Let $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ and $B$ be two matrices such that $A^{100} = 100B + I$. Then the sum of all the elements of $B^{100}$ is _______
Explanation:
Decompose Matrix A We can write matrix $A$ as the sum of the Identity matrix $I$ and a $\text {residual matrix } C:$
$ \begin{aligned} & \quad A=I+C=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{ll} 2 & -4 \\ 1 & -2 \end{array}\right], \quad \text { Let } C=\left[\begin{array}{ll} 2 & -4 \\ 1 & -2 \end{array}\right] . \\ & C^2=\left[\begin{array}{ll} 2 & -4 \\ 1 & -2 \end{array}\right]\left[\begin{array}{ll} 2 & -4 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{ll} 4-4 & -8+8 \\ 2-2 & -4+4 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0 \end{aligned} $
Since $C^2=0, C$ is a nilpotent matrix. Consequently, $C^n=0$ for all $n \geq 2$.
Find $A^{100}$ using Binomial Expansion :
$ A^{100}=(I+C)^{100}={ }^{100} C_0 I^{100}+{ }^{100} C_1 I^{99} C+{ }^{100} C_2 I^{98} C^2+\cdots+{ }^{100} C_{100} C^{100} $
Since $C^2=C^3=\cdots=C^{100}=0$, the expansion simplifies to :
$ \begin{gathered} A^{100}={ }^{100} C_0 I+{ }^{100} C_1 C \\ A^{100}=1 \cdot I+100 \cdot C \\ A^{100}=I+100 C \end{gathered} $
Find Matrix B :
The problem states that $A^{100}=100 B+I$. Comparing this to our result:
$ \begin{gathered} 100 B+I=100 C+I \\ 100 B=100 C \\ B=C \end{gathered} $
Calculate the Sum of Elements of $B^{100}$ :
Since $B=C$ and we know $C^2=0$, then: $\quad B^2=0$
Consequently, $B^{100}=0$ (the zero matrix).
$B^{100}=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
The sum of all the elements of $B^{100}$ is 0 .
The number of $3 \times 2$ matrices A , which can be formed using the elements of the set $\{-2,-1,0,1,2\}$ such that the sum of all the diagonal elements of $\mathrm{A}^{\mathrm{T}} \mathrm{A}$ is 5 , is
$\_\_\_\_$
Explanation:
Let matrix $A$ of order $3 \times 2$ is
$A=\left[\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right], a, b, c, d, e, f \in\{-2,-1,0,1,2\}$
$A^T=\left[\begin{array}{lll}a & c & e \\ b & d & f\end{array}\right]$
$ A^T A=\left[\begin{array}{lll} a & c & e \\ b & d & f \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \\ e & f \end{array}\right]=\left[\begin{array}{cc} a^2+c^2+e^2 & a b+c d+e f \\ a b+c d+e f & b^2+d^2+f^2 \end{array}\right] $
sum of diagonal elements of $A^T A=5$
$ a^2+c^2+e^2+b^2+d^2+f^2=5 $
possible values of squares is $\{0,1,4\}$
Case 1 : one element is 4,1 element is 1 and four elements are 0 .
one element is $4=2$ choices $\{-2,2\}$
one element is $1=2$ choices $\{-1,1\}$
4 element is zero $=1$ choice $\{0\}$
Selecting one element 4 out of 6 choice $\left\{a^2, b^2, c^2, d^2, e^2, f^2\right\}$, and have 2 choice $\{-2,2\}$, total $={ }^6 C_1 \times 2=12$
Selecting one element which is 1 from remaining 5 element $={ }^5 C_1$, and have two choice $\{-1,1\}$, total $={ }^5 C_1 \times 2=10$
Selecting 4 element which is zero from remaining 4 elements $={ }^4 C_4=1$ total matrices for Case $1=12 \times 10 \times 1=120$
Case 2 : five elements are 1 and one element is 0 .
