Matrices and Determinants
Consider the matrices : $A=\left[\begin{array}{cc}2 & -5 \\ 3 & m\end{array}\right], B=\left[\begin{array}{l}20 \\ m\end{array}\right]$ and $X=\left[\begin{array}{l}x \\ y\end{array}\right]$. Let the set of all $m$, for which the system of equations $A X=B$ has a negative solution (i.e., $x<0$ and $y<0$), be the interval $(a, b)$. Then $8 \int_\limits a^b|A| d m$ is equal to _________.
Explanation:
$\begin{aligned} & A X=B \\ & 2 x-5 y=20 \\ & 3 x+m y=m \\ & \Rightarrow 3\left(\frac{20+5 y}{2}\right)+m y=m \end{aligned}$
$\begin{aligned} & \Rightarrow 30+\frac{15}{2} y+m y=m \\ & \Rightarrow y\left(\frac{15}{2}+m\right)=m-30 \\ & \Rightarrow y=\frac{m-30}{\frac{15}{2}+m}<0 \Rightarrow m \in\left(-\frac{15}{2}, 30\right) \end{aligned}$
Similarly : $3 x+m\left(\frac{2 x-20}{5}\right)=m$
$\begin{aligned} \Rightarrow & 3 x+\frac{2 m x}{5}-\frac{20 m}{5}=m \\ \Rightarrow & \frac{15 x+2 m x}{5}=5 m \Rightarrow x=\frac{25 m}{15+2 m} \\ & x<0 \Rightarrow \frac{25 m}{15+2 m}<0 \Rightarrow m \in\left(-\frac{15}{2}, 0\right) \\ \therefore \quad & m \in\left(-\frac{15}{2}, 0\right) \\ & a=-\frac{15}{2}, b=0 \\ & 8 \int_\limits{-\frac{15}{2}}^0(2 m+15) d m=450 \\ \end{aligned}$
Let $A$ be a non-singular matrix of order 3. If $\operatorname{det}(3 \operatorname{adj}(2 \operatorname{adj}((\operatorname{det} A) A)))=3^{-13} \cdot 2^{-10}$ and $\operatorname{det}(3\operatorname{adj}(2 \mathrm{A}))=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}$, then $|3 \mathrm{~m}+2 \mathrm{n}|$ is equal to _________.
Explanation:
$|\operatorname{adj}(2 \operatorname{adj}(|A| A))|=3^{-13} \cdot 2^{-10}$
Let $|A| A=B \Rightarrow|B|=\| A|A|=|A|^3|A|=|A|^4$
$\begin{aligned} \Rightarrow \quad & \operatorname{adj}(|A| A)=(\operatorname{adj} B) \\ \Rightarrow \quad & 2 \operatorname{adj}(|A| A)=(2 \operatorname{adj} B)=C \text { (say) } \\ & |\operatorname{3adj}(C)|=3^3 \cdot|C|^2 \end{aligned}$
$\begin{aligned} & |C|=|(2 \operatorname{adj} B)|=2^3|B|^2=2^3 \cdot\left|A^4\right|^2=2^3 \cdot|A|^8 \\ & \Rightarrow|\operatorname{3adj} C|=3^3 \cdot\left(2^3|A|^8\right)^2=3^{-13} \cdot 2^{-10} \\ & \quad=2^6|A|^{16}=3^{-16} \cdot 2^{-10} \\ & \Rightarrow|A|^{16}=(3 \cdot 2)^{-16}=\left(\frac{1}{6}\right)^{16} \\ & \Rightarrow|A|= \pm \frac{1}{6} \end{aligned}$
$\begin{array}{r} \mid \text { 3adj }\left.2 A\left|=3^3\right| 2 A\right|^2=3^3 \cdot\left(2^3|A|\right)^2=3^3 \cdot 2^6|A|^2 \\ =3^3 \cdot 2^6 \cdot \frac{1}{36}=\frac{27 \times 64}{36}=48 \end{array}$
$ \begin{aligned} & \Rightarrow 2^m \cdot 3^n=2^4 \cdot 3^1 \Rightarrow m=4 \\ & \qquad n=1 \\ & \Rightarrow|3 \times 4+2 \times 1|=14 \\ \end{aligned}$
Let $A=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]$. If the sum of the diagonal elements of $A^{13}$ is $3^n$, then $n$ is equal to ________.
