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Let \(A=\left[\begin{array}{lll}0 & 1 & 2 \\ \mathrm{a} & 0 & 3 \\ 1 & \mathrm{c} & 0\end{array}\right]\), where \(\mathrm{a}, \mathrm{c} \in \mathbb{R}\). If \(\mathrm{A}^3=\mathrm{A}\) and the positive value of a belongs to the interval \((\mathrm{n}-1, \mathrm{n}]\), where \(\mathrm{n} \in \mathbb{N}\), then \(\mathrm{n}\) is equal to ___.
\(
A=\left[\begin{array}{lll}
0 & 1 & 2 \\
a & 0 & 3 \\
1 & c & 0
\end{array}\right]
\)
\(
\begin{aligned}
& A^3=A \\
& A^2=\left[\begin{array}{lll}
0 & 1 & 2 \\
a & 0 & 3 \\
1 & c & 0
\end{array}\right]\left[\begin{array}{lll}
0 & 1 & 2 \\
a & 0 & 3 \\
1 & c & 0
\end{array}\right]
\end{aligned}
\)
\(
\mathrm{A}^2=\left[\begin{array}{ccc}
\mathrm{a}+2 & 2 \mathrm{c} & 3 \\
3 & \mathrm{a}+3 \mathrm{c} & 2 \mathrm{a} \\
\mathrm{ac} & 1 & 2+3 \mathrm{c}
\end{array}\right]
\)
\(
A^3=\left[\begin{array}{ccc}
a+2 & 2 c & 3 \\
3 & a+3 c & 2 a \\
a c & a & 2+3 c
\end{array}\right]\left[\begin{array}{lll}
0 & 1 & 2 \\
a & 0 & 3 \\
1 & c & 0
\end{array}\right]
\)
\(
\mathrm{A}^3=\left[\begin{array}{ccc}
2 \mathrm{ac}+3 & \mathrm{a}+2+3 \mathrm{c} & 2 \mathrm{a}+4+6 \mathrm{c} \\
\mathrm{a}(\mathrm{a}+3 \mathrm{c})+2 \mathrm{a} & 3+2 \mathrm{ac} & 6+3 \mathrm{a}+9 \mathrm{c} \\
\mathrm{a}+2+3 \mathrm{c} & \mathrm{ac}+\mathrm{c}(2+3 \mathrm{c}) & 2 \mathrm{ac}+3
\end{array}\right]
\)
\(
\begin{aligned}
& \text { Given } A^3=A \\
& 2 a c+3=0 \ldots(1) \text { and } a+2+3 c=1 \\
& a+1+3 c=0 \\
& a+1-\frac{9}{2 a}=0 \\
& 2 a^2+2 a-9=0 \\
& f(1)<0, f(2)>0 \\
& a \in(1,2] \\
& n=2
\end{aligned}
\)
Let \(A=\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right]\). If \(B=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] A\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]\), then the sum of all the elements of the matrix \(\sum_{n=1}^{50} B^n\) is equal to
\(
\begin{aligned}
& \text { Let } C=\left[\begin{array}{cc}
1 & 2 \\
-1 & -1
\end{array}\right], D=\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right] \\
& D C=\left[\begin{array}{cc}
1 & 2 \\
-1 & -1
\end{array}\right]\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{B}=\mathrm{CAD} \\
& \mathrm{B}^{\mathrm{n}}=\underbrace{(\mathrm{CAD})(\mathrm{CAD})(\mathrm{CAD}) \ldots . .(\mathrm{CAD})}_{\text {n-times }}
\end{aligned}
\)
\(
\Rightarrow B^n=C A^n D \dots(1)
\)
\(
\begin{aligned}
& \mathrm{A}^2=\left[\begin{array}{cc}
1 & \frac{1}{51} \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & \frac{1}{51} \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & \frac{2}{51} \\
0 & 1
\end{array}\right] \\
& \mathrm{A}^3=\left[\begin{array}{cc}
1 & \frac{3}{51} \\
0 & 1
\end{array}\right] \\
& \text { Similarly } \mathrm{A}^{\mathrm{n}}=\left[\begin{array}{cc}
1 & \frac{\mathrm{n}}{51} \\
0 & 1
\end{array}\right] \\
& \mathrm{Bn}=\left[\begin{array}{cc}
1 & 2 \\
-1 & -1
\end{array}\right]\left[\begin{array}{cc}
1 & \frac{\mathrm{n}}{51} \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right]
\end{aligned}
\)
\(
\begin{aligned}
& =\left[\begin{array}{cc}
1 & \frac{\mathrm{n}}{51}+2 \\
-1 & -\frac{\mathrm{n}}{51}-1
\end{array}\right]\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
\frac{\mathrm{n}}{51}+1 & \frac{\mathrm{n}}{51} \\
-\frac{\mathrm{n}}{51} & 1-\frac{\mathrm{n}}{51}
\end{array}\right] \\
& \sum_{\mathrm{n}=1}^{50} \mathrm{~B}^{\mathrm{n}}=\left[\begin{array}{cc}
25-50 & 25 \\
-25 & -25-50
\end{array}\right]=\left[\begin{array}{cc}
75 & 25 \\
-25 & 25
\end{array}\right]
\end{aligned}
\)
Sum of the elements \(=100\)
Let \(D_k=\left|\begin{array}{ccc}1 & 2 k & 2 k-1 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2\end{array}\right|\). If \(\sum_{k=1}^n D_k=96\), then \(n\) is equal to
\(
\mathrm{D}_{\mathrm{k}}=\left|\begin{array}{ccc}
1 & 2 \mathrm{k} & 2 \mathrm{k}-1 \\
\mathrm{n} & \mathrm{n}^2+\mathrm{n}+2 & \mathrm{n}^2 \\