squares $a^2, b^2, c^2, d^2, e^2, f^2 \in\{1,1,1,1,1,0\}$
selection of 5 one from $a^2, b^2, c^2, d^2, e^2, f^2={ }^6 C_5$
and for 1 there are two choice for each elements $a, b, c, d, e, f=\{-1,1\}$, total $=2^5$
for 0 only 1 choices
total matrices of case $2={ }^6 C_5 \times 2^5=192$
total number of such matrices = Case $1+$ Case 2
$=120+192=312$
Let $A=\left[\begin{array}{ccc}0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0\end{array}\right]$ and $B$ be a matrix such that $B(I-A)=I+A$. Then the sum of the diagonal elements of $\mathrm{B}^{\mathrm{T}} \mathrm{B}$ is equal to $\_\_\_\_$
Explanation:
Observing properties of Matrix $A$
$ \begin{aligned} A & =\left[\begin{array}{ccc} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{array}\right] \\ A^T & =\left[\begin{array}{ccc} 0 & -2 & 3 \\ 2 & 0 & -1 \\ -3 & 1 & 0 \end{array}\right]=-A \end{aligned} $
Rearranging the given equation
$ \begin{aligned} & B(I-A)=I+A \\ & B=(I+A)(I-A)^{-1} \end{aligned} $
Finding $B^T$
$ \begin{aligned} B^T & =\left[(I+A)(I-A)^{-1}\right]^T \\ B^T & =\left((I-A)^{-1}\right)^T(I+A)^T \\ B^T & =\left(I-A^T\right)^{-1}\left(I+A^T\right) \end{aligned} $
Since $A^T=-A$ :
$ \begin{aligned} B^T & =(I+A)^{-1}(I-A) \\ B^T B & =\left[(I+A)^{-1}(I-A)\right] \cdot\left[(I+A)(I-A)^{-1}\right] \\ B^T B & =(I+A)^{-1}(I+A)(I-A)(I-A)^{-1} \\ B^T B & =I \cdot I=I \\ B^T B & =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned} $
the sum of diagonal elements $=1+1+1=3$
The sum of the diagonal elements of $B^T B$ is 3 .
Let $|\mathrm{A}|=6$, where A is a $3 \times 3$ matrix. If $\left|\operatorname{adj}\left(\operatorname{adj}\left(\mathrm{A}^2 \cdot \operatorname{adj}(2 \mathrm{~A})\right)\right)\right|=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}, \mathrm{m}, \mathrm{n} \in \mathbf{N}$, then $\mathrm{m}+\mathrm{n}$ is equal to
$\_\_\_\_$ .
Explanation:
$ \begin{aligned} & (2 \mathrm{~A})(\operatorname{adj} 2 \mathrm{~A})=|2 \mathrm{~A}| /_{3 \times 3} \\ & =2^3 \cdot|\mathrm{~A}| / 3_{\times 3} \\ & =48 \mathrm{I} \\ & \Rightarrow \operatorname{Aadj}(2 \mathrm{~A})=24 \mathrm{I} \\ & \Rightarrow \mathrm{~A}^2(\operatorname{adj} 2 \mathrm{~A})=24 \mathrm{~A} \\ & |\operatorname{adj}(3 \operatorname{adj}(24 \mathrm{~A}))|=|3 \operatorname{adj}(24 \mathrm{~A})|^{3-1} \\ & =\left(3^3\right)^2|(\operatorname{adj}(24 \mathrm{~A}))|^2 \\ & =3^6 \cdot\left(|24 \mathrm{~A}|^2\right)^2 \\ & =3^6|24 \mathrm{~A}|^4 \\ & =3^6 \cdot\left[(24)^3|\mathrm{~A}|\right]^4 \\ & =3^6 \cdot 24^{12} \cdot|\mathrm{~A}|^4=3^6 \cdot 24^{12} \cdot 6^4 \\ & =3^6 \cdot\left(2^3\right)^{12} \cdot 3^{12} \cdot 3^4 \cdot 2^4 \\ & =2^{40} \cdot 3^{22} \\ & \Rightarrow m+n=62 \end{aligned} $
Let A be a $3 \times 3$ matrix such that $\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\mathrm{O}$. If $\mathrm{A}\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{l}3 \\ 3 \\ 2\end{array}\right], \mathrm{A}^2\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}-3 \\ 19 \\ -24\end{array}\right]$ and $\operatorname{det}(\operatorname{adj}(2 \operatorname{adj}(\mathrm{~A}+\mathrm{I})))=(2)^\alpha \cdot(3)^\beta \cdot(11)^\gamma, \alpha, \beta, \gamma$ are non-negative integers, then $\alpha+\beta+\gamma$ is equal to $\_\_\_\_$