Explanation:
$\begin{aligned} & A=\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array}\right] \\ & A^2=\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array}\right] \\ & A^2=\left[\begin{array}{cc} 3 & -3 \\ 3 & 0 \end{array}\right]=3\left[\begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array}\right] \\ & A^4=9\left[\begin{array}{ll} 0 & -1 \\ 1 & -1 \end{array}\right] \\ & A^8=81\left[\begin{array}{ll} -1 & 1 \\ -1 & 0 \end{array}\right] \\ & A^{12}=729\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & A^{13}=729\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array}\right] \\ & A^{13}=\left[\begin{array}{cc} 1458 & -729 \\ 729 & 729 \end{array}\right] \end{aligned}$
$\begin{aligned} & \text { Sum }=2187=3^n \\ & 3^7=3^n \\ & n=7 \end{aligned}$
If the system of equations
$\begin{aligned} & 2 x+7 y+\lambda z=3 \\ & 3 x+2 y+5 z=4 \\ & x+\mu y+32 z=-1 \end{aligned}$
has infinitely many solutions, then $(\lambda-\mu)$ is equal to ______ :
Explanation:
To determine if the system of equations:
$\begin{aligned} 2x + 7y + \lambda z = 3 \\ 3x + 2y + 5z = 4 \\ x + \mu y + 32z = -1 \end{aligned}$
has infinitely many solutions, we must use Cramer's rule.
The determinants are calculated as follows:
$\begin{aligned} \Delta &= -2\lambda + 3\lambda\mu - 10\mu - 509 \\ \Delta_1 &= 2\lambda + 3\lambda\mu - 15\mu - 739 \\ \Delta_2 &= -7\lambda - 7 \\ \Delta_3 &= \mu + 39 \end{aligned}$
To have infinitely many solutions, the determinants must satisfy:
$\begin{aligned} \Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0 \end{aligned}$
Solving these equations, we find:
$\lambda = -1, \mu = -39$
Thus, the value of $ \lambda - \mu $ is:
$ \lambda - \mu = 38 $
Let $\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in \mathbb{R}$. If $x(\alpha, 1,2)+y(1, \beta, 2)+z(2,3, \gamma)=(0,0,0)$ for some $x, y, z \in \mathbb{R}, x y z \neq 0$, then $6 \alpha+4 \beta+\gamma$ is equal to _________.
Explanation:
Given that $\alpha \beta \gamma = 45$ and $\alpha, \beta, \gamma \in \mathbb{R}$, consider the equation $x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0)$ for some $x, y, z \in \mathbb{R}$ where $x y z \neq 0$. To find the value of $6 \alpha + 4 \beta + \gamma$, follow these steps:
- Express the given equation in matrix form:
$ \begin{aligned} & \alpha x + y + 2 z = 0 \\ & x + \beta y + 3 z = 0 \\ & 2 x + 2 y + \gamma z = 0 \end{aligned} $
- Since $x, y, z \neq 0$, the determinant of the coefficients matrix must be zero:
$ \left|\begin{array}{ccc} \alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma \end{array}\right| = 0 $
- Calculate the determinant of the matrix:
$ \alpha \beta \gamma - 6 \alpha - 4 \beta - \gamma + 10 = 0 $
- Given $\alpha \beta \gamma = 45$, substitute this value into the equation:
$ 45 - 6 \alpha - 4 \beta - \gamma + 10 = 0 $
- Simplify the equation:
$ 45 + 10 = 6 \alpha + 4 \beta + \gamma $
- Thus,
$ 6 \alpha + 4 \beta + \gamma = 55 $
So, the value of $6 \alpha + 4 \beta + \gamma$ is 55.
Let $A$ be a $2 \times 2$ symmetric matrix such that $A\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right]$ and the determinant of $A$ be 1 . If $A^{-1}=\alpha A+\beta I$, where $I$ is an identity matrix of order $2 \times 2$, then $\alpha+\beta$ equals _________.