\mathrm{n} & \mathrm{n}^2+\mathrm{n} & \mathrm{n}^2+\mathrm{n}+2
\end{array}\right|
\)
\(
\sum_{k=1}^n D_k=96 \Rightarrow
\)
\(
\left|\begin{array}{ccc}
\sum_{\mathrm{k}=1}^{\mathrm{n}} 1 & \sum 2 \mathrm{k} & \sum(2 \mathrm{k}-1) \\
\mathrm{n} & \mathrm{n}^2+\mathrm{n}+2 & \mathrm{n}^2 \\
\mathrm{n} & \mathrm{n}^2+\mathrm{n} & \mathrm{n}^2+\mathrm{n}+2
\end{array}\right|=96
\)
\(
\Rightarrow\left|\begin{array}{ccc}
\mathrm{n} & \mathrm{n}^2+\mathrm{n} & \mathrm{n}^2 \\
\mathrm{n} & \mathrm{n}^2+\mathrm{n}+2 & \mathrm{n}^2 \\
\mathrm{n} & \mathrm{n}^2+\mathrm{n} & \mathrm{n}^2+\mathrm{n}+2
\end{array}\right|=96
\)
\(
R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3-R_1
\)
\(
\left|\begin{array}{ccc}
\mathrm{n} & \mathrm{n}^2+\mathrm{n} & \mathrm{n}^2 \\
0 & 2 & 0 \\
0 & 0 & \mathrm{n}+2
\end{array}\right|=96
\)
\(
\Rightarrow \mathrm{n}(2 \mathrm{n}+4)=96 \Rightarrow \mathrm{n}(\mathrm{n}+2)=48 \Rightarrow \mathrm{n}=6
\)
Let \(A\) be a \(3 \times 3\) matrix such that \(|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))|=12^4\) Then \(\left|A^{-1} \operatorname{adj} A\right|\) is equal to
\(
\begin{aligned}
|\operatorname{adj}(\operatorname{adj} \operatorname{adj} \mathrm{A})| & =|\mathrm{A}|^{(\mathrm{n}-1)^3}=12^4 \\
& |\mathrm{~A}|^8=(12)^4 \\
& |\mathrm{~A}|=(12)^{\frac{1}{2}} \\
\therefore\left|\mathrm{A}^{-1} \cdot \operatorname{adj}(\mathrm{A})\right|= & \left|\mathrm{A}^{-1}\right| \times|\operatorname{adj} \mathrm{A}| \\
& =\frac{1}{|\mathrm{~A}|} \times|\mathrm{A}|^{\mathrm{n}-1} \\
& =\frac{1}{|\mathrm{~A}|} \times|\mathrm{A}|^2=|\mathrm{A}|=\sqrt{12}=2 \sqrt{3}
\end{aligned}
\)
Let \(\alpha\) be a root of the equation \((a-c) x^2+(b-a) x+(c-b)=0\) where \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are distinct real numbers such that the matrix \(\left[\begin{array}{ccc}\alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{array}\right]\) is singular. Then, the value of \(\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}\) is
\(
(a-c) x^2+(b-a) x+(c-b)=0 \quad(a \neq c)
\)
\(\mathrm{x}=1\) is one root \& other root is \(\frac{\mathrm{c}-\mathrm{b}}{\mathrm{a}-\mathrm{c}} \dots(1)\)
\(
\text { now }\left|\begin{array}{ccc}
\alpha^2 & \alpha & 1 \\
1 & 1 & 1 \\
\mathrm{a} & \mathrm{b} & \mathrm{c}
\end{array}\right| \text { is singular }
\)
\(
\Rightarrow\left|\begin{array}{ccc}
\alpha^2 & \alpha & 1 \\
1 & 1 & 1 \\
a & b & c
\end{array}\right|=0 \Rightarrow \alpha^2(c-b)-\alpha(c-a)+(b-a)=0
\)
\(
\Rightarrow \alpha^2(\mathrm{c}-\mathrm{b})+\alpha(\mathrm{a}-\mathrm{c})+(\mathrm{b}-\mathrm{a})=0
\)
\(
\text { satisfied by } \alpha=1 \text { or } \alpha=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{c}-\mathrm{b}} \dots(2)
\)
Now, if \(\alpha=1\) then \(\forall \mathrm{a} \neq \mathrm{b} \neq \mathrm{c}\)
\(
\begin{aligned}
& \sum \frac{(a-c)^2}{(b-a)(c-b)}=\frac{\sum(a-c)^3}{(a-b)(b-c)(c-a)} \\
& =\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)} \\
& =3
\end{aligned}
\)
\(
\text { [if } \mathrm{A}+\mathrm{B}+\mathrm{C}=0 \quad \Rightarrow \quad \mathrm{A}^3+\mathrm{B}^3+\mathrm{C}^3=3 \mathrm{ABC} \text { ] }
\)
Let \(\mathrm{N}\) denote the number that turns up when a fair die is rolled. If the probability that the system of equations
\(
\begin{aligned}
& x+y+z=1 \\
& 2 x+\mathrm{Ny}+2 z=2 \\
& 3 x+3 y+\mathrm{N} z=3
\end{aligned}
\)
has unique solution is \(\frac{k}{6}\), then the sum of value of \(\mathrm{k}\) and all possible values of \(\mathrm{N}\) is
For unique solution
\(
\begin{aligned}
& \Delta \neq 0 \\
& \left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & \mathrm{~N} & 2 \\
3 & 3 & \mathrm{~N}
\end{array}\right| \neq 0 \\
& \Rightarrow\left(\mathrm{N}^2-6\right)-(2 \mathrm{~N}-6)+(6-3 \mathrm{~N}) \neq 0 \\
& \Rightarrow \mathrm{N}^2-5 \mathrm{~N}+6 \neq 0 \\
& \Rightarrow \mathrm{N} \neq 3 \quad \& \quad \mathrm{~N} \neq 2
