Multiple choice questions with one correct answer
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Let \(A\) be a \(3 \times 3\) matrix such that \(A\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]\) Then \(A^{-1}\) is:
\(
\text { (a) Given } A\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 3 \\
0 & 1 & 1
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right]
\)
Applying \(\mathrm{C}_1 \leftrightarrow \mathrm{C}_3\)
\(\mathrm{A}\left[\begin{array}{lll}3 & 2 & 1 \\ 3 & 2 & 0 \\ 1 & 1 & 0\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\)
Again Applying \(\mathrm{C}_2 \leftrightarrow \mathrm{C}_3\)
\(\mathrm{A}\left[\begin{array}{lll}3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)
\(
\text { pre-multiplying both sides by } \mathrm{A}^{-1}
\)
\(
\mathrm{A}^{-1} \mathrm{~A}\left[\begin{array}{lll}
3 & 1 & 2 \\
3 & 0 & 2 \\
1 & 0 & 1
\end{array}\right]=\mathrm{A}^{-1}\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\)
\(
\mathrm{I}\left[\begin{array}{lll}
3 & 1 & 2 \\
3 & 0 & 2 \\
1 & 0 & 1
\end{array}\right]=\mathrm{A}^{-1} \quad \mathrm{I}=\mathrm{A}^{-1}
\)
\(
\left(\because \mathrm{A}^{-1} \mathrm{~A}=\mathrm{I} \text { and } \mathrm{I}=\text { Identity matrix }\right)
\)
\(
\text { Hence, } A^{-1}=\left[\begin{array}{lll}
3 & 1 & 2 \\
3 & 0 & 2 \\
1 & 0 & 1
\end{array}\right]
\)
If \(P=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]\) is the adjoint of a \(3 \times 3\) matrix \(A\) and \(|A|=4\), then \(\alpha\) is equal to :
(b)
\(
|\mathrm{P}|=1(12-12)-\alpha(4-6)+3(4-6)=2 \alpha-6
\)
\(
\begin{aligned}
\text { Now, } \operatorname{adj} A=P & \Rightarrow|\operatorname{adj} A|=|P| \\
& \Rightarrow|A|^2=|P| \\
& \Rightarrow|P|=16
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow 2 a-6=16 \\
& \Rightarrow a=11
\end{aligned}
\)
Let \(P\) and \(Q\) be \(3 \times 3\) matrices \(P \neq Q\). If \(P^3=Q^3\) and \(P^2 Q=Q^2 P\) then determinant of \(\left(P^2+Q^2\right)\) is equal to :
(c) Given \(P^3=Q^3 \dots(1) \)
and \(P^2 Q=\mathrm{Q}^2 P \dots(2) \)
Subtracting (1) and (2), we get
\(
\begin{aligned}
& P^3-P^2 Q=Q^3-Q^2 P \\
\Rightarrow & P^2(P-Q)+Q^2(P-Q)=0 \\
\Rightarrow & \left(P^2+Q^2\right)(P-Q)=0
\end{aligned}
\)
If \(\left|P^2+Q^2\right| \neq 0\) then \(P^2+Q^2\) is invertible.
\(
\Rightarrow P-Q=0 \Rightarrow P=Q
\)
Which gives a contradiction \((\because P \neq Q)\)
Hence \(\left|P^2+Q^2\right|=0\)
Let \(A=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)\). If \(u_1\) and \(u_2\) are column matrices such that \(A u_1=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)\) and \(A u_2=\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)\), then \(u_1+u_2\) is equal to :
(d) Let \(A u_1=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)\) and \(A u_2=\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)\)
Then, \(A u_1+A u_2=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)+\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)\)
\(\Rightarrow A\left(u_1+u_2\right)=\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right) \dots(1)\)
Also, \(A=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)\)
\(
\Rightarrow|A|=1(1)-0(2)+0(4-3)=1
\)
We know,
\(
\begin{aligned}
& A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\
& \Rightarrow A^{-1}=\operatorname{adj}(A) \quad(\because|A|=1)
\end{aligned}
\)
Now, from equation (1), we have
\(
\begin{aligned}
& u_1+u_2=A^{-1}\left(\begin{array}{l}
1 \\
1 \\
0
