Jee Math Test Series Quiz-5

Hint

System has intfinitely many solution
\(
\begin{aligned}
& \Rightarrow\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
3 & 2 & \lambda
\end{array}\right|=0 \\
& \Rightarrow \lambda=1 \\
& D_1=\left|\begin{array}{ccc}
6 & 1 & 1 \\
10 & 2 & 3 \\
\mu & 2 & 1
\end{array}\right|=0 \\
& \mu=14 \\
& \mu-\lambda^2=13
\end{aligned}
\)

Hint

For non-zero solution
\(
\left|\begin{array}{ccc}
2 & 2 \mathrm{a} & \mathrm{a} \\
2 & 3 \mathrm{~b} & \mathrm{~b} \\
2 & 4 \mathrm{c} & \mathrm{c}
\end{array}\right|=0, \Rightarrow\left|\begin{array}{ccc}
1 & 2 \mathrm{a} & \mathrm{a} \\
0 & 3 \mathrm{~b}-2 \mathrm{a} & \mathrm{b}-\mathrm{a} \\
0 & 4 \mathrm{c}-2 \mathrm{a} & \mathrm{c}-\mathrm{a}
\end{array}\right|=0
\)
\(
\begin{aligned}
& \Rightarrow(3 \mathrm{~b}-2 \mathrm{a})(\mathrm{c}-\mathrm{a})-(\mathrm{b}-\mathrm{a})(4 \mathrm{c}-2 \mathrm{a})=0 \\
& \Rightarrow 2 \mathrm{ac}=\mathrm{bc}+\mathrm{ab} \\
& \Rightarrow \frac{2}{\mathrm{~b}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}} \text { Hence } \frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}, \frac{1}{\mathrm{c}} \text { are in A.P. }
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \mathrm{D}=\left|\begin{array}{ccc}
\lambda & 3 & 2 \\
2 \lambda & 3 & 5 \\
4 & \lambda & 6
\end{array}\right|=(\lambda+8)(2-\lambda) \\
& \text { for } \lambda=2 ; \mathrm{D}_1 \neq 0
\end{aligned}
\)

Hence, no solution for \(\lambda=2\)

Hint

\(
\begin{aligned}
& 2 \times(\mathrm{ii})-2 \times(\mathrm{i})-(\mathrm{iii}): \\
& 0=2 \mu-2-\delta \\
& \Rightarrow \delta=2(\mu-1)
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 7 x+6 y-2 z=0 \\
& 3 x+4 y+2 z=0 \\
& x-2 y-6 z=0
\end{aligned}
\)
\(
\Delta=\left|\begin{array}{ccc}
7 & 6 & -2 \\
3 & 4 & 2 \\
1 & -2 & -6
\end{array}\right|=0 \Rightarrow \text { infinite solutions }
\)
Now (1) \(+(2) \Rightarrow y=-x\) put in (1), (2) & (3) all will lead to \(x=2 z\)

Hint

\(
\begin{aligned}
& 2 x-y+2 z=2 \\
& x-2 y+\lambda z=-4 \\
& x+\lambda y+z=4
\end{aligned}
\)
For no solution :
\(
\begin{aligned}
& \mathrm{D}=\left|\begin{array}{ccc}
2 & -1 & 2 \\
1 & -2 & \lambda \\
1 & \lambda & 1
\end{array}\right|=0 \\
& \Rightarrow 2\left(-2-\lambda^2\right)+1(1-\lambda)+2(\lambda+2)=0 \\
& \Rightarrow-2 \lambda^2+\lambda+1=0 \\
& \Rightarrow \lambda=1,-\frac{1}{2} \\
& \mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc}
2 & -1 & 2 \\
-4 & 2 & \lambda \\
4 & \lambda & 1
\end{array}\right|=2\left|\begin{array}{ccc}
1 & -1 & 2 \\
-2 & -2 & \lambda \\
\lambda & \lambda & 1
\end{array}\right| \\
& =2(1+\lambda)
\end{aligned}
\)
whichis not equal to zero for
\(
\lambda=1,-\frac{1}{2}
\)

Hint

\(
\Delta=\left|\begin{array}{ccc}
1 & -2 & 5 \\
-2 & 4 & 1 \\
-7 & 14 & 9
\end{array}\right|=0
\)
Let \(\mathrm{x}=\mathrm{k}\)
\(
\begin{aligned}
& \Rightarrow \text { Put in (1) \& (2) } \\
& k-2 y+5 z=0 \\
& -2 k+4 y+z=0
\end{aligned}
\)
\(
\mathrm{z}=0, \mathrm{y}=\frac{\mathrm{k}}{2}
\)
\(\therefore \mathrm{x}, \mathrm{y}, \mathrm{z}\) are integer
\(\Rightarrow \mathrm{k}\) is even integer
Now \(x=k, y=\frac{k}{2}, z=0\) put in condition
\(
\begin{aligned}
& 15 \leq \mathrm{k}^2+\left(\frac{\mathrm{k}}{2}\right)^2+0 \leq 150 \\
& 12 \leq \mathrm{k}^2 \leq 120 \\
& \Rightarrow \mathrm{k}= \pm 4, \pm 6, \pm 8, \pm 10 \\
& \Rightarrow \text { Number of element in } \mathrm{S}=8 .
\end{aligned}
\)

Hint

\(\mathrm{D}=\left|\begin{array}{ccc}1 & -2 & 3 \\ 2 & 1 & 1 \\ 1 & -7 & \mathrm{a}\end{array}\right|=0 \Rightarrow \mathrm{a}=8\)
also, \(D_1=\left|\begin{array}{ccc}9 & -2 & 3 \\ b & 1 & 1 \\ 24 & -7 & 8\end{array}\right|=0 \Rightarrow b=3\)
hence, \(a-b=8-3=5\)