\end{aligned}
\)
Hence \(N\) can be \(\{1,4,5,6\}\) Fav case : \(\frac{4}{6}=\frac{k}{6} \Rightarrow k=4\)
\(
\text { Sum }=20
\)
If \(A\) and \(B\) are two non-zero \(n \times n\) matrics such that \(A^2+B=A^2 B\), then
\(
\begin{aligned}
& A^2+B=A^2 B \dots(1) \\
& A^2-A^2 B+B=0 \\
& A^2-A^2 B-(I-B)=-I \\
& \left(I-A^2\right)(I-B)=I
\end{aligned}
\)
So, \(\left(I-A^2\right) \&(I-B)\) are inverses of each other So
\(
\begin{aligned}
& (I-B)\left(I-A^2\right)=I \\
& I-B-A^2+B A^2=I \\
& B A^2=B+A^2 \dots(2)
\end{aligned}
\)
So, from (1) & (2)
\(
\mathrm{A}^2 \mathrm{~B}=\mathrm{BA}^2
\)
If \(\mathrm{A}=\frac{1}{2}\left[\begin{array}{cc}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{array}\right]\), then :
\(
\begin{aligned}
& \mathrm{A}=\frac{1}{2}\left[\begin{array}{cc}
1 & \sqrt{3} \\
-\sqrt{3} & 1
\end{array}\right] \\
& \mathrm{A}=\left[\begin{array}{cc}
\cos 60^{\circ} & \sin 60^{\circ} \\
-\sin 60^{\circ} & \cos 60^{\circ}
\end{array}\right] \\
& \text { If } \mathrm{A}=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \text { Here } \alpha=\frac{\pi}{3} \\
& \mathrm{~A}^2=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \\
& =\left[\begin{array}{cc}
\cos 2 \alpha & \sin 2 \alpha \\
-\sin 2 \alpha & \cos 2 \alpha
\end{array}\right] \\
& \mathrm{A}^{30}=\left[\begin{array}{cc}
\cos 30 \alpha & \sin 30 \alpha \\
-\sin 30 \alpha & \cos 30 \alpha
\end{array}\right] \\
& \mathrm{A}^{30}=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]=\mathrm{I} \\
& \mathrm{A}^{25}=\left[\begin{array}{cc}
\cos 25 \alpha & \sin 25 \alpha \\
-\sin 25 \alpha & \cos 25 \alpha
\end{array}\right]=\left[\begin{array}{cc}
\frac{1}{2} & \frac{\sqrt{3}}{2} \\
\frac{-\sqrt{3}}{2} & \frac{1}{2}
\end{array}\right] \\
& \mathrm{A}^{25}=\mathrm{A} \\
& \mathrm{A}^{25}-\mathrm{A}=0
\end{aligned}
\)
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?
\(
\left|\begin{array}{lll}
\alpha & 1 & 1 \\
1 & \alpha & 1 \\
1 & 1 & \alpha
\end{array}\right|=0
\)
\(
\begin{aligned}
& \alpha\left(\alpha^2-1\right)-1(\alpha-1)+1(1-\alpha)=0 \\
& \alpha^3-3 \alpha+2=0 \\
& \alpha^2(\alpha-1)+\alpha(\alpha-1)-2(\alpha-1)=0 \\
& (\alpha-1)\left(\alpha^2+\alpha-2\right)=0 \\
& \alpha=1, \alpha=-2,1
\end{aligned}
\)
For \(\alpha=1, \beta=1\)
\(
\left.\begin{array}{l}
x+y+z=1 \\
x+y+z=b
\end{array}\right\} \text { infinite solution }
\)
For \(\alpha=2, \beta=1\)
\(
\Delta=4
\)
\(
\Delta_1=\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}\right|=3-1-1 \quad \Rightarrow \mathrm{x}=\frac{1}{4}
\)
\(
\Delta_2=\left|\begin{array}{lll}
2 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 2
\end{array}\right|=2-1=1 \quad \Rightarrow \mathrm{y}=\frac{1}{4}
\)
\(
\Delta_3=\left|\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 1
\end{array}\right|=2-1=1 \quad \Rightarrow z=\frac{1}{4}
\)
\(
\text { For } \alpha=2 \Rightarrow \text { unique solution }
\)
Let \(S\) denote the set of all real values of \(\lambda\) such that the system of equations
\(
\begin{aligned}
& \lambda x+y+z=1 \\
& x+\lambda y+z=1 \\
& x+y+\lambda z=1
\end{aligned}
\)
is inconsistent, then \(\sum_{\lambda \in S}\left(|\lambda|^2+|\lambda|\right)\) is equal to
Given system of equation is inconsistent
\(
\begin{aligned}
& \Rightarrow \quad \Delta=0 \\
& \left|\begin{array}{lll}
\lambda & 1 & 1 \\
1 & \lambda & 1 \\
1 & 1 & \lambda
\end{array}\right|=0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \lambda^3-3 \lambda+2=0 \\
& \Rightarrow(\lambda-1)^2(\lambda+2)=0 \\
& \Rightarrow \lambda=1,-2
\end{aligned}
\)
But for \(\lambda=1\) all planes are same Then \(\lambda=-2\)
\(\sum_{\lambda \in S}\left(|\lambda|^2+|\lambda|\right)=4+2=6\)
Let \(\mathrm{A}=\left[\begin{array}{cc}\frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{array}\right]\) and \(\mathrm{B}=\left[\begin{array}{cc}1 & -\mathrm{i} \\ 0 & 1\end{array}\right]\), where \(\mathrm{i}=\sqrt{-1}\).
