha^{6}+\ldots\right] I$
$=\frac{-\alpha^{2}\left(1-\left(-\alpha^{2}\right)^{49}\right)}{1+\alpha^{2}} \cdot 1$
$I-M^{2}=\left(1+\alpha^{2}\right) I$
$\left(I-M^{2}\right) N=-\alpha^{2}\left(\alpha^{98}+1\right)=-2$
$\therefore \alpha=1$
If the system of linear equations
$2x - 3y = \gamma + 5$,
$\alpha x + 5y = \beta + 1$, where $\alpha$, $\beta$, $\gamma$ $\in$ R has infinitely many solutions then the value
of | 9$\alpha$ + 3$\beta$ + 5$\gamma$ | is equal to ____________.
Explanation:
If 2x $-$ 3y = $\gamma$ + 5 and $\alpha$x + 5y = $\beta$ + 1 have infinitely many solutions then
${2 \over \alpha } = {{ - 3} \over 5} = {{\gamma + 5} \over {\beta + 1}}$
$ \Rightarrow \alpha = - {{10} \over 3}$ and $3\beta + 5\gamma = - 28$
So $|9\alpha + 3\beta + 5\gamma | = | - 30 - 28| = 58$
Let $A = \left( {\matrix{ {1 + i} & 1 \cr { - i} & 0 \cr } } \right)$ where $i = \sqrt { - 1} $. Then, the number of elements in the set { n $\in$ {1, 2, ......, 100} : An = A } is ____________.
Explanation:
$\therefore$ ${A^2} = \left[ {\matrix{ {1 + i} & 1 \cr { - i} & 0 \cr } } \right]\left[ {\matrix{ {1 + i} & 1 \cr { - 1} & 0 \cr } } \right] = \left[ {\matrix{ i & {1 + i} \cr {1 - i} & { - i} \cr } } \right]$
${A^4} = \left[ {\matrix{ i & {1 + i} \cr {1 - i} & { - i} \cr } } \right]\left[ {\matrix{ i & {1 + i} \cr {1 - i} & { - i} \cr } } \right] = I$
So A5 = A, A9 = A and so on.
Clearly n = 1, 5, 9, ......, 97
Number of values of n = 25
The positive value of the determinant of the matrix A, whose
Adj(Adj(A)) = $\left( {\matrix{ {14} & {28} & { - 14} \cr { - 14} & {14} & {28} \cr {28} & { - 14} & {14} \cr } } \right)$, is _____________.
Explanation:
$\left| {adj(adj(A))} \right| = {\left| A \right|^{{2^2}}} = {\left| A \right|^4}$
$\therefore$ ${\left| A \right|^4} = \left| {\matrix{ {14} & {28} & { - 14} \cr { - 14} & {14} & {28} \cr {28} & { - 14} & {14} \cr } } \right|$
$ = {(14)^3}\left| {\matrix{ 1 & 2 & { - 1} \cr { - 1} & 1 & 2 \cr 2 & { - 1} & 1 \cr } } \right|$
$ = {(14)^3}(3 - 2( - 5) - 1( - 1))$
${\left| A \right|^4} = {(14)^4} \Rightarrow \left| A \right| = 14$
Let $X = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right],\,Y = \alpha I + \beta X + \gamma {X^2}$ and $Z = {\alpha ^2}I - \alpha \beta X + ({\beta ^2} - \alpha \gamma ){X^2}$, $\alpha$, $\beta$, $\gamma$ $\in$ R. If ${Y^{ - 1}} = \left[ {\matrix{ {{1 \over 5}} & {{{ - 2} \over 5}} & {{1 \over 5}} \cr 0 & {{1 \over 5}} & {{{ - 2} \over 5}} \cr 0 & 0 & {{1 \over 5}} \cr } } \right]$, then ($\alpha$ $-$ $\beta$ + $\gamma$)2 is equal to ____________.
Explanation:
$\because$ $X = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right]$
$\therefore$ ${X^2} = \left[ {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$
$\therefore$ $Y = \alpha I + \beta X + \gamma {X^2}\left[ {\matrix{ \alpha & \beta & \gamma \cr 0 & \alpha & \beta \cr 0 & 0 & \alpha \cr } } \right]$
$\because$ $Y\,.\,{Y^{ - 1}} = I$
$\therefore$ $\left[ {\matrix{ \alpha & \beta & \gamma \cr 0 & \alpha & \beta \cr 0 & 0 & \alpha \cr } } \right]\left[ {\matrix{ {{1 \over 5}} & {{{ - 2} \over 5}} & {{1 \over 5}} \cr 0 & {{1 \over 5}} & {{{ - 2} \over 5}} \cr 0 & 0 & {{1 \over 5}} \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$
$\therefore$ $\left[ {\matrix{ {{\alpha \over 5}} & {{{\beta - 2\alpha } \over 5}} & {{{\alpha - 2\beta + \gamma } \over 5}} \cr 0 & {{\alpha \over 5}} & {{{\beta - 2\alpha } \over 5}} \cr 0 & 0 & {{\alpha \over 5}} \cr } } \right] = \left[ {\matrix{ 0 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$
$\therefore$ $\alpha$ = 5, $\beta$ = 10, $\gamma$ =15
$\therefore$ ($\alpha$ $-$ $\beta$ + $\gamma$)2 = 100