Hint

For infinite solutions
\(
\Delta=\Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}} \quad=0
\)
Now \(\Delta=0 \Rightarrow\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 4 & -1 \\ 3 & 2 & \lambda\end{array}\right|=0\)
\(
\begin{aligned}
& \Rightarrow \lambda=\frac{9}{2} \\
& \Delta_{x=0} \Rightarrow\left|\begin{array}{rrr}
2 & 1 & 1 \\
6 & 4 & -1 \\
\mu & 2 & -\frac{9}{2}
\end{array}\right|=0 \\
& \Rightarrow \mu=5
\end{aligned}
\)
For \(\lambda=\frac{9}{2} \& \mu=5, \Delta_y=\Delta_z=0\)
Now check option \(2 \lambda+\mu=14\)

Hint

\(
\begin{aligned}
&\mathrm{C}_3 \rightarrow \mathrm{C}_3-\left(\mathrm{C}_1-\mathrm{C}_2\right)\\
&f(\theta)=\left|\begin{array}{ccc}
-\sin ^2 \theta & -1-\sin ^2 \theta & 0 \\
-\cos ^2 \theta & -1-\cos ^2 \theta & 0 \\
12 & 10 & -4
\end{array}\right|
\end{aligned}
\)
\(
\begin{aligned}
& =-4\left[\left(1+\cos ^2 \theta\right) \sin ^2 \theta-\cos ^2 \theta\left(1+\sin ^2 \theta\right)\right] \\
& =-4\left[\sin ^2 \theta+\sin ^2 \theta \cos ^2 \theta-\cos ^2 \theta-\cos ^2 \theta \sin ^2 \theta \right.
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{f}(\theta)=4 \cos 2 \theta \\
& \theta \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right] \\
& 2 \theta \in\left[\frac{\pi}{2}, \pi\right] \\
& f(\theta) \in[-4,0] \\
& (\mathrm{m}, \mathrm{M})=(-4,0)
\end{aligned}
\)

Hint

\(
\mathrm{D}=\left|\begin{array}{ccc}
2 & -4 & \lambda \\
1 & -6 & 1 \\
\lambda & -10 & 4
\end{array}\right|
\)
\(
\begin{aligned}
& =2(3 \lambda+2)(\lambda-3) \\
& D_1=-2(\lambda-3) \\
& D_2=-2(\lambda+1)(\lambda-3) \\
& D_3=-2(\lambda-3)
\end{aligned}
\)
When \(\lambda=3\), then
\(
\mathrm{D}=\mathrm{D}_1=\mathrm{D}_2=\mathrm{D}_3=0
\)
\(\Rightarrow\) Infinite many solution
when \(\lambda=-\frac{2}{3}\) then \(D_1, D_2, D_3\) none of them is zero so equations are inconsistant
\(
\therefore \lambda=-\frac{2}{3}
\)

Hint

\(
\begin{aligned}
& x+y+3 z=0 \dots(i) \\
& x+3 y+k^2 z=0 \dots(ii) \\
& 3 x+y+3 z=0 \dots(iii)
\end{aligned}
\)
\(
\begin{aligned}
& \left|\begin{array}{ccc}
1 & 1 & 3 \\
1 & 3 & \mathrm{k}^2 \\
3 & 1 & 3
\end{array}\right|=0 \\
& \Rightarrow 9+3+3 \mathrm{k}^2-27-\mathrm{k}^2-3=0 \\
& \Rightarrow \mathrm{k}^2=9
\end{aligned}
\)
(i) – (iii) \(\Rightarrow-2 \mathrm{x}=0 \Rightarrow \mathrm{x}=0\)

Now from (i) \(\Rightarrow y+3 z=0\)
\(
\begin{aligned}
& \Rightarrow \frac{y}{z}=-3 \\
& x+\frac{y}{z}=-3
\end{aligned}
\)

Hint

\(
a+x=b+y=c+z+1
\)
\(
\left|\begin{array}{lll}
x & a+y & x+a \\
y & b+y & y+b \\
z & c+y & z+c
\end{array}\right| \quad C_3 \rightarrow C_3-C_1
\)
\(
\left|\begin{array}{lll}
x & a+y & a \\
y & b+y & b \\
z & c+y & c
\end{array}\right| \quad \quad C_2 \rightarrow C_2-C_3
\)
\(
\left|\begin{array}{ccc}
x & y & a \\
y & y & b \\
z & y & c
\end{array}\right| \quad R_3 \rightarrow R_3-R_1, R_2 \rightarrow R_2-R_1
\)
\(
\left|\begin{array}{ccc}
x & y & a \\
y-x & 0 & b-a \\
z-x & 0 & c-a
\end{array}\right|
\)
\(
\begin{aligned}
& =(-y)[(y-x)(c-a)-(b-a)(z-x)] \\
& =(-y)[(a-b)(c-a)+(a-b)(a-c-1)] \\
& =(-y)[(a-b)(c-a)+(a-b)(a-c)+b-a) \\
& =-y(b-a)=y(a-b)
\end{aligned}
\)

Hint

For infinite many solutions
\(
\mathrm{D}=\mathrm{D}_1=\mathrm{D}_2=\mathrm{D}_3=0
\)
Now \(D=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda\end{array}\right|=0\)
\(
\begin{aligned}
& 1 .(2 \lambda-9)-1 .(\lambda-3)+1 .(3-2)=0 \\
& \therefore \lambda=5
\end{aligned}
\)
Now \(D_1=\left|\begin{array}{lll}2 & 1 & 1 \\ 5 & 2 & 3 \\ \mu & 3 & 5\end{array}\right|=0\)
\(
\begin{aligned}
& 2(10-9)-1(25-3 \mu)+1(15-2 \mu)=0 \\
& \mu=8
\end{aligned}
\)

Hint

\(
\left|\begin{array}{ccc}
\cos ^2 x & 1+\sin ^2 x & \sin 2 x \\
1+\cos ^2 x & \sin ^2 x & \sin 2 x \\
\cos ^2 x & \sin ^2 x & 1+\sin 2 x
\end{array}\right|
\)
\(
\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2, \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3
\)
\(
\left|\begin{array}{ccc}
-1 & 1 & 0 \\
1 & 0 & -1 \\
\cos ^2 x & \sin ^2 x & 1+\sin 2 x
\end{array}\right|
\)
\(
\begin{aligned}
& =-1\left(\sin ^2 \mathrm{x}\right)-1\left(1+\sin 2 \mathrm{x}+\cos ^2 \mathrm{x}\right) \\
& =-\sin 2 \mathrm{x}-2 \\
& \mathrm{~m}=-3, \mathrm{M}=-1
\end{aligned}
\)