Explanation:
$ \begin{aligned} & A=-A^T \\ & A=\left[\begin{array}{ccc} 0 & \alpha & \beta \\ -\alpha & 0 & \gamma \\ -\beta & -\gamma & 0 \end{array}\right] \\ & {\left[\begin{array}{ccc} 0 & \alpha & \beta \\ -\alpha & 0 & \gamma \\ -\beta & -\gamma & 0 \end{array}\right]\left[\begin{array}{c} 1 \\ -1 \\ 0 \end{array}\right]=\left[\begin{array}{l} 3 \\ 3 \\ 2 \end{array}\right]} \\ & \alpha=-3,-\beta+\gamma=2 \\ & {\left[\begin{array}{ccc} 0 & \alpha & \beta \\ -\alpha & 0 & \gamma \\ -\beta & -\gamma & 0 \end{array}\right]\left[\begin{array}{l} 3 \\ 3 \\ 2 \end{array}\right]=\left[\begin{array}{c} -3 \\ 19 \\ \beta=3 \end{array}\right]} \end{aligned} $
$ \begin{aligned} & |A+I|=\left|\begin{array}{ccc} 1 & -3 & 3 \\ 3 & 1 & 5 \\ -3 & -5 & 1 \end{array}\right|=44 \\ & \operatorname{det}(\operatorname{adj}(2 \operatorname{adj}(A+I)))=2^6 \operatorname{det}(\operatorname{adj}(\operatorname{adj}(A+I)))=2^6(44)^4=2^{14} \times 11^4 \end{aligned} $
For some $\alpha, \beta \in \mathbf{R}$, let $A=\left[\begin{array}{ll}\alpha & 2 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 1 \\ 1 & \beta\end{array}\right]$ be such that $A^2-4 A+2 I=B^2-3 B+I=O$. Then $\left(\operatorname{det}\left(\operatorname{adj}\left(A^3-B^3\right)\right)\right)^2$ is equal to $\_\_\_\_$ .
Explanation:
$ \begin{aligned} &\text { Using characteristic equation, }\\ &\begin{aligned} & \left|\begin{array}{cc} \alpha-\lambda & 2 \\ 1 & 2-\lambda \end{array}\right|=0 \\ & (\lambda-2)(\lambda-\alpha)-2=0 \\ & \lambda^2+\lambda(-2-\alpha)+2 \alpha-2=0 \\ & \Rightarrow A^2+A(-2-\alpha)+(2 \alpha-2) I=0 \\ & \Rightarrow-2-\alpha=-4 \Rightarrow \alpha=2 \end{aligned} \end{aligned} $
$ \begin{aligned} &\text { Similarly, }\\ &\begin{aligned} &\left|\begin{array}{cc} 1-K & 2 \\ 1 & \beta-K \end{array}\right|=0 \\ & \Rightarrow(K-\beta)(K-1)-1=0 \\ & K^2+K(-1-\beta)+\beta-1=0 \\ & \Rightarrow-1-\beta=-3 \Rightarrow \beta=2 \end{aligned} \end{aligned} $
$ \begin{aligned} & \Rightarrow A=\left[\begin{array}{ll} 2 & 2 \\ 1 & 2 \end{array}\right], B=\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right] \\ & \text { Let } A^3-B^3=\left[\begin{array}{cc} 15 & 20 \\ 6 & 7 \end{array}\right]=C, \operatorname{det}(C)=-15 \\ & \mid \text { adj }\left.C\right|^2=|C|^2=225 \end{aligned} $
The number of singular matrices of order 2 , whose elements are from the set $\{2,3,6,9\}$, is __________.
Explanation:
Let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
for $A$ to be singular matrix
$a d=b c$
Case 1: exactly 1 number is used $\Rightarrow{ }^4 C_1$ ways
Case 2 : exactly 2 numbers is used $\Rightarrow{ }^4 C_2$ ways
Case 3 : exactly 3 numbers used $\Rightarrow$ none will be singular.