Explanation:
Let $A=\left[\begin{array}{ll}a & b \\ b & c\end{array}\right]$
$|A|=1 \Rightarrow a c-b^2=0 \quad \text{... (i)}$
$\text { Given }\left[\begin{array}{ll} a & b \\ b & c \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \end{array}\right]=\left[\begin{array}{l} 3 \\ 7 \end{array}\right]$
$\begin{aligned} & \Rightarrow \quad a+b=3 \quad \text{... (ii)}\\ & \text { and } b+c=7 \quad \text{... (iii)} \end{aligned}$
from (i), (ii) and (iii) $a=1, b=2, c=5$
$\begin{aligned} \Rightarrow & A=\left[\begin{array}{ll} 1 & 2 \\ 2 & 5 \end{array}\right] \Rightarrow A^{-1}=\left[\begin{array}{cc} 5 & -2 \\ -2 & 1 \end{array}\right] \\ & \text { Given } A^{-1}=\alpha A+\beta I \\ \Rightarrow & {\left[\begin{array}{cc} 5 & -2 \\ -2 & 1 \end{array}\right]=\alpha\left[\begin{array}{ll} 1 & 2 \\ 2 & 5 \end{array}\right]+\beta\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] } \\ \Rightarrow & \alpha=-1 \text { and } \beta=6 \\ & \alpha+\beta=5 \end{aligned}$
Let $A$ be a square matrix of order 2 such that $|A|=2$ and the sum of its diagonal elements is $-$3 . If the points $(x, y)$ satisfying $\mathrm{A}^2+x \mathrm{~A}+y \mathrm{I}=\mathrm{O}$ lie on a hyperbola, whose transverse axis is parallel to the $x$-axis, eccentricity is $\mathrm{e}$ and the length of the latus rectum is $l$, then $\mathrm{e}^4+l^4$ is equal to ________.
Explanation:
$|A|=2 \sum \mathrm{dia}=-3$
$\therefore \quad$ character equation : $A^2+3 A+2 I=0$
$\Rightarrow x=3 \quad y=2$
$\because$ We are getting only one point $(3,2)$ but its given many points satisfy this equation.
Moreover hyperbola whose transverse axis is $x$ axis and passing through $(3,2)$ is not unique.
$\therefore$ multiple value of '$e$' and $L(L R)$ is possible.
We'll not get a unique result.
Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=3\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$. Then the maximum value of $\operatorname{det}(\mathrm{A})$ is _________.
Explanation:
Let $A = \left[ {\matrix{ {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr } } \right]$
Now
$A\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=3\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]$
$\left[ {\matrix{ {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = \left[ {\matrix{ 3 \cr 3 \cr 3 \cr } } \right]$
$\begin{aligned} & a_{11}+a_{12}+a_{13}=3 \\ & a_{21}+a_{22}+a_{23}=3 \\ & a_{31}+a_{32}+a_{33}=3 \end{aligned}$
Now for maximum value of $\operatorname{det}(A)=a_{i j}\left\{\begin{array}{ll}0 & i \neq j \\ 3 & i=j\end{array}\right\}$
$\therefore|A|=27$
Explanation:
$\begin{aligned} & \mathrm{AX}=\lambda \mathrm{X} \\\\ & \mathrm{A}^2 \mathrm{X}=\lambda \mathrm{AX} \\\\ & \mathrm{X}=\lambda(\lambda \mathrm{X}) \\\\ & \mathrm{X}=\lambda^2 \mathrm{X} \\\\ & \mathrm{X}\left(\lambda^2-1\right)=0 \\\\ & \lambda^2=1 \\\\ & \lambda= \pm 1\end{aligned}$
Sum of square of all possible values $=2$
Let A be a $3 \times 3$ matrix and $\operatorname{det}(A)=2$. If $n=\operatorname{det}(\underbrace{\operatorname{adj}(\operatorname{adj}(\ldots . .(\operatorname{adj} A))}_{2024-\text { times }}))$, then the remainder when $n$ is divided by 9 is equal to __________.