Hint

\(
\begin{aligned}
& (\lambda-1) \mathrm{x}+(3 \lambda+1) \mathrm{y}+2 \lambda \mathrm{z}=0 \\
& (\lambda-1) \mathrm{x}+(4 \lambda-2) \mathrm{y}+(\lambda+3) \mathrm{z}=0 \\
& 2 \mathrm{x}+(3 \lambda+1) \mathrm{y}+(3 \lambda-3) \mathrm{z}=0 \\
& \left|\begin{array}{ccc}
\lambda-1 & 3 \lambda+1 & 2 \lambda \\
\lambda-1 & 4 \lambda-2 & \lambda+3 \\
2 & 3 \lambda+1 & 3 \lambda-3
\end{array}\right|=0 \\
& \mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2 \& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3 \\
& \left|\begin{array}{ccc}
0 & 3-\lambda & \lambda-3 \\
\lambda-3 & \lambda-3 & -2(\lambda-3) \\
2 & 3 \lambda+1 & 3 \lambda-3
\end{array}\right|=0 \\
& (\lambda-3)^2\left|\begin{array}{ccc}
0 & -1 & 1 \\
1 & 1 & -2 \\
2 & 3 \lambda+1 & 3 \lambda-3
\end{array}\right|=0 \\
& (\lambda-3)^2[(3 \lambda+1)+(3 \lambda-1)]=0 \\
& 6 \lambda(\lambda-3)^2=0 \Rightarrow \lambda=0,3 \\
& \text { Sum }=3 \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \left|\begin{array}{ccc}
x & 1 & 1 \\
1 & y & 1 \\
1 & 1 & z
\end{array}\right| \geq 0 \\
& x y z-x-y-z+2 \geq 0
\end{aligned}
\)
\(
\begin{aligned}
& x y z+2 \geq x+y+z \geq 3(x y z)^{1 / 3} \\
& x y z+2-3(x y z)^{1 / 3} \geq 0 \\
& u t(x y z)=t^3 \\
& t^3-3 t+2 \geq 0 \\
& (t+2)(t-1)^2 \geq 0 \\
& {[t=-2] t^3=-8}
\end{aligned}
\)

Hint

Let \(f(\theta)=\left|\begin{array}{ccc}1 & \cos \theta & 1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1\end{array}\right|\)
\(
\begin{aligned}
& =(1+\sin \theta \cos \theta)-\cos \theta(-\sin \theta-\cos \theta)+1\left(-\sin ^2 \theta+1\right) \\
& =1+\sin \theta \cos \theta+\sin \theta \cos \theta+\cos ^2 \theta-\sin ^2 \theta+1 \\
& =2+2 \sin \theta \cos \theta+\cos 2 \theta \\
& =2+\sin 2 \theta+\cos 2 \theta \dots(1)
\end{aligned}
\)
Now, the maximum value of (1) is \(2+\sqrt{1^2+1^2}=2+\sqrt{2}\) and the minimum value of (1) is
\(
2-\sqrt{1^2+1^2}=2-\sqrt{2} \text {. }
\)

Hint

\(
\begin{aligned}
& \operatorname{det}\left[(\mathrm{I}+\mathrm{B})^{50}-50 \mathrm{~B}\right] \\
& =\operatorname{det}\left[{ }^{50} \mathrm{C}_0 \mathrm{I}+{ }^{50} \mathrm{C}_1 \mathrm{~B}+{ }^{50} \mathrm{C}_2 \mathrm{~B}^2+{ }^{50} \mathrm{C}_3 \mathrm{~B}^3+\ldots\right. \\
& \left.\quad+{ }^{50} \mathrm{C}_{50} \mathrm{~B}^{50} \mathrm{~B}^{50}-50 \mathrm{~B}\right]
\end{aligned}
\)
{All terms having \(\mathrm{B}^{\mathrm{n}}, 2 \leq n \leq 50\)
will be zero because given that \(\mathrm{B}^2=0\) \}
\(
=\operatorname{det}[\mathrm{I}+50 \mathrm{~B}-50 \mathrm{~B}]=\operatorname{det}[\mathrm{I}]=1
\)

Hint

(d) The matrices in the form
\(
\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right], a_{i j} \in\{0,1,2\}, a_{11}=a_{12} \text { are }
\)
\(
\left[\begin{array}{cc}
0 & 0 / 1 / 2 \\
0 / 1 / 2 & 0
\end{array}\right],\left[\begin{array}{cc}
1 & 0 / 1 / 2 \\
0 / 1 / 2 & 1
\end{array}\right],\left[\begin{array}{cc}
2 & 0 / 1 / 2 \\
0 / 1 / 2 & 2
\end{array}\right]
\)
At any place, \(0 / 1 / 2\) means 0,1 or 2 will be the element at that place.

Hence there aretotal \(27=3 \times 3+3 \times 3+3 \times 3\) ) matrices of the above form. Out of which the matrices which are singular are
\(
\left[\begin{array}{cc}
0 & 0 / 1 / 2 \\
0 & 0
\end{array}\right],\left[\begin{array}{cc}
0 & 0 \\
1 / 2 & 0
\end{array}\right],\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right],\left[\begin{array}{ll}
2 & 2 \\
2 & 2
\end{array}\right]
\)
Hence there are total \(7(=3+2+1+1)\) singular matrices. Therefore number of all non-singular matrices in the given form \(=27-7=20\)

Hint

\(
\begin{aligned}
& {\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]} \\
& {\left[\begin{array}{ll}
a^2+b c & a b+b d \\
a c+c d & b c+d^2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]}
\end{aligned}
\)
\(
\begin{aligned}
& b(a+d)=0, b=0 \text { or } a=-d \dots(1)\\
& c(a+d)=0, c=0 \text { or } a=-d \dots(2) \\
& a^2+b c=1, b c+d^2=1 \dots(3)
\end{aligned}
\)
‘ \(a\) ‘ and ‘ \(d\) ‘ are diagonal elements \(a+d=0\) statement- 1 is correct.