Case 4: exactly 4 numbers is used
$\begin{aligned} & \Rightarrow a b=c d \Rightarrow 2 \times 9=3 \times 6 \\ & \Rightarrow{ }^4 C_1 \times 2!=8 \text { matrix } . \end{aligned}$
$\therefore$ Total ways $\Rightarrow 4+6 \times 4+8=36$ matrices.
Let $A=\left[\begin{array}{ccc}\cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right]$. If for some $\theta \in(0, \pi), A^2=A^T$, then the sum of the diagonal elements of the matrix $(\mathrm{A}+\mathrm{I})^3+(\mathrm{A}-\mathrm{I})^3-6 \mathrm{~A}$ is equal to _________ .
Explanation:
Note that $A$ is orthogonal:
$A A^T=A^T A=I \text { and } A^T=A^{-1}$
Given $A^2=A^T$, then:
$\begin{aligned} & A^3=I \\ & \operatorname{Tr}(A+I)^3+(A-l)^3-6 A=\operatorname{Tr}\left(2 A^3+6 A-6 A\right) \\ & =\operatorname{Tr}\left(2 A^3\right)=\operatorname{Tr}(2 I) \\ & \left(\text { Using }(A+l)^3+(A-I)^3=2 A^3+6 A \text { and } 2 A^3=2 I\right)= \\ & 6 \end{aligned}$
Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A=\left[\begin{array}{ccc}\lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2\end{array}\right],|A|=-1$. Let $B$ be the inverse of the matrix $\operatorname{adj}\left(\operatorname{Aadj}\left(A^2\right)\right)$. Then $|(\lambda \mathrm{B}+\mathrm{I})|$ is equal to______
Explanation:
$\begin{aligned} & B=\left[\operatorname{adj}\left(A \operatorname{adj}\left(A^2\right)\right)\right]^{-1} \\ & \operatorname{Adj}\left(A^2\right)=(\operatorname{adj} A)^2 \Rightarrow A \operatorname{adj}\left(A^2\right)=A \operatorname{adj}(A) \cdot(\operatorname{adj} A) \\ & =A\left(|A| A^{-1}\right)^2=|A|^2\left(A^{-1}\right)=A^{-1} \\ & \Rightarrow B=\left(\operatorname{adj}\left(A^{-1}\right)\right)^{-1}=\left(\left|\left(A^{-1}\right)\right| A\right)^{-1}=\frac{A^{-1}}{-1}=-A^{-1} \\ & \Rightarrow B=-A^{-1} \end{aligned}$
$\begin{aligned} & |A|=-1=\left|\begin{array}{lll} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{array}\right|=-1 \Rightarrow \lambda=3 \\ & |3 B+I|=\left|I-3 A^{-1}\right|=\frac{|A|\left|I-3 A^{-1}\right|}{|A|}=\frac{|A-3 I|}{|A|} \\ & =\frac{|A-3 I|}{-1}=\frac{\left|\begin{array}{ccc} 0 & 2 & 3 \\ 4 & 2 & 6 \\ 7 & -1 & -1 \end{array}\right|}{-1}=38 \\ & \Rightarrow|3 B+I|=38 \end{aligned}$
Let $S=\left\{m \in \mathbf{Z}: A^{m^2}+A^m=3 I-A^{-6}\right\}$, where $A=\left[\begin{array}{cc}2 & -1 \\ 1 & 0\end{array}\right]$. Then $n(S)$ is equal to __________.