If \(\mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA}\), then the inverse of the matrix \(\mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}\) is
\(
\text { Now, } M^2=\left(A^{\top} B A\right)\left(A^{\top} B A\right)=A^{\top} B^2 A, \quad A A^{\top}=I
\)
\(
\Rightarrow M^{2023}=A^{\top} B^{2023} A
\)
\(
\begin{array}{ll}
\text { Let } \mathrm{D}=\mathrm{AM}^{2023} \mathrm{~A}^{\top}=\mathrm{AA}^{\top} \mathrm{B}^{2023} \mathrm{AA}^{\top}, & \mathrm{AA}^{\top}=\mathrm{I} \\
\mathrm{D}=\mathrm{B}^{2023} &
\end{array}
\)
\(
\text { Now, } B^2=\left[\begin{array}{cc}
1- & i \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
1- & i \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & -2 i \\
0 & 1
\end{array}\right]
\)
Now, \(B^{2023}=\left[\begin{array}{cc}1 & -2023 i \\ 0 & 1\end{array}\right]\)
\(
D^{-1}=\left[\begin{array}{cc}
1 & 2023 i \\
0 & 1
\end{array}\right]
\)
Let \(S_1\) and \(S_2\) be respectively the sets of all \(a \in \mathbb{R}-\{0\}\) for which the system of linear equations
\(
\begin{aligned}
& a x+2 a y-3 a z=1 \\
& (2 a+1) x+(2 a+3) y+(a+1) z=2 \\
& (3 a+5) x+(a+5) y+(a+2) z=3
\end{aligned}
\)
has unique solution and infinitely many solutions. Then
\(
\begin{aligned}
& \Delta=\left|\begin{array}{ccc}
\mathrm{a} & 2 \mathrm{a} & -3 \mathrm{a} \\
2 \mathrm{a}+1 & 2 \mathrm{a}+3 & \mathrm{a}+1 \\
3 \mathrm{a}+5 & \mathrm{a}+5 & \mathrm{a}+2
\end{array}\right| \\
& \Delta=\mathrm{a}\left(15 \mathrm{a}^2+31 \mathrm{a}+36\right)=0 \\
& \quad \mathrm{a}=0 \\
& \Delta \neq 0 \text { for all } a \in \mathbb{R}-\{0\} \\
& \therefore \mathrm{S}_1=\mathbb{R}-\{0\}, \mathrm{S}_2=\Phi
\end{aligned}
\)
Let \(x, y, z>1\) and \(A=\left[\begin{array}{ccc}1 & \log _x y & \log _x z \\ \log _y x & 2 & \log _y z \\ \log _z x & \log _z y & 3\end{array}\right]\). Then \(\left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right|\) is equal to
\(
\left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right|=\left|\mathrm{A}^2\right|^{(3-1)^2}=|\mathrm{A}|^8
\)
\(
|A|=\left|\begin{array}{ccc}
1 & \frac{\ln y}{\ln } & \frac{\ln z}{\ln x} \\
\frac{\ln x}{\ln y} & 2 & \frac{\ln z}{\ln y} \\
\frac{\ln x}{\ln z} & \frac{\ln y}{\ln z} & 3
\end{array}\right|
\)
\(
=\frac{1}{\ln x \ln y \ln z}\left|\begin{array}{ccc}
\ln x & \ln y & \ln z \\
\ln x & 2 \ln y & \ln z \\
\ln x & \ln y & 3 \ln z
\end{array}\right|
\)
\(
\begin{aligned}
& =\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 3
\end{array}\right| \\
& |\mathrm{A}|=2
\end{aligned}
\)
\(
\therefore|\operatorname{adj}(\operatorname{adj} A)|=2^8
\)
The set of all values of \(t \in \mathbb{R}\), for which the matrix \(\left[\begin{array}{ccc}e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^t & e^{-t} \cos t & e^{-t} \sin t\end{array}\right]\) is invertible, is
\(
|A|=\left|\begin{array}{ccc}
e^t & e^{-t}(s-2 c) & e^{-t}(-2 s-c) \\
e^t & e^{-t}(2 s+c) & e^{-t}(s-2 c) \\
e^t & e^{-t} c & e^{-t} s