Explanation:
$\begin{aligned} & |\mathrm{A}|=2 \\ & \underbrace{\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \ldots . .(\mathrm{a})))}_{2024 \text { times }}=|\mathrm{A}|^{(\mathrm{n}-1)^{2024}} \\ & =|\mathrm{A}|^{2024} \\ & =2^{2^{2024}} \end{aligned}$
$\begin{aligned} & 2^{2024}=\left(2^2\right) 2^{2022}=4(8)^{674}=4(9-1)^{674} \\ & \Rightarrow 2^{2024} \equiv 4(\bmod 9) \\ & \Rightarrow 2^{2024} \equiv 9 \mathrm{~m}+4, \mathrm{~m} \leftarrow \text { even } \\ & 2^{9 \mathrm{~m}+4} \equiv 16 \cdot\left(2^3\right)^{3 \mathrm{~m}} \equiv 16(\bmod 9) \\ & \quad \equiv 7 \end{aligned}$
Let for any three distinct consecutive terms $a, b, c$ of an A.P, the lines $a x+b y+c=0$ be concurrent at the point $P$ and $Q(\alpha, \beta)$ be a point such that the system of equations
$\begin{aligned} & x+y+z=6, \\ & 2 x+5 y+\alpha z=\beta \text { and } \end{aligned}$
$x+2 y+3 z=4$, has infinitely many solutions. Then $(P Q)^2$ is equal to _________.
Explanation:
$\because \mathrm{a}, \mathrm{b}, \mathrm{c}$ and in A.P
$\Rightarrow 2 b=a+c \Rightarrow a-2 b+c=0$
$\therefore \mathrm{ax}+\mathrm{by}+\mathrm{c}$ passes through fixed point $(1,-2)$
$\therefore \mathrm{P}=(1,-2)$
For infinite solution,
$\begin{aligned} & D=D_1=D_2=D_3=0 \\ & D:\left|\begin{array}{lll} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{array}\right|=0 \\ & \Rightarrow \alpha=8 \\ & D_1:\left|\begin{array}{lll} 6 & 1 & 1 \\ \beta & 5 & \alpha \\ 4 & 2 & 3 \end{array}\right|=0 \Rightarrow \beta=6 \\ & \therefore Q=(8,6) \\ & \therefore Q^2=113 \end{aligned}$
Let $A$ be a $2 \times 2$ real matrix and $I$ be the identity matrix of order 2. If the roots of the equation $|\mathrm{A}-x \mathrm{I}|=0$ be $-1$ and 3, then the sum of the diagonal elements of the matrix $\mathrm{A}^2$ is
Explanation:
$|A-x I|=0$
Roots are $-$1 and 3
Sum of roots $=\operatorname{tr}(A)=2$
Product of roots $=|\mathrm{A}|=-3$
Let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
We have $\mathrm{a}+\mathrm{d}=2$
$\mathrm{ad}-\mathrm{bc}=-3$
$A^2=\left[\begin{array}{ll}a & b \\ c & d \end{array}\right] \times\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll} a^2+b c & a b+b d \\ a c+c d & b c+d^2 \end{array}\right]$
We need $a^2+b c+b c+d^2$
$\begin{aligned} & =a^2+2 b c+d^2 \\ & =(a+d)^2-2 a d+2 b c \\ & =4-2(a d-b c) \\ & =4-2(-3) \\ & =4+6 \\ & =10 \end{aligned}$
$ \mathrm{AB}_1=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right], \mathrm{AB}_2=\left[\begin{array}{l} 2 \\ 3 \\ 0 \end{array}\right], \quad \mathrm{AB}_3=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] $
If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^3+\beta^3$ is equal to ____________.