Explanation:
$\begin{aligned} &\begin{aligned} & A=\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \end{array}\right] \\ & A^2=\left[\begin{array}{ll} 3 & -2 \\ 2 & -1 \end{array}\right], A^3=\left[\begin{array}{ll} 4 & -3 \\ 3 & -2 \end{array}\right], A^4=\left[\begin{array}{ll} 5 & -4 \\ 4 & -3 \end{array}\right] \end{aligned}\\ &\text { and so on }\\ &\begin{aligned} & \mathrm{A}^6=\left[\begin{array}{ll} 7 & -6 \\ 6 & -5 \end{array}\right] \\ & \mathrm{A}^{\mathrm{m}}=\left[\begin{array}{cc} \mathrm{m}+1 & -\mathrm{m} \\ \mathrm{~m} & -\mathrm{m}+1 \end{array}\right], \\ & \mathrm{A}^{\mathrm{m}^2}=\left[\begin{array}{cc} \mathrm{m}^2+1 & -\mathrm{m}^2 \\ \mathrm{~m}^2 & -\left(\mathrm{m}^2-1\right) \end{array}\right] \\ & \mathrm{A}^{\mathrm{m}^2}+\mathrm{A}^{\mathrm{m}}=3 \mathrm{I}-\mathrm{A}^{-6} \\ & {\left[\begin{array}{cc} \mathrm{m}^2+1 & -\mathrm{m}^2 \\ \mathrm{~m}^2 & -\left(\mathrm{m}^2-1\right) \end{array}\right]+\left[\begin{array}{cc} \mathrm{m}+1 & -\mathrm{m} \\ \mathrm{~m} & -\mathrm{m}+1 \end{array}\right]} \end{aligned} \end{aligned}$
$\begin{aligned} & =3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} -5 & 6 \\ -6 & 7 \end{array}\right] \\ & =\left[\begin{array}{ll} 8 & -6 \\ 6 & -4 \end{array}\right] \\ & =\mathrm{m}^2+1+\mathrm{m}+1=8 \\ & =\mathrm{m}^2+\mathrm{m}-6=0 \Rightarrow \mathrm{~m}=-3,2 \\ & \mathrm{n}(\mathrm{~s})=2 \end{aligned}$
Let M denote the set of all real matrices of order $3 \times 3$ and let $\mathrm{S}=\{-3,-2,-1,1,2\}$. Let
$\begin{aligned} & \mathrm{S}_1=\left\{\mathrm{A}=\left[a_{\mathrm{ij}}\right] \in \mathrm{M}: \mathrm{A}=\mathrm{A}^{\mathrm{T}} \text { and } a_{\mathrm{ij}} \in \mathrm{~S}, \forall \mathrm{i}, \mathrm{j}\right\}, \\ & \mathrm{S}_2=\left\{\mathrm{A}=\left[a_{\mathrm{ij}}\right] \in \mathrm{M}: \mathrm{A}=-\mathrm{A}^{\mathrm{T}} \text { and } a_{\mathrm{ij}} \in \mathrm{~S}, \forall \mathrm{i}, \mathrm{j}\right\}, \\ & \mathrm{S}_3=\left\{\mathrm{A}=\left[a_{\mathrm{ij}}\right] \in \mathrm{M}: a_{11}+a_{22}+a_{33}=0 \text { and } a_{\mathrm{ij}} \in \mathrm{~S}, \forall \mathrm{i}, \mathrm{j}\right\} . \end{aligned}$
If $n\left(S_1 \cup S_2 \cup S_3\right)=125 \alpha$, then $\alpha$ equls __________.