\end{array}\right|
\)
\(
\Rightarrow=e^t \cdot e^{-t} \cdot e^{-t}\left|\begin{array}{ccc}
1 & s-2 c & -2 s-c \\
1 & 2 s+c & s-2 c \\
1 & c & s
\end{array}\right|
\)
\(
\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2 \& \quad \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3
\)
\(
=e^t\left|\begin{array}{ccc}
0 & -s-3 c & -3 s-c \\
0 & 2 s & -2 c \\
1 & c & s
\end{array}\right|
\)
\(
\begin{aligned}
& \Rightarrow \mathrm{e}^{-\mathrm{t}}\left[1\left(2 \mathrm{sc}+6 \mathrm{c}^2+6 \mathrm{~s}^2+2 \mathrm{sc}\right)\right] \\
& \Rightarrow \mathrm{e}^{-\mathrm{t}}\left[4 \mathrm{sc}+6\left(\mathrm{c}^2+\mathrm{s}^2\right)\right]=\mathrm{e}^{-\mathrm{t}}(6+2 \sin 2 \mathrm{t}) \\
& \because 2 \sin 2 \mathrm{t} \in[-2,2] \\
& \therefore \mathrm{e}^{-\mathrm{t}}(6+2 \sin 2 \mathrm{t}) \neq 0 \quad \forall \mathrm{t} \in \mathrm{R}
\end{aligned}
\)
Let \(A\) be a symmetric matrix such that \(|A|=2\) and \(\left[\begin{array}{ll}2 & 1 \\ 3 & \frac{3}{2}\end{array}\right] A-\left[\begin{array}{ll}1 & 2 \\ \alpha & \beta\end{array}\right]\). If the sum of the diagonal elements of \(A\) is \(s\), then \(\frac{\beta s}{\alpha^2}\) is equal to
\(
\left[\begin{array}{cc}
2 & 1 \\
3 & \frac{3}{2}
\end{array}\right]\left[\begin{array}{ll}
\mathrm{a} & \mathrm{b} \\
\mathrm{b} & \mathrm{c}
\end{array}\right]=\left[\begin{array}{cc}
1 & 2 \\
\alpha & \beta
\end{array}\right]
\)
Now \(a c-b^2=2\) and \(2 a+b=1\) and \(2 b+c=2\)
solving all these above equations we get
\(
\begin{aligned}
& \frac{1-b}{2} \times\left(\frac{2-2 b}{1}\right)-b^2=2 \\
& \Rightarrow(1-b)^2-b^2=2 \\
& \Rightarrow 1-2 b=2 \\
& \Rightarrow b=-\frac{1}{2} \text { and } \\
& a=\frac{3}{4} \text { and } c=3
\end{aligned}
\)
Hence \(\alpha=3 a+\frac{3 b}{2}=\frac{9}{4}-\frac{3}{4}=\frac{3}{2}\) and \(\beta=3 b+\frac{3 c}{2}=-\frac{3}{2}+\frac{9}{2}=3\) also \(\mathrm{s}=\mathrm{a}+\mathrm{c}=\frac{15}{4}\)
\(
\therefore \frac{\beta \mathrm{s}}{\alpha^2}=\frac{3 \times 15}{4 \times \frac{9}{4}}=5
\)
Let \(\alpha\) and \(\beta\) be real numbers. Consider a \(3 \times 3\) matrix A such that \(A^2=3 A+\alpha I\). If \(A^4=21 A+\beta I\), then
\(
\begin{aligned}
& \mathrm{A}^2=3 \mathrm{~A}+\alpha \mathrm{I} \\
& \mathrm{A}^3=3 \mathrm{~A}^2+\alpha \mathrm{A} \\
& \mathrm{A}^3=3(3 \mathrm{~A}+\alpha \mathrm{I})+\alpha \mathrm{A} \\
& \mathrm{A}^3=9 \mathrm{~A}+\alpha \mathrm{A}+3 \alpha \mathrm{I} \\
& \mathrm{A}^4=(9+\alpha) \mathrm{A}^2+3 \alpha \mathrm{A} \\
& =(9+\alpha)(3 \mathrm{~A}+\alpha \mathrm{I})+3 \alpha \mathrm{A} \\
& =\mathrm{A}(27+6 \alpha)+\alpha(9+\alpha) \mathrm{I} \\
& \Rightarrow 27+6 \alpha=21 \Rightarrow \alpha=-1 \\
& \Rightarrow \beta=\alpha(9+\alpha)=-8
\end{aligned}
\)
Consider the following system of equations
\(
\begin{aligned}
& \alpha x+2 y+z=1 \\
& 2 \alpha x+3 y+z=1 \\
& 3 x+\alpha y+2 z=\beta
\end{aligned}
\)
for some \(\alpha, \beta \in \mathbb{R}\). Then which of the following is NOT correct.
