Matrices and Determinants
Let $A=\left[\begin{array}{lll}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{array}\right]$, where $a, c \in \mathbb{R}$. If $A^{3}=A$ and the positive value of $a$ belongs to the interval $(n-1, n]$, where $n \in \mathbb{N}$, then $n$ is equal to ___________.
Explanation:
$ \begin{aligned} A^2 & =\left[\begin{array}{lll} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{array}\right] \\\\ & =\left[\begin{array}{ccc} a+2 & 2 c & 3 \\ 3 & a+3 c & 2 a \\ a c & 1 & 2+3 c \end{array}\right] \end{aligned} $
$ \begin{aligned} A^3 & =\left[\begin{array}{ccc} a+2 & 2 c & 3 \\ 3 & a+3 c & 2 a \\ a c & 1 & 2+3 c \end{array}\right]\left[\begin{array}{ccc} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{array}\right] \\\\ & =\left[\begin{array}{ccc} 2 a c+3 & a+2+3 c & 2 a+4+6 c \\ a(a+3 c)+2 a & 3+2 a c & 6+3 a+9 c \\ a+2+3 c & a c+2 c+3 c^2 & 2 a c+3 \end{array}\right] \end{aligned} $
$ \begin{aligned} & A^3 =A [Given]\\\\ & 2 a c+3= 0 \text { and } a+2+3 c=1 \\\\ & a^2+2 a+3 a c =a \\\\ & \Rightarrow a^2 +a+3\left(-\frac{3}{2}\right)=0\\\\ & \Rightarrow 2 a^2+2 a-9=0 \end{aligned} $
When, $a=1,2 a^2+2 a-9<0$ and
When, $a=2,2 a^2+2 a-9>0$
$\therefore$ Positive value of $a \in(1,2]$
Hence, $n=2$
Let $\mathrm{S}$ be the set of values of $\lambda$, for which the system of equations
$6 \lambda x-3 y+3 z=4 \lambda^{2}$,
$2 x+6 \lambda y+4 z=1$,
$3 x+2 y+3 \lambda z=\lambda$ has no solution. Then $12 \sum_\limits{i \in S}|\lambda|$ is equal to ___________.
Explanation:
Therefore for the given set of equations
$ \Delta=\left|\begin{array}{ccc} 6 \lambda & -3 & 3 \\ 2 & 6 \lambda & 4 \\ 3 & 2 & 3 \lambda \end{array}\right|=0 $
$ \begin{aligned} &\Rightarrow6 \lambda\left(18 \lambda^2-8\right)+3(6 \lambda-12)+3(4-18 \lambda)=0 \\\\ &\Rightarrow18 \lambda^3-14 \lambda-4=0 \\\\ &\Rightarrow(\lambda-1)(3 \lambda+1)(3 \lambda+2)=0 \\\\ &\Rightarrow \lambda=1,-\frac{1}{3},-\frac{2}{3} \end{aligned} $
Also for each values of $\lambda=1, \frac{-1}{3}, \frac{-2}{3}$, we have
$ \left|\begin{array}{ccc} 6 \lambda & -3 & 4 \lambda^2 \\ 2 & 6 \lambda & 1 \\ 3 & 2 & \lambda \end{array}\right| \neq 0 $
which implies that, for each values of $\lambda$, the given system of equations has no solution.