Explanation:
$\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]$
No. of elements in $\mathrm{S}_1: \mathrm{A}=\mathrm{A}^{\mathrm{T}} \Rightarrow 5^3 \times 5^3$
No. of elements in $S_2: A=-A^T \Rightarrow 0$
since no zero in $\mathrm{S}_2$
No. of elements in $\mathrm{S}_3 \Rightarrow$
$\left.\begin{array}{c}a_{11}+a_{22}+a_{33}=0 \Rightarrow(1,2,-3) \Rightarrow 31 \\ \text { or } \\ (1,1,-2) \Rightarrow 3 \\ \text { or } \\ (-1,-1,2) \Rightarrow 3\end{array}\right\} \Rightarrow 12 \times 5^6$
$\begin{aligned} & \mathrm{n}\left(\mathrm{~S}_1 \cap \mathrm{~S}_3\right)=12 \times 5^3 \\ & \mathrm{n}\left(\mathrm{~S}_1 \cup \mathrm{~S}_2 \cup \mathrm{~S}_3\right)=5^6(1+12)-12 \times 5^3 \\ & \quad \Rightarrow 5^3 \times\left[13 \times 5^3-12\right]=125 \alpha \\ & \quad \alpha=1613 \end{aligned}$
Let A be a $3 \times 3$ matrix such that $\mathrm{X}^{\mathrm{T}} \mathrm{AX}=\mathrm{O}$ for all nonzero $3 \times 1$ matrices $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$. If $\mathrm{A}\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{c}1 \\ 4 \\ -5\end{array}\right], \mathrm{A}\left[\begin{array}{l}1 \\ 2 \\ 1\end{array}\right]=\left[\begin{array}{c}0 \\ 4 \\ -8\end{array}\right]$, and $\operatorname{det}(\operatorname{adj}(2(\mathrm{~A}+\mathrm{I})))=2^\alpha 3^\beta 5^\gamma, \alpha, \beta, \gamma \in N$, then $\alpha^2+\beta^2+\gamma^2$ is
Explanation:
$\begin{aligned} & X^T A X=0 \\ & (x y z)\left(\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=0 \\ & (x y z)\left(\begin{array}{l} a_1 x+a_2 y+a_3 z \\ b_1 x+b_2 y+b_3 z \\ c_1 x+c_2 y+c_3 z \end{array}\right)=0 \end{aligned}$
$\begin{aligned} & x\left(a_1 x+a_2 y+a_3 z\right)+y\left(b_1 x+b_2 y+b_3 z\right) \\ & +z\left(c_1 x+c_2 y+c_3 z\right)=0 \\ & a_1=0, b_2=0 c_3=0 \\ & a_2+b_1=0, a_3+c_1=0, b_3=c_2=0 \end{aligned}$
$\begin{aligned} & A=\text { skew symm matrix } \\ & A=\left(\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right) ; \quad A=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)=\left(\begin{array}{l} 1 \\ 4 \\ -5 \end{array}\right) \\ & \Rightarrow A=\left(\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right)\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)=\left(\begin{array}{l} 1 \\ 4 \\ -5 \end{array}\right) \end{aligned}$
$\begin{aligned} & x+y=1 \\ & -x+z=4 \\ & y+z=5 \\ & \left(\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right)\left(\begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right)=\left(\begin{array}{l} 1 \\ 4 \\ -8 \end{array}\right) \end{aligned}$
$\begin{array}{ll} 2 x+y=0 & x=-1 \\ -x+z=4 & y=2 \\ -y-2 z=-8 & z=3 \end{array}$
$\begin{aligned} & A=\left(\begin{array}{ccc} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{array}\right) \\ & 2(\mathrm{~A}+\mathrm{I})=\left(\begin{array}{ccc} 2 & -2 & 4 \\ 2 & 2 & 6 \\ -2 & -6 & 2 \end{array}\right) \\ & 2(\mathrm{~A}+\mathrm{I})=120 \Rightarrow \operatorname{det}|\operatorname{adi}(2(\mathrm{~A}+\mathrm{I}))| \\ & =120^2=2^6 \cdot 3^2 \cdot 5^2 \\ & \alpha=6, \beta=2, \gamma=2 \end{aligned}$
Let $A$ be a square matrix of order 3 such that $\operatorname{det}(A)=-2$ and $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))=2^{m+n} \cdot 3^{m n}, m>n$. Then $4 m+2 n$ is equal to __________.
Explanation:
$\begin{aligned} & \text { As } A \operatorname{adj} A=|A| I, \operatorname{det}(\lambda A)=\lambda^n \operatorname{det} A \\\\ & \operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))=3^3 \operatorname{det}(\operatorname{adj}(-6 \operatorname{adj}(3 A))) \\\\ & =3^3(-6 \operatorname{adj}(3 A))^2 \\\\ & =3^3(-6)^6|3 A|^4 \\\\ & =3^9 2^6 \cdot 3^{12} \cdot(-2)^4 \\\\ & =3^{21} \cdot 2^{10}\end{aligned}$
Now comparing with given condition
$ \begin{aligned} & 2^{m+n} 3^{m n}=2^{10} \cdot 3^{21} \\\\ & m+n=10, m n=21 \\\\ & \Rightarrow m=7, n=3(m>n) \\\\ & \therefore 4 m+2 n=28+6=34 \end{aligned} $
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 $\mathrm{D}_{\mathrm{k}}=\left|\begin{array}{ccc}1 & 2 k & 2 k-1 \\ n & n^{2}+n+2 & n^{2} \\ n & n^{2}+n & n^{2}+n+2\end{array}\right|$. 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} $
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
Let $A = \left( {\matrix{ 2 & { - 2} \cr 1 & { - 1} \cr } } \right)$ and $B = \left( {\matrix{ { - 1} & 2 \cr { - 1} & 2 \cr } } \right)$. Then the number of elements in the set {(n, m) : n, m $\in$ {1, 2, .........., 10} and nAn + mBm = I} is ____________.