\(
\begin{aligned}
& \because \mathrm{D}=\left|\begin{array}{ccc}
\alpha & 2 & 1 \\
2 \alpha & 3 & 1 \\
3 & \alpha & 2
\end{array}\right| \\
& \mathrm{D}=\alpha(6-\alpha)+2(3-4 \alpha)+1\left(2 \alpha^2-9\right)
\end{aligned}
\)
\(
\begin{aligned}
& =6 \alpha-\alpha^2+6-8 \alpha+2 \alpha^2-9 \\
& D=\alpha^2-2 \alpha-3 \\
& \text { for no solution, } \mathrm{D}=0 \\
& \Rightarrow \quad \alpha^2-2 \alpha-3=0 \\
& \quad(\alpha+1)(\alpha-3)=0 \\
& \Rightarrow \quad \alpha=-1, \alpha=3
\end{aligned}
\)
Now,
\(
D_1=\left|\begin{array}{lll}
1 & 2 & 1 \\
1 & 3 & 1 \\
\beta & \alpha & 2
\end{array}\right|, D_2=\left|\begin{array}{ccc}
\alpha & 1 & 1 \\
2 \alpha & 1 & 1 \\
3 & \beta & 2
\end{array}\right| \text { and } D_3=\left|\begin{array}{ccc}
\alpha & 2 & 1 \\
2 \alpha & 3 & 1 \\
3 & \alpha & \beta
\end{array}\right|
\)
if \(\alpha=-1\) then
\(
D_1=\left|\begin{array}{ccc}
1 & 2 & 1 \\
1 & 3 & 1 \\
\beta & -1 & 2
\end{array}\right|, D_2=\left|\begin{array}{ccc}
-1 & 1 & 1 \\
-2 & 1 & 1 \\
3 & \beta & 2
\end{array}\right|, D_3=\left|\begin{array}{ccc}
-1 & 2 & 1 \\
-2 & 3 & 1 \\
3 & -1 & \beta
\end{array}\right|
\)
\(
\Rightarrow \text { only for } \beta=2, D_1=0, D_2=0, D_3=0
\)
\(\therefore\) It has no solution if \(\alpha=-1\) and \(\beta \neq 2\) if \(\alpha=3\)
\(
\begin{aligned}
& D_1=\left|\begin{array}{lll}
1 & 2 & 1 \\
1 & 3 & 1 \\
\beta & 3 & 2
\end{array}\right|, D_2=\left|\begin{array}{lll}
3 & 1 & 1 \\
6 & 1 & 1 \\
3 & \beta & 2
\end{array}\right|, D_3=\left|\begin{array}{lll}
3 & 2 & 1 \\
6 & 3 & 1 \\
3 & 3 & \beta
\end{array}\right| \\
& \Rightarrow \text { Only for } \beta=2, D_1=D_2=D_3=0 \\
& \Rightarrow \text { It has no solution for } \beta \neq 2 \\
& \therefore \text { It has no solution for } \alpha=3 \text { and for all } \beta \neq 2
\end{aligned}
\)
For \(\alpha, \beta \in \mathbb{R}\), suppose the system of linear equations
\(
\begin{aligned}
& x-y+z=5 \\
& 2 x+2 y+\alpha z=8 \\
& 3 x-y+4 z=\beta
\end{aligned}
\)
has infinitely many solutions. Then \(\alpha\) and \(\beta\) are the roots of
\(
\begin{aligned}
& \left|\begin{array}{ccc}
1 & -1 & 1 \\
2 & 2 & \alpha \\
3 & -1 & 4
\end{array}\right|=0 \\
& 1(8+\alpha)+1(8-3 \alpha)+1(-2-6)=0 \\
& \Rightarrow 8+\alpha+8-3 \alpha-8=0 \\
& -2 \alpha=-8 \\
& \alpha=4 \\
& D_1=0 \\
& \Rightarrow\left|\begin{array}{ccc}
5 & -1 & 1 \\
8 & 2 & 4 \\
\beta & -1 & 4
\end{array}\right|=0 \\
& 5(8+4)+1(32-4 \beta)+1(-8-2 \beta)=0 \\
& 60+32-4 \beta-8-2 \beta=0 \\
& \Rightarrow-6 \beta=-84 \\
& \beta=14
\end{aligned}
\)
Equation having roots as \(\alpha \& \beta\)
\(
x^2-18 x+56=0
\)
If \(\mathrm{P}\) is a \(3 \times 3\) real matrix such that \(\mathrm{P}^{\mathrm{T}}=\mathrm{aP}+(\mathrm{a}-1) \mathrm{I}\), where \(\mathrm{a}>1\), then
\(
\mathrm{P}^{\mathrm{T}}=\mathrm{aP}+(\mathrm{a}-1) \mathrm{I}
\)
\(
\begin{aligned}
& \Rightarrow \mathrm{P}=\mathrm{aP}^{\mathrm{T}}+(\mathrm{a}-1) \mathrm{I} \\
& \Rightarrow \mathrm{P}^{\mathrm{T}}-\mathrm{P}=\mathrm{a}\left(\mathrm{P}-\mathrm{P}^{\mathrm{T}}\right)
\end{aligned}
\)
\(\Rightarrow \mathrm{P}=\mathrm{P}^{\mathrm{T}}\), as \(\mathrm{a} \neq-1\)
Now, \(\mathrm{P}=\mathrm{aP}+(\mathrm{a}-1) \mathrm{I}\)
\(
\begin{aligned}
& \Rightarrow \mathrm{P}=-\mathrm{I} \Rightarrow|\mathrm{P}|=1 \\
& \Rightarrow|\operatorname{Adj} \mathrm{P}|=1
\end{aligned}
\)
Let the system of linear equations
\(
\begin{aligned}
& x+y+k z=2 \\
& 2 x+3 y-z=1 \\
& 3 x+4 y+2 z=k
\end{aligned}
\)
have infinitely many solutions. Then the system
\(
\begin{aligned}
& (k+1) x+(2 k-1) y=7 \\
& (2 k+1) x+(k+5) y=10
\end{aligned}
\)
has:
\(
\begin{aligned}
& x+y+k z=2 \\
& 2 x+3 y-z=1 \\
& 3 x+4 y+2 z=k
\end{aligned}
\)
Have Infinitely many solution then
\(
\begin{aligned}
& \left|\begin{array}{ccc}
1 & 1 & \mathrm{k} \\
2 & 3 & -1 \\
3 & 4 & 2
\end{array}\right|=0 \\
& 1(10)-1(7)+\mathrm{k}(8-9)=0 \\
& \Rightarrow 10-7-\mathrm{k}=0 \\
& \Rightarrow \mathrm{k}=3 \\
& \text { For } \mathrm{k}=3 \\
& 4 \mathrm{x}+5 \mathrm{y}=7 \\
& 7 \mathrm{x}+8 \mathrm{y}=10
\end{aligned}
\)
has unique solution and solution is \((-2,3)\).