Explanation:
$\mathrm{A}=\left[\begin{array}{lll} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right] \quad \mathrm{B}=\left[\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3\right]$
$\mathrm{B}_1=\left[\begin{array}{l} \mathrm{x}_1 \\ \mathrm{y}_1 \\ \mathrm{z}_1 \end{array}\right], \quad \mathrm{B}_2=\left[\begin{array}{l} \mathrm{x}_2 \\ \mathrm{y}_2 \\ \mathrm{z}_2 \end{array}\right], \quad \mathrm{B}_3=\left[\begin{array}{l} \mathrm{x}_3 \\ \mathrm{y}_3 \\ \mathrm{z}_3 \end{array}\right]$
$\mathrm{AB}_1=\left[\begin{array}{lll} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]\left[\begin{array}{l} \mathrm{x}_1 \\ \mathrm{y}_1 \\ \mathrm{z}_1 \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right]$
$\begin{gathered} \mathrm{x}_1=1, \mathrm{y}_1=-1, \mathrm{z}_1=-1 \\ \mathrm{AB}_2=\left[\begin{array}{lll} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]\left[\begin{array}{l} \mathrm{x}_2 \\ \mathrm{y}_2 \\ \mathrm{z}_2 \end{array}\right]=\left[\begin{array}{l} 2 \\ 3 \\ 0 \end{array}\right] \\ \mathrm{x}_2=2, \mathrm{y}_2=1, \mathrm{z}_2=-2 \\ \mathrm{AB}_3=\left[\begin{array}{lll} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]\left[\begin{array}{l} \mathrm{x}_3 \\ \mathrm{y}_3 \\ \mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] \\ \mathrm{x}_3=2, \mathrm{y}_3=0, \mathrm{z}_3=-1 \\ \mathrm{~B}=\left[\begin{array}{ccc} 1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1 \end{array}\right] \\ \alpha=|\mathrm{B}|=3 \\ \beta=1 \\ \alpha^3+\beta^3=27+1=28 \end{gathered}$
Let $B=\left[\begin{array}{ll}1 & 3 \\ 1 & 5\end{array}\right]$ and $A$ be a $2 \times 2$ matrix such that $A B^{-1}=A^{-1}$. If $B C B^{-1}=A$ and $C^4+\alpha C^2+\beta I=O$, then $2 \beta-\alpha$ is equal to
Let $\lambda, \mu \in \mathbf{R}$. If the system of equations
$\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7 x+11 y-9 z=2 \\ & 97 x+155 y-189 z=\mu \end{aligned}$
has infinitely many solutions, then $\mu+2 \lambda$ is equal to :
If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$, then $\frac{\mathrm{a}}{\alpha-\mathrm{a}}+\frac{\mathrm{b}}{\beta-\mathrm{b}}+\frac{\gamma}{\gamma-\mathrm{c}}$ is equal to :
If the system of equations $x+4 y-z=\lambda, 7 x+9 y+\mu z=-3,5 x+y+2 z=-1$ has infinitely many solutions, then $(2 \mu+3 \lambda)$ is equal to :
Let $A=\left[\begin{array}{lll}2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b\end{array}\right]$. If $A^3=4 A^2-A-21 I$, where $I$ is the identity matrix of order $3 \times 3$, then $2 a+3 b$ is equal to
If $A$ is a square matrix of order 3 such that $\operatorname{det}(A)=3$ and $\operatorname{det}\left(\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 \mathrm{~A})^{-1}\right)\right)\right)\right)\right)=2^{\mathrm{m}} 3^{\mathrm{n}}$, then $\mathrm{m}+2 \mathrm{n}$ is equal to :
For $\alpha, \beta \in \mathbb{R}$ and a natural number $n$, let $A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$ is
The values of $m, n$, for which the system of equations
$\begin{aligned} & x+y+z=4, \\ & 2 x+5 y+5 z=17, \\ & x+2 y+\mathrm{m} z=\mathrm{n} \end{aligned}$
has infinitely many solutions, satisfy the equation :
Let $\alpha \beta \neq 0$ and $A=\left[\begin{array}{rrr}\beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha\end{array}\right]$. If $B=\left[\begin{array}{rrr}3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta\end{array}\right]$ is the matrix of cofactors of the elements of $A$, then $\operatorname{det}(A B)$ is equal to :
Let A and B be two square matrices of order 3 such that $\mathrm{|A|=3}$ and $\mathrm{|B|=2}$. Then $|\mathrm{A}^{\mathrm{T}} \mathrm{A}(\operatorname{adj}(2 \mathrm{~A}))^{-1}(\operatorname{adj}(4 \mathrm{~B}))(\operatorname{adj}(\mathrm{AB}))^{-1} \mathrm{AA}^{\mathrm{T}}|$ is equal to :