Explanation:
${A^2} = \left[ {\matrix{ 2 & { - 2} \cr 1 & { - 1} \cr } } \right]\left[ {\matrix{ 2 & { - 2} \cr 1 & { - 1} \cr } } \right] = \left[ {\matrix{ 2 & { - 2} \cr 1 & { - 1} \cr } } \right] = A$
$ \Rightarrow {A^K} = A,\,K \in I$
${B^2} = \left[ {\matrix{ { - 1} & 2 \cr { - 1} & 2 \cr } } \right]\left[ {\matrix{ { - 1} & 2 \cr { - 1} & 2 \cr } } \right] = \left[ {\matrix{ { - 1} & 2 \cr { - 1} & 2 \cr } } \right] = B$
So, ${B^K} = B,\,K \in I$
$n{A^n} + m{B^m} = nA + mB$
$ = \left[ {\matrix{ {2n - 2n} \cr {n - n} \cr } } \right] + \left[ {\matrix{ { - m} & {2m} \cr { - m} & {2m} \cr } } \right]$
$ = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$
So, $2n - m = 1,\, - n + m = 0,\,2m - n = 1$
So, $(m,n) = (1,1)$
Let $S = \left\{ {\left( {\matrix{ { - 1} & a \cr 0 & b \cr } } \right);a,b \in \{ 1,2,3,....100\} } \right\}$ and let ${T_n} = \{ A \in S:{A^{n(n + 1)}} = I\} $. Then the number of elements in $\bigcap\limits_{n = 1}^{100} {{T_n}} $ is ___________.
Explanation:
$\therefore$ b must be equal to 1
$\therefore$ In this case $\mathrm{A}^2$ will become identity matrix and a can take any value from 1 to 100
$\therefore$ Total number of common element will be 100 .
Explanation:
$ \Rightarrow 3A(I - A) = 0$ or ${A^2} = A$
$ \Rightarrow \left[ {\matrix{ {{a^2}} & {ab + bd} \cr 0 & {{d^2}} \cr } } \right] = \left[ {\matrix{ a & b \cr 0 & d \cr } } \right]$
$ \Rightarrow {a^2} = a,b(a + d - 1) = 0,{d^2} = d$
If b $\ne$ 0, a + d = 1 $\Rightarrow$ 4 ways
If b = 0, a = 0, 1 & d = 0, 1 $\Rightarrow$ 4 ways
$\Rightarrow$ Total 8 matrices
2x + y $-$ z = 3
x $-$ y $-$ z = $\alpha$
3x + 3y + $\beta$z = 3
has infinitely many solution, then $\alpha$ + $\beta$ $-$ $\alpha$$\beta$ is equal to _____________.
Explanation:
$-$ (1 + $\beta$)z = 3 $-$ $\alpha$
For infinitely many solution
$\beta$ + 1 = 0 = 3 $-$ $\alpha$ $\Rightarrow$ ($\alpha$, $\beta$) = (3, $-$1)
Hence, $\alpha$ + $\beta$ $-$ $\alpha$$\beta$ = 5
Explanation:
$\Rightarrow$ adj(adj (2A)) = adj(4 adjA) = 16 adj (adj A)
= 16 | A | A
$\Rightarrow$ adj (32 | A | A) = (32 | A |)2 adj A
12(32| A |)2 |adj A | = 23 (32 | A |)6 | adj A |
23 . 230 | A |6 . | A |2 = 241
| A |8 = 28 $\Rightarrow$ | A | = $\pm$2
| A |2 = | A |2 = 4