Hence the solution is unique and satisfying \(x+y=1\)
Let \(A=\left(\begin{array}{cc}m & n \\ p & q\end{array}\right), d=|A| \neq 0\) and \(|A-d(\operatorname{Adj} A)|=0\). Then
\(
\mathrm{A}=\left[\begin{array}{ll}
\mathrm{m} & \mathrm{n} \\
\mathrm{p} & \mathrm{q}
\end{array}\right], \mathrm{d}=|\mathrm{A}|=\mathrm{mq}-\mathrm{np}
\)
\(
A-d(\operatorname{Adj} \cdot A)=\left[\begin{array}{ll}
m & n \\
p & q
\end{array}\right]-d\left[\begin{array}{cc}
q & -n \\
-p & m
\end{array}\right]
\)
\(
=\left[\begin{array}{cc}
m-d q & n+d n \\
p+p d & q-d m
\end{array}\right]
\)
\(
\begin{aligned}
& |A-d(\operatorname{Adj} A)|=(m-d q)(q-d m)-(n+d n)(p+p d)=0 \\
& \Rightarrow m q-m^2 d-d q^2+d^2 q m=n p(1+d)^2 \\
& \Rightarrow\left(m q-m^2 d-d q^2+d^2 q m\right)=(m q-d)(1+d)^2
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow m q-m^2 d-d q^2+d^2 q m=m q+m q d^2+2 m q d-d(1+d)^2 \\
& \Rightarrow d(1+d)^2=m^2 d+d q^2+2 m q d \\
& \Rightarrow(1+d)^2=(m+q)^2
\end{aligned}
\)
If a point \(\mathrm{P}(\alpha, \beta, \gamma)\) satisfying
\(
(\alpha \beta \gamma)\left(\begin{array}{ccc}
2 & 10 & 8 \\
9 & 3 & 8 \\
8 & 4 & 8
\end{array}\right)=(000)
\)
lies on the plane \(2 x+4 y+3 z=5\), then \(6 \alpha+9 \beta+7 \gamma\) is equal to :
\(
\begin{aligned}
& 2 \alpha+9 \beta+8 \gamma=0 \dots(1) \\
& 10 \alpha+3 \beta+4 \gamma=0 \dots(2) \\
& 8 \alpha+8 \beta+8 \gamma=0 \dots(3) \\
& \alpha+\beta+\gamma=0 \\
& \gamma=-\alpha-\beta \\
& 2 \alpha+9 \beta-8 \alpha-8 \beta=0 \\
& \beta=6 \alpha \\
& \gamma=-\alpha-6 \alpha=-7 \alpha
\end{aligned}
\)
\((\alpha, 6 \alpha,-7 \alpha)\) Satisfies the above system of equation
\(
\begin{aligned}
& 2 \alpha+4(6 \alpha)+3(-7 \alpha)=5 \\
& 5 \alpha=5 \\
& \alpha=1 \\
& \beta=6
\end{aligned}
\)
\(
\begin{aligned}
& \gamma=-7 \\
& \quad 6 \alpha+9 \beta+7 \gamma=6+54-49=11
\end{aligned}
\)
Let \(\mathrm{A}=\left[\mathrm{a}_{i j}\right], \mathrm{a}_{i j} \in Z \cap[0,4], 1 \leq i, j \leq 2\). The number of matrices \(\mathrm{A}\) such that the sum of all entries is a prime number \(p \in(2,13)\) is
\(
A=\left[a_{i j}\right], \text { aij } \in Z \cap[0,4] \text {, so } a_{i j}=\{0,1,2,3,4\}
\)
\(
A=\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}
\)
\(
\begin{aligned}
& a_{11}+a_{12}+a_{21}+a_{22}=P \quad \text { where } P \text { is a prime number } \\
& a_{11}+a_{12}+a_{21}+a_{22}=3,5,7,11
\end{aligned}
\)
if sum \(=3\)
\(
\begin{aligned}
& \left(1+x+x^2+\ldots .+x^4\right)^4 \rightarrow x^3 \\
& \left(1-x^5\right)^4(1-x)^{-4} \rightarrow x^3
\end{aligned}
\)
\(
\left(1-4 x^5+6 x^{10}-4 x^{15}+x^{20}\right)\left(1+{ }^4 C_1 x+{ }^5 C_2 x^2+{ }^6 C_3 x^3+{ }^7 C_4 x^7 \ldots\right)
\)
For \(3{ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{6}=20\).
\(
\begin{aligned}
& \text { If sum }=5 \\
& \left(1-4 x^5\right)(1-x)^{-4} \rightarrow x^5 \\
& \Rightarrow{ }^{4+5-1} C_5-4 x^{4.4+0-1} C_0={ }^8 C_5-4=52
\end{aligned}
\)
For \(7 \quad(-4){ }^5 \mathrm{C}_2+{ }^{10} \mathrm{C}_7=-40+120=80\).