Now, \(\operatorname{det}(A)=a d – b c\)
Now, from (3) \(a^2+b c=1\) and \(d^2+b c=1\)
So, \(a^2-d^2=0\)
Adding \(a^2+d^2+2 b c=2\)
\(\Rightarrow(a+d)^2-2 a d+2 b c=2\)
or \(0-2(a d-b c)=2\)
So, \(a d-b c=1 \Rightarrow \operatorname{det}(A)=-1\)
So, statement -2 is also true.
But statement -2 is not the correct explanation of statement-1.

Hint

(d) Statement-1: Determinant of skew symmetric matrix of odd order is zero.
Statement-2 \(: \operatorname{det}\left(A^T\right)=\operatorname{det}(\mathrm{A})\).
\(\operatorname{det}(-\mathrm{A})=-(-1)^{\mathrm{n}} \operatorname{det}(\mathrm{A})\).
where \(\mathrm{A}\) is a \(n \times n\) order matrix.

Hint

\(
\text { (b) Let } A=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \text { where a, b, c, d } \neq 0
\)
\(
A^2=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)
\)
\(
\Rightarrow A^2=\left(\begin{array}{ll}
a^2+b c & a b+b d \\
a c+c d & b c+d^2
\end{array}\right)
\)
\(
\begin{aligned}
& \Rightarrow a^2+b c=1, b c+d^2=1 \\
& a b+b d=a c+c d=0 \\
& c \neq 0 \text { and } b \neq 0 \Rightarrow a+d=0 \\
& |A|=a d-b c=-a^2-b c=-1
\end{aligned}
\)

Hint

(d) Let \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) then \(A^2=\mathrm{I}\)
\(
\Rightarrow a^2+b c=1 \text { and } a b+b d=0
\)
\(a c+c d=0\) and \(b c+d^2=1\)
From these four relations,
\(
a^2+b c=b c+d^2 \Rightarrow a^2=d^2
\)
and
\(
b(a+d)=0=c(a+d) \Rightarrow a=-d
\)
We can take \(a=1, b=0, c=0, d=-1\) as one possible set of values, then \(A=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]\)
Clearly \(A \neq I\) and \(A \neq-I\) and \(\operatorname{det} A=-1\)
\(\therefore\) Statement 1 is true.
Also if \(A \neq I\) then \(\operatorname{tr}(A)=0\)
\(\therefore\) Statement 2 is false.

Hint

(a) Given \(A=\left[\begin{array}{ccc}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{array}\right]\) and \(\left|A^2\right|=25\)
\(
\begin{aligned}
& \therefore A^2=\left[\begin{array}{ccc}
5 & 5 \alpha & \alpha \\
0 & \alpha & 5 \alpha \\
0 & 0 & 5
\end{array}\right]\left[\begin{array}{ccc}
5 & 5 \alpha & \alpha \\
0 & \alpha & 5 \alpha \\
0 & 0 & 5
\end{array}\right] \\
& =\left[\begin{array}{ccc}
25 & 25 \alpha+5 \alpha^2 & 5 \alpha+25 \alpha^2+5 \alpha \\
0 & \alpha^2 & 5 \alpha^2+25 \alpha \\
0 & 0 & 25
\end{array}\right] \\
& \therefore\left|A^2\right|=25\left(25 \alpha^2\right) \\
& \therefore 25=25\left(25 \alpha^2\right) \Rightarrow|\alpha|=\frac{1}{5}
\end{aligned}
\)

Hint

\(
\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
1 & \omega^n & \omega^{2 n} \\
\omega^n & \omega^{2 n} & 1 \\
\omega^{2 n} & 1 & \omega^n
\end{array}\right| \\
& =1\left(\omega^{3 n}-1\right)-\omega^n\left(\omega^{2 n}-\omega^{2 n}\right)+\omega^{2 n}\left(\omega^n-\omega^{4 n}\right) \\
& =\omega^{3 n}-1-0+\omega^{3 n}-\omega^{6 n} \\
& =1-1+1-1=0\left[\because \omega^{3 n}=1\right]
\end{aligned}
\)

Hint

(a) We have
\(
\begin{aligned}
& \quad \frac{1}{2}\left|\begin{array}{ccc}
\mathrm{k} & -3 \mathrm{k} & 1 \\
5 & \mathrm{k} & 1 \\
-\mathrm{k} & 2 & 1
\end{array}\right|=28 \\
& \Rightarrow 5 \mathrm{k}^2+13 \mathrm{k}-46=0 \\
& \text { or } 5 \mathrm{k}^2+13 \mathrm{k}+66=0 \\
& \text { Now, } 5 \mathrm{k}^2+13 \mathrm{k}-46=0 \\
& \Rightarrow \quad \mathrm{k}=\frac{-13 \pm \sqrt{1089}}{10} \\
& \therefore \quad \mathrm{k}=\frac{-23}{5} ; \mathrm{k}=2
\end{aligned}
\)
since \(\mathrm{k}\) is an integer, \(\therefore \mathrm{k}=2\)
Also \(5 k^2+13 k+66=0\)
\(
\Rightarrow \mathrm{k}=\frac{-13 \pm \sqrt{-1151}}{10}
\)
So no real solution exist
For orthocentre
\(
\begin{aligned}
& \mathrm{BH} \perp \mathrm{AC} \\
& \therefore \quad\left(\frac{\beta-2}{\alpha-5}\right)\left(\frac{8}{-4}\right)=-1 \\
& \Rightarrow \quad \alpha-2 \beta=1 \dots(a)
\end{aligned}
\)
Also \(\mathrm{CH} \perp \mathrm{AB}\)
\(
\begin{aligned}
& \therefore \quad\left(\frac{\beta-2}{\alpha+2}\right)\left(\frac{8}{3}\right)=-1 \\
& \Rightarrow \quad 3 \alpha+8 \beta=10 \dots(b)
\end{aligned}
\)
Solving (a) and (b), we get
\(
\alpha=2, \beta=\frac{1}{2}
\)
orthocentre is \(\left(2, \frac{1}{2}\right)\)