$ \begin{aligned} & \text { Therefore } S \in\left\{1, \frac{-1}{3}, \frac{-2}{3}\right\} \text { and } \\\\ &12 \sum_{\lambda \in S}|\lambda| \\\\ & =12\left(|1|+\left|\frac{-1}{3}\right|+\left|\frac{-2}{3}\right|\right) \\\\ & =12\left(1+\frac{1}{3}+\frac{2}{3}\right)=12\left(\frac{6}{3}\right)=24 \end{aligned} $
Explanation:
$\Rightarrow 2^{n \cdot(n-1)}\left|\operatorname{adj}\left(2 A^{-1}\right)\right|^{(n-1)}=2^{84}$
$\Rightarrow 2^{n(n-1)}\left|2 A^{-1}\right|^{(n-1)^{2}}=2^{84}$
$\Rightarrow 2^{n(n-1)} \cdot 2^{n(n-1)^{2}} \cdot \frac{1}{|A|^{(n-1)^{2}}}=2^{84}$
$\Rightarrow 2^{n(n-1)+n(n-1)^{2}-(n-1)^{2}}=2^{84}\{\because|4|=2\}$
$\therefore n(n-1)+(n-1)^{3}=84$
$\therefore n=5$
Let A be a symmetric matrix such that $\mathrm{|A|=2}$ and $\left[ {\matrix{ 2 & 1 \cr 3 & {{3 \over 2}} \cr } } \right]A = \left[ {\matrix{ 1 & 2 \cr \alpha & \beta \cr } } \right]$. If the sum of the diagonal elements of A is $s$, then $\frac{\beta s}{\alpha^2}$ is equal to __________.
Explanation:
$A = \left( {\matrix{ a & c \cr c & b \cr } } \right)$
$|A| = ab - {c^2} = 2$ ...... (1)
$\left( {\matrix{ 2 & 1 \cr 3 & {{3 \over 2}} \cr } } \right)\left( {\matrix{ a & c \cr c & b \cr } } \right) = \left( {\matrix{ 1 & 2 \cr \alpha & \beta \cr } } \right)$
$2a + c = 1$ ..... (2)
$2c + b = 2$ ..... (3)
$3a + {3 \over 2}c = \alpha $ .... (4)
$3c + {3 \over 2}b = \beta $ ..... (5)
From (1), (2) and (3)
$a = {3 \over 4},b = 3,c = - {1 \over 2}$
$\Rightarrow$ Now $\alpha = {6 \over 4}$
$\beta = 3$
$s = {{15} \over 4}$
${{\beta s} \over {{\alpha ^2}}} = {{3 \times {{15} \over 4}} \over {{{\left( {{6 \over 4}} \right)}^2}}} = {{{{45} \over 4}} \over {{9 \over 4}}} = 5$
Let $\mathrm{A_1,A_2,A_3}$ be the three A.P. with the same common difference d and having their first terms as $\mathrm{A,A+1,A+2}$, respectively. Let a, b, c be the $\mathrm{7^{th},9^{th},17^{th}}$ terms of $\mathrm{A_1,A_2,A_3}$, respective such that $\left| {\matrix{ a & 7 & 1 \cr {2b} & {17} & 1 \cr c & {17} & 1 \cr } } \right| + 70 = 0$.
If $a=29$, then the sum of first 20 terms of an AP whose first term is $c-a-b$ and common difference is $\frac{d}{12}$, is equal to ___________.
Explanation:
$ \begin{aligned} & b=A+8 d+1 \\\\ & c=A+16 d+2 \\\\ & \left|\begin{array}{ccc} a & 7 & 1 \\ 26 & 17 & 1 \\ c & 17 & 1 \end{array}\right|=-70 \\\\ & \Rightarrow\left|\begin{array}{ccc} A+6 d & 7 & 1 \\ 2 A+16 d+2 & 17 & 1 \\ A+16 d+2 & 17 & 1 \end{array}\right|=-70 \\\\ & R_{3} \rightarrow R_{3}-R_{2}, \quad R_{2} \rightarrow R_{2}-R_{1} \\\\ & \Rightarrow\left|\begin{array}{ccc} A+6 d & 7 & 1 \\ A+10 d+2 & 10 & 0 \\ -A & 0 & 0 \end{array}\right|=-70 \end{aligned} $
$ \begin{aligned} \Rightarrow \quad & A=-7 \\\\ & a=A+6 d=29 \Rightarrow d=6 \\\\ & b=-7+48+1=42 \\\\ & c=-7+96+2=91 \\\\ & c-a-b=91-29-42=20 \\\\ & \text { Sum }=\frac{20}{2}\left[2 \times 20+19 \times \frac{6}{12}\right]=10\left[40+\frac{19}{2}\right]=495 \end{aligned} $
satisfy $4 m+n=22$ and $17 m+4 n=93$.