For \(11{ }^{14} \mathrm{C}_{11}+(-4){ }^9 \mathrm{C}_6+6\left({ }^4 \mathrm{C}_1\right)=364-336+24=52\)
\(
\therefore \text { Total matrices }=20+52+80+52=204
\)
Let \(\mathrm{A}\) be a \(n \times n\) matrix such that \(|\mathrm{A}|=2\). If the determinant of the matrix \(\operatorname{Adj}\left(2 \cdot \operatorname{Adj}\left(2 \mathrm{~A}^{-1}\right)\right) \cdot\) is \(2^{84}\), then \(\mathrm{n}\) is equal to
\(
|A|=2
\)
\(\left|\operatorname{Adj}\left(2 \operatorname{Adj}\left(2 \mathrm{~A}^{-1}\right)\right)\right|=2^{84}\)
\(
\left|2 \operatorname{Adj}\left(2 A^{-1}\right)\right|^{n-1}=2^{84}
\)
\(
\left(2^{\mathrm{n}}\left|\operatorname{Adj}\left(2 \mathrm{~A}^{-1}\right)\right|\right)^{\mathrm{n}-1}=2^{84}
\)
\(
\left(2^{\mathrm{n}}\left|2^{\mathrm{n}-1} \operatorname{Adj}\left(\mathrm{A}^{-1}\right)\right|\right)^{\mathrm{n}-1}=2^{84}
\)
\(
\left(2^{\mathrm{n}} \times\left(2^{\mathrm{n}-1}\right)^{\mathrm{n}}\left|\operatorname{Adj}\left(\mathrm{A}^{-1}\right)\right|\right)^{\mathrm{n}-1}=2^{84}
\)
\(
\left(2^{\mathrm{n}} \times 2^{\mathrm{n}(\mathrm{n}-1)} \times\left|\mathrm{A}^{-1}\right|\right)^{\mathrm{n}-1}=2^{84}
\)
\(
\left(2^{\mathrm{n}} \times 2^{\mathrm{n}(\mathrm{n}-1)} \times\left(\frac{1}{2}\right)^{\mathrm{n}-1}\right)^{\mathrm{n}-1}=2^{84}
\)
\(
\begin{aligned}
& \left(2^{n+n^2-n-n+1}\right)^{n-1}=2^{84} \\
& \left(2^{n^2-n+1}\right)^{n-1}=2^{84} \\
& \left(n^2-n+1\right)(n+1)=84
\end{aligned}
\)
For the system of linear equations
\(
\begin{aligned}
& x+y+z=6 \\
& \alpha x+\beta y+7 z=3 \\
& x+2 y+3 z=14
\end{aligned}
\)
which of the following is NOT true?
\(
\begin{aligned}
& x+y+z=6 \dots(1) \\
& \alpha x+\beta y+7 z=3 \dots(2) \\
& x+2 y+3 z=14 \dots(3)
\end{aligned}
\)
equation (3) – equation (1)
\(
\begin{aligned}
& y+2 z=8 \\
& y=8-2 z
\end{aligned}
\)
From (1) \(x=-2+z\)
Value of \(x\) and \(y\) put in equation (2)
\(
\begin{aligned}
& \alpha(-2+z)+\beta(8-2 z)+7 z=3 \\
& -2 \alpha+\alpha z+8 \beta-2 \beta z+7 z=3 \\
& (\alpha-2 \beta+7) z=2 \alpha-8 \beta+3
\end{aligned}
\)
if \(\alpha-2 \beta+7 \neq 0\) then system has unique solution
if \((\alpha-2 \beta+7=0)\) and \(2 \alpha-8 \beta+3 \neq 0\) then system has no solution
if \((\alpha-2 \beta+7=0)\) and \(2 \alpha-8 \beta+3=0\) then system has infinite solution
Let \(A=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right)\). Then the sum of the diagonal elements of the matrix \((A+I)^{11}\) is equal to
\(
\begin{aligned}
A & =\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 4 & -1 \\
0 & 12 & -3
\end{array}\right] \\
A^2 & =\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 4 & -1 \\
0 & 12 & -3
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 4 & -1 \\
0 & 12 & -3
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 4 & -1 \\
0 & 12 & -3
\end{array}\right]
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{A}^3=\mathrm{A}^4=\mathrm{A}^5 \ldots=\mathrm{A} \\
& (\mathrm{A}+\mathrm{I})^{11}={ }^{11} \mathrm{C}_0 \mathrm{~A}^{11}+{ }^{11} \mathrm{C}_1 \mathrm{~A}^{10}+{ }^{11} \mathrm{C}_2 \mathrm{~A}^9+\ldots{ }^{11} \mathrm{C}_{11} \mathrm{I} \\
& ==\left({ }^{11} \mathrm{C}_0+{ }^{11} \mathrm{C}_1+{ }^{11} \mathrm{C}_2+\ldots{ }^{11} \mathrm{C}_{10}\right) \mathrm{A}+\mathrm{I} \\
& =\left(2^{11}-1\right) \mathrm{A}+\mathrm{I} \\
& =2047 \mathrm{~A}+\mathrm{I}
\end{aligned}
\)
Sum of diagonal element \(=2047(1+4-3)+3\)
\(
=4097
\)