Hint

(b) Given \(2 \omega+1=\mathrm{z}\);
\(
\begin{aligned}
& z=\sqrt{3} i \\
& \Rightarrow \omega=\frac{\sqrt{3} i-1}{2}
\end{aligned}
\)
\(\Rightarrow \omega\) is complex cube root of unity Applying \(\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3\)
\(
\begin{aligned}
& =\left|\begin{array}{ccc}
3 & 0 & 0 \\
1 & -\omega^2-1 & \omega^2 \\
1 & \omega^2 & \omega
\end{array}\right| \\
& =3(-1-\omega-\omega)=-3(1+2 \omega)=-3 \mathrm{z} \\
& \Rightarrow \mathrm{k}=-\mathrm{z}
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
& \left|\begin{array}{lll}
\cos x & \sin x & \sin x \\
\sin x & \cos x & \sin x \\
\sin x & \sin x & \cos x
\end{array}\right|=0 \\
& R_1 \rightarrow R_1-R_2 \\
& R_2 \rightarrow R_2-R_3
\end{aligned}
\)
\(
\begin{aligned}
& \left|\begin{array}{ccc}
\cos x-\sin x & \sin x-\cos x & 0 \\
0 & \cos x-\sin x & \sin x-\cos x \\
\sin x & \sin x & \cos x
\end{array}\right|=0 \\
& C_2 \rightarrow C_2+C_3 \\
& \left|\begin{array}{ccc}
\cos x-\sin x & \sin x-\cos x & 0 \\
0 & 0 & \sin x-\cos x \\
\sin x & \sin x & \cos x
\end{array}\right|=0 \\
& \text { Expanding using second row }
\end{aligned}
\)
\(
\begin{aligned}
& 2 \sin x(\sin x-\cos x)^2=0 \\
& \sin x=0 \text { or } \sin x=\cos x \\
& x=0 \text { or } x=\frac{\pi}{4}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \left|\begin{array}{ccc}
3 & 1+f(1) & 1+f(2) \\
1+f(1) & 1+f(2) & 1+f(3) \\
1+f(2) & 1+f(3) & 1+f(4)
\end{array}\right| \\
& =\left|\begin{array}{ccc}
1+1+1 & 1+\alpha+\beta & 1+\alpha^2+\beta^2 \\
1+\alpha+\beta & 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 \\
1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 & 1+\alpha^4+\beta^4
\end{array}\right| \\
& =\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & \alpha & \beta \\
1 & \alpha^2 & \beta^2
\end{array}\right| \times\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & \alpha & \beta \\
1 & \alpha^2 & \beta^2
\end{array}\right| \\
& =\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & \alpha & \beta \\
1 & \alpha^2 & \beta^2
\end{array}\right|=[(1-\alpha)(1-\beta)(\alpha-\beta)]^2
\end{aligned}
\)
\(
\text { So, } K=1
\)

Hint

\(
\sum_{r=1}^{n-1} r=1+2+3+\ldots+(n-1)=\frac{n(n-1)}{2}
\)
\(
\sum_{r=1}^{n-1}(2 r-1)=1+3+5+\ldots+[2(n-1)-2]=(n-1)^2
\)
\(
\sum_{r=1}^{n-1}(3 r-2)=1+4+7+\ldots+(3 n-3-2)=\frac{(n-1)(3 n-4)}{2}
\)
\(
\therefore \sum_{r=1}^{n-1} \Delta_r=\left|\begin{array}{ccc}
\Sigma r & \Sigma(2 r-1) & \Sigma(3 r-2) \\
\frac{n}{2} & n-1 & a \\
\frac{n(n-1)}{2} & (n-1)^2 & \frac{(n-1)(3 n-4)}{2}
\end{array}\right|
\)
\(\sum_{r=1}^{n-1} \Delta_r\) consists of \((n-1)\) determinants in L.H.S. and in R.H.S every constituent of first row consists of \((n-1)\) elements and hence it can be splitted into sum of \((n-1)\) determinants.
\(
\therefore \sum_{r=1}^{n-1} \Delta_r=\left|\begin{array}{ccc}
\frac{n(n-1)}{2} & (n-1)^2 & \frac{(n-1)(3 n-4)}{2} \\
\frac{n}{2} & n-1 & a \\
\frac{n(n-1)}{2} & (n-1)^2 & \frac{(n-1)(3 n-4)}{2}=0
\end{array}\right|
\)
\(\left(\because \mathrm{R}_1\right.\) and \(\mathrm{R}_3\) are identical)
Hence, value of \(\sum_{r=1}^{n-1} \Delta_{\mathrm{r}}\) is independent of both ‘ \(a\) ‘ and ‘ \(n\) ‘.

Hint

\(
\text { Let } \Delta=\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
(a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\
(a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2
\end{array}\right|
\)
\(
\text { Apply } R_2 \rightarrow R_2-R_3
\)
\(
\Delta=\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
(a+\lambda)^2-(a-\lambda)^2 & (b+\lambda)^2-(b-\lambda)^2 & (c+\lambda)^2-(c-\lambda)^2 \\
(a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2
\end{array}\right|
\)
\(
=\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
4 a \lambda & 4 b \lambda & 4 c \lambda \\
(a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2
\end{array}\right|
\)
\(
\left(\because(x+y)^2-(x-y)^2=4 x y\right)
\)
Taking out 4 common from \(\mathrm{R}_2\)
\(
=4\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
a \lambda & b \lambda & c \lambda \\
a^2+\lambda^2-2 a \lambda & b^2+\lambda^2-2 b \lambda & c^2+\lambda^2-2 c \lambda
\end{array}\right|
\)
Apply \(R_3 \rightarrow\left[R_3-\left(R_1-2 R_2\right)\right]\)
\(
=4\left|\begin{array}{lll}
a^2 & b^2 & c^2 \\
a \lambda & b \lambda & c \lambda \\
\lambda^2 & \lambda^2 & \lambda^2
\end{array}\right|
\)
Taking out \(\lambda\) common from \(\mathrm{R}_2\) and \(\lambda^2\) from \(\mathrm{R}_3\).
\(
=4 \lambda\left(\lambda^2\right)\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
a & b & c \\
1 & 1 & 1
\end{array}\right|
\)
\(
\begin{aligned}
& =k \lambda\left|\begin{array}{lll}
a^2 & b^2 & c^2 \\
a & b & c \\
1 & 1 & 1
\end{array}\right| \\
& \Rightarrow k=4 \lambda^2
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \left|\begin{array}{ccc}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|=\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
b & c & a \\
c & a & b
\end{array}\right| \\
= & (a+b+c)\left|\begin{array}{ccc}
1 & 1 & 1 \\
b & c & a \\
c & a & b
\end{array}\right| \\
= & (a+b+c)\left|\begin{array}{ccc}
0 & 0 & 1 \\
b-c & c-a & a \\
c-a & a-b & b
\end{array}\right| \\
= & (a+b+c)\left[a b+b c+c a-a^2-b^2-c^2\right] \\
= & -(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]
\end{aligned}
\)
Since \(a, b, c\) are sides of \(a\) scalene triangle, therefore at least two of the \(a, b, c\) will be unequal.
\(
(a-b)^2+(b-c)^2+(c-a)^2>0
\)
Also \(a+b+c>0\)
\(
-(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]<0
\)