If $\operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A)))=3^{a} 5^{b} 6^{c}$ then $a+b+c$ is equal to :
Let for $A = \left[ {\matrix{ 1 & 2 & 3 \cr \alpha & 3 & 1 \cr 1 & 1 & 2 \cr } } \right],|A| = 2$. If $\mathrm{|2\,adj\,(2\,adj\,(2A))| = {32^n}}$, then $3n + \alpha $ is equal to
If the system of equations
$2 x+y-z=5$
$2 x-5 y+\lambda z=\mu$
$x+2 y-5 z=7$
has infinitely many solutions, then $(\lambda+\mu)^{2}+(\lambda-\mu)^{2}$ is equal to
For the system of linear equations
$2 x+4 y+2 a z=b$
$x+2 y+3 z=4$
$2 x-5 y+2 z=8$
which of the following is NOT correct?
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
$ \begin{aligned} & x+2 y+z=7 \\\\ & x+\alpha z=11 \\\\ & 2 x-3 y+\beta z=\gamma \end{aligned} $
Match each entry in List-I to the correct entries in List-II.
| List - I | List - II |
|---|---|
| (P) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has | (1) a unique solution |
| (Q) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has | (2) no solution |
| (R) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$, then the system has | (3) infinitely many solutions |
| (S) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has | (4) $x=11, y=-2$ and $z=0$ as a solution |
| (5) $x=-15, y=4$ and $z=0$ as a solution |
The correct option is:
Then the number of invertible matrices in $R$ is :
Explanation:
$\begin{gathered}R=\left[\begin{array}{lll}a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0\end{array}\right] \\\\ a, b, c, d, \in\{0,3,5,7,11,13,17,19\}\end{gathered}$
Number of invertible matrices $=$ (Total matrices $)-$ (Non Invertible matrices)
$\begin{aligned} & \text { Total matrices }=\begin{array}{cccc}a, & b, & c, & d \\ \downarrow& \downarrow & \downarrow & \downarrow \\ 8 & 8 & 8 & 8\end{array} \\\\ & =8 \times 8 \times 8 \times 8=8^4=4096 \\ & \end{aligned}$
For Non-invertible matrices,
$ \begin{aligned} & |R|=0 \\\\ & |R|=-5(a d-b c)=0 \end{aligned} $
Cases when both side are zero.
(i) All four $a, b, c, d$ are zero.
$ a d=b c=0 \quad 1 \text { ways } $
(ii) Three zero and one different digit used for $a, b$, $c, d$.
$ \Rightarrow a d=b c $
Select three from four $a, b, c, d$ assign them zero.
$ \text { i.e., }{ }^4 C_3 \times 1 \times 7=28 \text { ways } $
(iii) Two zero and two different digits
Hence $2 \times 7 \times 2 \times 7=196$ ways
Case II: When both side are same but non zero number.
$ a d=b c \neq 0 $
(i) All four $a, b, c, d$ are same.
i.e., $a d=b c$ ( 7 ways)
(ii) Two alike & two alike of another.
$ a d=b c $
$ { }^7 \mathrm{C}_1 \times{ }^6 \mathrm{C}_1 \times 2 !=84 \text { ways } $
Total number of non invertible matrices are
$ \begin{aligned} & =1+28+196+7+84 \\\\ & =316 \end{aligned} $
Hence number of invertible matric
$ \begin{aligned} & =8^4-316 \\\\ & =3780 \end{aligned} $
If $X_{4 \times 3}, Y_{4 \times 3}$ and $P_{2 \times 3}$ are the matrices, then the order of the matrix $\left[P\left(X^T Y\right)^{-1} P^T\right]^T$ is
$4 \times 3$
$3 \times 4$
$3 \times 3$
$2 \times 2$