If the system of equations
$\begin{array}{r} 11 x+y+\lambda z=-5 \\ 2 x+3 y+5 z=3 \\ 8 x-19 y-39 z=\mu \end{array}$
has infinitely many solutions, then $\lambda^4-\mu$ is equal to :
Let $A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$ and $B=I+\operatorname{adj}(A)+(\operatorname{adj} A)^2+\ldots+(\operatorname{adj} A)^{10}$. Then, the sum of all the elements of the matrix $B$ is:
Let $\alpha \in(0, \infty)$ and $A=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]$. If $\operatorname{det}\left(\operatorname{adj}\left(2 A-A^T\right) \cdot \operatorname{adj}\left(A-2 A^T\right)\right)=2^8$, then $(\operatorname{det}(A))^2$ is equal to:
If the system of equations
$\begin{aligned} & x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\ & x+(\cos \alpha) y+(\sin \alpha) z=0 \\ & x+(\sin \alpha) y-(\cos \alpha) z=0 \end{aligned}$
has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :
$ \begin{aligned} & 2 x+3 y-z=5 \\\\ & x+\alpha y+3 z=-4 \\\\ & 3 x-y+\beta z=7 \end{aligned} $
has infinitely many solutions, then $13 \alpha \beta$ is equal to :
Let $A$ be a $3 \times 3$ real matrix such that
$A\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right)=2\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right)=4\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)=2\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) \text {. }$
Then, the system $(A-3 I)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$ has :
If the system of linear equations
$\begin{aligned} & x-2 y+z=-4 \\ & 2 x+\alpha y+3 z=5 \\ & 3 x-y+\beta z=3 \end{aligned}$
has infinitely many solutions, then $12 \alpha+13 \beta$ is equal to
Let $R=\left(\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right)$ be a non-zero $3 \times 3$ matrix, where $x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right) \neq 0, \theta \in(0,2 \pi)$. For a square matrix $M$, let trace $(M)$ denote the sum of all the diagonal entries of $M$. Then, among the statements:
(I) Trace $(R)=0$
(II) If trace $(\operatorname{adj}(\operatorname{adj}(R))=0$, then $R$ has exactly one non-zero entry.
Consider the system of linear equations $x+y+z=5, x+2 y+\lambda^2 z=9, x+3 y+\lambda z=\mu$, where $\lambda, \mu \in \mathbb{R}$. Then, which of the following statement is NOT correct?
Consider the system of linear equations $x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda^2 z=\mu^2+15$ where $\lambda, \mu \in \mathbf{R}$. Which one of the following statements is NOT correct ?
Let $A=\left[\begin{array}{ccc}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{array}\right]$ and $P=\left[\begin{array}{lll}1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5\end{array}\right]$. The sum of the prime factors of $\left|P^{-1} A P-2 I\right|$ is equal to
$\text { Let } A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{array}\right] \text { and }|2 \mathrm{~A}|^3=2^{21} \text { where } \alpha, \beta \in Z \text {, Then a value of } \alpha \text { is }$
Let $\mathrm{A}$ be a square matrix such that $\mathrm{AA}^{\mathrm{T}}=\mathrm{I}$. Then $\frac{1}{2} A\left[\left(A+A^T\right)^2+\left(A-A^T\right)^2\right]$ is equal to
The values of $\alpha$, for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$, lie in the interval
Given below are two statements :
Statement I : $ f(-x)$ is the inverse of the matrix $f(x)$.
Statement II : $f(x) f(y)=f(x+y)$.
In the light of the above statements, choose the correct answer from the options given below :
Let $\mathrm{D}_{\mathrm{k}}=\left|\begin{array}{ccc}1 & 2 k & 2 k-1 \\ n & n^{2}+n+2 & n^{2} \\ n & n^{2}+n & n^{2}+n+2\end{array}\right|$. If $\sum_\limits{k=1}^{n} \mathrm{D}_{\mathrm{k}}=96$, then $n$ is equal to _____________.
Explanation:
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?