Hint

Let \(\Delta=\left|\begin{array}{ccc}b^2+c^2 & a b & a c \\ a b & c^2+a^2 & b c \\ a c & b c & a^2+b^2\end{array}\right|\)
Multiply \(C_1\) by \(a, C_2\) by b and \(C_3\) by \(\mathrm{c}\) and hence divide by \(a b c\).
\(
=\frac{1}{a b c}\left|\begin{array}{ccc}
a\left(b^2+c^2\right) & a b^2 & a c^2 \\
a^2 b & b\left(c^2+a^2\right) & b c^2 \\
a^2 c & b^2 c & c\left(a^2+b^2\right)
\end{array}\right|
\)
Take out \(a, b, c\) common from \(R_1, R_2\) and \(R_3\) respectively.
\(
\Delta=\frac{a b c}{a b c}\left|\begin{array}{ccc}
b^2+c^2 & b^2 & c^2 \\
a^2 & c^2+a^2 & c^2 \\
a^2 & b^2 & a^2+b^2
\end{array}\right|
\)
Apply \(C_1 \rightarrow C_1-C_2-C_3\)
\(
\begin{aligned}
& \Delta=\left|\begin{array}{ccc}
0 & b^2 & c^2 \\
-2 c^2 & c^2+a^2 & c^2 \\
-2 b^2 & b^2 & a^2+b^2
\end{array}\right| \\
& =-2\left|\begin{array}{ccc}
0 & b^2 & c^2 \\
c^2 & c^2+a^2 & c^2 \\
b^2 & b^2 & a^2+b^2
\end{array}\right|
\end{aligned}
\)
Apply \(C_2-C_1\) and \(C_3-C_1\)
\(
\begin{aligned}
& =-2\left|\begin{array}{ccc}
0 & b^2 & c^2 \\
c^2 & a^2 & 0 \\
b^2 & 0 & a^2
\end{array}\right| \\
& =-2\left[-b^2\left(c^2 a^2\right)+c^2\left(-a^2 b^2\right)\right] \\
& =2 a^2 b^2 c^2+2 a^2 b^2 c^2=4 a^2 b^2 c^2 \\
& \text { But } \Delta=k a^2 b^2 c^2 \therefore k=4
\end{aligned}
\)

Hint

Let \(\Delta=\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right|\) Applying \(C_1+C_3\) and \(C_2+C_3\) \(\Delta=\left|\begin{array}{ccc}-a+c & 2 a+b+c & a+c \\ 2 b+a+c & -b+c & b+c \\ a-c & b-c & -2 c\end{array}\right|\) Now, applying \(R_1+R_3\) and \(R_2+R_3\)
\(
\Delta=\left|\begin{array}{ccc}
0 & 2(a+b) & a-c \\
2(a+b) & 0 & b-c \\
a-c & b-c & -2 c
\end{array}\right|
\)
On expanding, we get
\(
\begin{aligned}
\Delta= & -2(a+b)\{-2 c[2(a+b)] \\
& \quad-(a-c)(b-c)\} \\
& +(a-c)[2(a+b)(b-c)] \\
\Delta= & 8 \mathrm{c}(a+b)(a+b) \\
& \quad+4(a+b)(a-c)(b-c) \\
= & 4(a+b)\left[2 a c+2 b c+a b-b c-a c+c^2\right] \\
= & 4(a+b)\left[a c+b c+a b+c^2\right] \\
= & 4(a+b)[c(a+c)+b(a+c)] \\
= & 4(a+b)(b+c)(c+a) \\
= & \alpha(a+b)(b+c)(c+a)
\end{aligned}
\)
Hence, \(\alpha=4\)

Hint

(b) Vertices of triangle in complex form is \(z, i z, z+i z\)
In cartesian form vertices are \((x, y),(-y, x)\) and \((x-y, x+y)\)
\(
\begin{aligned}
& \therefore \text { Area of triangle }=\frac{1}{2}\left|\begin{array}{ccc}
x & y & 1 \\
-y & x & 1 \\
x-y & x+y & 1
\end{array}\right| \\
& =\frac{1}{2}[x(x-x-y)-y(-y-x+y)+1 \\
& \left.\left(-y x-y^2-x^2+x y\right)\right]
\end{aligned}
\)
\(
=\frac{1}{2}\left[-x y+x y-y^2-x^2\right]=\frac{1}{2}\left(x^2+y^2\right)
\)
( \(\because\) Area can not be negative)
\(
=\frac{1}{2}|z|^2 \quad\left(\because z=x+i y,|z|^2=x^2+y^2\right)
\)

Hint

(b) Given parabola is \(x^2=8 y\)
\(
\Rightarrow 4 a=8 \Rightarrow a=2
\)
To find: Area of \(\triangle A B C\)
\(
\begin{aligned}
& A=(-2 a, a)=(-4,2) \\
& B=(2 a, a)=(4,2) \\
& \mathrm{C}=(0,0)
\end{aligned}
\)
\(
\therefore \text { Area }=\frac{1}{2}\left|\begin{array}{ccc}
-4 & 2 & 1 \\
4 & 2 & 1 \\
0 & 0 & 1
\end{array}\right|
\)

\(
\begin{aligned}
& =\frac{1}{2}[-4(2)-2(4)+1(0)] \\
& =\frac{-16}{2}=-8 \approx 8 \text { sq. unit } \\
& (\because \text { area cannot be negative) }
\end{aligned}
\)

Hint

\(
\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+
\)
\(
\left|\begin{array}{ccc}
a+1 & b+1 & c-1 \\
a-1 & b-1 & c+1 \\
(-1)^{n+2} a & (-1)^{n+1} b & (-1)^n c
\end{array}\right|=0
\)
\(
\Rightarrow\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+\left|\begin{array}{ccc}
a+1 & a-1 & (-1)^{n+2} a \\
b+1 & b-1 & (-1)^{n+1} b \\
c-1 & c+1 & (-1)^n c
\end{array}\right|=0
\)
(Taking transpose of second determinant)
\(
C_1 \Leftrightarrow C_3
\)
\(
\Rightarrow\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|-\left|\begin{array}{ccc}
(-1)^{n+2} a & a-1 & a+1 \\
(-1)^{n+2}(-b) & b-1 & b+1 \\
(-1)^{n+2} c & c+1 & c-1
\end{array}\right|=0
\)
\(
C_2 \Leftrightarrow C_3
\)
\(
\Rightarrow\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+(-1)^{n+2}\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|=0
\)
\(
\Rightarrow\left[1+(-1)^{n+2}\right]\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|=0
\)
\(
C_2-C_1, C_3-C_1
\)
\(
\Rightarrow\left[1+(-1)^{n+2}\right]\left|\begin{array}{ccc}
a & 1 & -1 \\
-b & 2 b+1 & 2 b-1 \\
c & -1 & 1
\end{array}\right|=0; R_1+R_3
\)
\(
\begin{aligned}
& \Rightarrow\left[1+(-1)^{n+2}\right]\left|\begin{array}{ccc}
a+c & 0 & 0 \\
-b & 2 b+1 & 2 b-1 \\
c & -1 & 1
\end{array}\right|=0 \\
& \Rightarrow\left[1+(-1)^{n+2}\right](a+c)(2 b+1+2 b-1)=0 \\
& \Rightarrow 4 b(a+c)\left[1+(-1)^{n+2}\right]=0 \\
& \Rightarrow 1+(-1)^{n+2}=0 \text { as } b(a+c) \neq 0
\end{aligned}
\)
\(\Rightarrow n\) should be an odd integer.

Hint

(d) Given, \(D=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|\)
Apply \(R_2 \rightarrow R_2-R_1\) and \(R \rightarrow R_3-R_1\)
\(
\therefore D=\left|\begin{array}{lll}
1 & 1 & 1 \\
0 & x & 0 \\
0 & 0 & y
\end{array}\right|=x y
\)
Hence, \(D\) is divisible by both \(x\) and \(y\)

Hint

\(\because a_1, a_2, a_3, \ldots \ldots\) are in G.P.
\(\therefore\) Using \(a_n=a r^{n-1}\), we get the given determinant, as
\(
\left|\begin{array}{ccc}
\log a r^{n-1} & \log a r^n & \log a r^{n+1} \\
\log a r^{n+2} & \log a r^{n+3} & \log a r^{n+4} \\
\log a r^{n+5} & \log a r^{n+6} & \log a r^{n+7}
\end{array}\right|
\)
Operating \(C_3-C_2\) and \(C_2-C_1\) and using
\(
\log m-\log n=\log \frac{m}{n} \text { we get }
\)
\(
=\left|\begin{array}{lll}
\log a r^{n-1} & \log r & \log r \\
\log a r^{n+2} & \log r & \log r \\
\log a r^{n+5} & \log r & \log r
\end{array}\right|
\)
\(=0\) (two columns being identical)

Hint

\(
\text { Applying, } C_1 \rightarrow C_1+C_2+C_3 \text { we get }
\)
\(
f(x)=\left|\begin{array}{ccc}
1+\left(a^2+b^2+c^2+2\right) x & \left(1+b^2\right) x & \left(1+c^2\right) x \\
1+\left(a^2+b^2+c^2+2\right) x & 1+b^2 x & \left(1+c^2 x\right) \\
1+\left(a^2+b^2+c^2+2\right) x & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right|
\)
\(
=\left|\begin{array}{ccc}
1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\
1 & 1+b^2 x & \left(1+c^2 x\right) \\
1 & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right|
\)
[As given that \(a^2+b^2+c^2=-2\) ]
\(
\therefore a^2+b^2+c^2+2=0
\)
Applying \(R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3\)
\(
\therefore f(x)=\left|\begin{array}{ccc}
0 & x-1 & 0 \\
0 & 1-x & x-1 \\
1 & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right|
\)
\(
f(x)=(x-1)^2
\)
Hence degree \(=2\).

Hint

(d) Let \(r\) be the common ratio, then
\(
\left|\begin{array}{ccc}
\log a_n & \log a_{n+1} & \log a_{n+2} \\
\log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\
\log a_{n+6} & \log a_{n+7} & \log a_{n+8}
\end{array}\right|
\)
\(
=\left|\begin{array}{ccc}
\log a_1 r^{n-1} & \log a_1 r^n & \log a_1 r^{n+1} \\
\log a_1 r^{n+2} & \log a_1 r^{n+3} & \log a_1 r^{n+4} \\
\log a_1 r^{n+5} & \log a_1 r^{n+6} & \log a_1 r^{n+7}
\end{array}\right|
\)
\(
=\left|\begin{array}{lll}
\log a_1+(n-1) \log r & \log a_1+n \log r & \log a_1+(n+1) \log r \\
\log a_1+(n+2) \log r & \log a_1+(n+3) \log r & \log a_1+(n+4) \log r \\
\log a_1+(n+5) \log r & \log a_1+(n+6) \log r & \log a_2+(n+7) \log r
\end{array}\right|
\)
\(
\left.=0 \quad \text { Apply } c_2 \rightarrow c_2-\frac{1}{2} c_1-\frac{1}{2} c_3\right]
\)

Hint

\(
\text { We have }\left|\begin{array}{ccc}
a & b & a x+b \\
b & c & b x+c \\
a x+b & b x+c & 0
\end{array}\right|
\)
\(
\begin{aligned}
& \text { By } R_3 \rightarrow R_3-\left(x R_1+R_2\right) \\
& =\left|\begin{array}{ccc}
a & b & a x+b \\
b & c & b x+c \\
0 & 0 & -\left(a x^2+2 b x+c\right)
\end{array}\right| \\
& =\left(a x^2+2 b x+c\right)\left(b^2-a c\right)=(+)(-)=-\mathrm{ve} .
\end{aligned}
\)

Hint

(d)
\(
\begin{aligned}
& l=\mathrm{AR}^{p-1} \Rightarrow \log 1=\log A+(p-1) \log R \\
& m=\mathrm{AR}^{q-1} \Rightarrow \log m=\log A+(q-1) \log \mathrm{R} \\
& n=\mathrm{AR}^{r-1} \Rightarrow \log n=\log A+(r-1) \log \mathrm{R}
\end{aligned}
\)

Now,
\(
\left|\begin{array}{lll}
\log l & p & 1 \\
\log m & q & 1 \\
\log n & r & 1
\end{array}\right|=\left|\begin{array}{lll}
\log A+(p-1) \log R & p & 1 \\
\log A+(q-1) \log R & q & 1 \\
\log A+(r-1) \log R & r & 1
\end{array}\right|
\)

Operating
\(
\begin{aligned}
& \mathrm{C}_1-(\log \mathrm{R}) \mathrm{C}_2+(\log \mathrm{R}-\log \mathrm{A}) \mathrm{C}_3 \\
& =\left|\begin{array}{lll}
0 & p & 1 \\
0 & q & 1 \\
0 & r & 1
\end{array}\right|=0
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
& \text { We have } A=\left[\begin{array}{cc}
2 & -3 \\
-4 & 1
\end{array}\right] \\
& \Rightarrow A^2=\left[\begin{array}{cc}
16 & -9 \\
-12 & 13
\end{array}\right] \\
& \Rightarrow 3 A^2=\left[\begin{array}{cc}
48 & -27 \\
-36 & 39
\end{array}\right] \\
& \text { Also } 12 A=\left[\begin{array}{cc}
24 & -36 \\
-48 & 12
\end{array}\right] \\
& \therefore 3 A^2+12 A=\left[\begin{array}{cc}
48 & -27 \\
-36 & 39
\end{array}\right]+\left[\begin{array}{cc}
24 & -36 \\
-48 & 12
\end{array}\right]=\left[\begin{array}{cc}
72 & -63 \\
-84 & 51
\end{array}\right] \\
& \operatorname{adj}\left(3 A^2+12 A\right)=\left[\begin{array}{ll}
51 & 63 \\
84 & 72
\end{array}\right]
\end{aligned}
\)

Hint

Consider the properties of matrices.
\(
\begin{aligned}
A^{-1} & =\frac{1}{|A|} \cdot \operatorname{adj}(A) \\
\operatorname{adj}(A) & =|A| A^{-1} \dots(1)
\end{aligned}
\)
Also,
\(
|\operatorname{adj}(A)|=|A|^{n-1}=|A|^2 \dots(2)
\)
From equations (1) and (2).
\(
\operatorname{adj}(\operatorname{adj}(A))=|A|^2(\operatorname{adj}(A))^{-1}
\)
So, 1,2 and 3 are correct option and option 4 occurs only when \(|A|=1\).
Hence, the question is asking for the relation which is not true, so option 4 is correct.

Hint

(d)
\(
\begin{aligned}
& \mathrm{A}(\operatorname{adj} \mathrm{A})=\mathrm{AA}^{\mathrm{T}} \\
& \Rightarrow \mathrm{A}^{-1} \mathrm{~A}(\operatorname{adj} \mathrm{A})=\mathrm{A}^{-1} \mathrm{AA}^{\mathrm{T}}
\end{aligned}
\)
\(
\operatorname{adj} A=A^T
\)
\(
\Rightarrow\left\lfloor\begin{array}{cc}
2 & b \\
-3 & 5 a
\end{array}\right\rfloor=\left\lfloor\begin{array}{cc}
5 a & 3 \\
-b & 2
\end{array}\right\rfloor
\)
\(
\begin{aligned}
& \Rightarrow \mathrm{a}=\frac{2}{5} \text { and } \mathrm{b}=3 \\
& \Rightarrow 5 \mathrm{a}+\mathrm{b}=5
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
& A^2-5 A=-7 I \\
& A A A^{-1}-5 A A^{-1}=-7 I A^{-1} \\
& A I-5 I=-7 A^{-1} \\
& A-5 I=-7 A^{-1} \\
& A^{-1}=\frac{1}{7}(5 I-A) \\
& A^3-2 A^2-3 A+I \\
& =A(5 A-7 I)-2 A^2-3 A+I \\
& =5 A^2-7 A-2 A^2-3 A+I \\
& =3 A^2-10 A+I \\
& =3(5 A-7 I)-10 A+I \\
& =5 A-20 I \\
& =5(A-4 I)
\end{aligned}
\)

Hint

\(
|5 \cdot \operatorname{adj} A|=5 \Rightarrow 5^3 \cdot|A|^{3-1}=5
\)
\(
\Rightarrow 125|\mathrm{~A}|^2=5 \Rightarrow|\mathrm{A}|= \pm \frac{1}{5}
\)

Hint

(d)

\(
B B^{\prime}=B\left(A^{-1} A^{\prime}\right)^{\prime}=B\left(A^{\prime}\right)^{\prime}\left(A^{-1}\right)^{\prime}
\)
\(
\begin{aligned}
& =B A\left(A^{-1}\right)^{\prime} \\
& =\left(A^{-1} A^{\prime}\right)\left(A\left(A^{-1}\right)^{\prime}\right) \\
& =A^{-1} A \cdot A^{\prime} \cdot\left(A^{-1}\right)^{\prime} \quad\left\{\text { as } A A^{\prime}=A^{\prime} A\right\} \\
& =I\left(A^{-1} A\right)^{\prime}=I \cdot I=I^2=I
\end{aligned}
\)