\end{array}\right) \\
& =\left[\begin{array}{ccc}
1 & 0 & 0 \\
-2 & 1 & 0 \\
1 & -2 & 1
\end{array}\right]\left(\begin{array}{l}
1 \\
1 \\
0
\end{array}\right)=\left[\begin{array}{c}
1 \\
-1 \\
-1
\end{array}\right]
\end{aligned}
\)
If \(A^T\) denotes the transpose of the matrix \(A=\left[\begin{array}{ccc}0 & 0 & a \\ 0 & b & c \\ d & e & f\end{array}\right]\), where \(a, b, c, d, e\) and \(f\) are integers such that \(a b d \neq 0\), then the number of such matrices for which \(A^{-1}=A^T\) is
\(
\text { (c) } A=\left[\begin{array}{lll}
0 & 0 & a \\
0 & b & c \\
d & e & f
\end{array}\right],|A|=-a b d \neq 0
\)
\(
\begin{aligned}
& c_{11}=+(b f-c e), c_{12}=-(-a d)=c d, c_{13}=+(-b d)=-b d \\
& c_{21}=-(-e a)=a e, c_{22}=+(-a d)=-a d, c_{23}=-(0)=0 \\
& c_{31}=+(-a b)=-a b, c_{32}=-(0)=0, c_{33}=0
\end{aligned}
\)
\(
\operatorname{Adj} A=\left[\begin{array}{ccc}
(b f-c e) & a e & -a b \\
c d & -a d & 0 \\
-b d & 0 & 0
\end{array}\right]
\)
\(
A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{a b d}\left[\begin{array}{ccc}
b f-c e & a e & -a b \\
c d & -a d & 0 \\
-b d & 0 & 0
\end{array}\right]
\)
\(
\begin{aligned}
& A^T=\left[\begin{array}{lll}
0 & 0 & d \\
0 & b & e \\
a & c & f
\end{array}\right] \\
& \text { Now } A^{-1}=A^T \\
& \Rightarrow \frac{1}{-a b d}\left[\begin{array}{ccc}
b f-c e & a e & -a b \\
c d & -a d & 0 \\
-b d & 0 & 0
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & d \\
0 & b & e \\
a & c & f
\end{array}\right]
\end{aligned}
\)
\(
\Rightarrow\left[\begin{array}{ccc}
b f-c e & a e & -a b \\
c d & -a d & 0 \\
-b d & 0 & 0
\end{array}\right]=\left[\begin{array}{ccc}
0 & 0 & -a b d^2 \\
0 & -a b^2 d & -a b d e \\
-a^2 b d & -a b c d & -a b d f
\end{array}\right]
\)
\(
\begin{aligned}
& \therefore b f-c e=a e=c d=0 \dots(i) \\
& a b d^2=a b, a b^2 d=a d, a^2 b d=b d \dots(ii) \\
& a b d e=a b c d=a b d f=0 \dots(iii)
\end{aligned}
\)
\(
\begin{aligned}
& \text { From (ii), } \\
& \left(a b d^2\right) \cdot\left(a b^2 d\right) \cdot\left(a^2 b d\right)=a b \cdot a d \cdot b d \\
& \Rightarrow(a b d)^4-(a b d)^2=0 \\
& \Rightarrow(a b d)^2\left[(a b d)^2-1\right]=0
\end{aligned}
\)
\(
\because a b d \neq 0, \therefore a b d= \pm 1 \dots(iv)
\)
From (iii) and (iv),
\(
e=c=f=0 \dots(v)
\)
From (i) and (v), \(b f=a e=c d=0 \dots(vi)\)
From (iv), (v), and (vi), it is clear that \(a, b, d\) can be any non-zero integer such that \(a b d= \pm 1\). But it is only possible if \(a=b=d= \pm 1\)
Hence, there are 2 choices for each \(a, b\) and \(d\). there fore, there are \(2 \times 2 \times 2\) choices for \(a, b\) and \(d\). Hence number of required matrices \(=2 \times 2 \times 2=(2)^3\)
Let \(A\) and \(B\) be real matrices of the form \(\left[\begin{array}{ll}\alpha & 0 \\ 0 & \beta\end{array}\right]\) and \(\left[\begin{array}{ll}0 & \gamma \\ \delta & 0\end{array}\right]\), respectively.
Statement 1: \(A B-B A\) is always an invertible matrix.
Statement 2: \(A B-B A\) is never an identity matrix
\(
\text { Let } A \text { and } B \text { be real matrices such that } A=\left[\begin{array}{ll}
\alpha & 0 \\
0 & \beta
\end{array}\right]
\)
and \(B=\left[\begin{array}{ll}0 & \gamma \\ \delta & 0\end{array}\right]\)
Now, \(A B=\left[\begin{array}{cc}0 & \alpha \gamma \\ \beta \delta & 0\end{array}\right]\)
and \(B A=\left[\begin{array}{cc}0 & \gamma \beta \\ \delta \alpha & 0\end{array}\right]\)
Statement-1:
\(
\begin{aligned}
& A B-B A=\left[\begin{array}{cc}
0 & \gamma(\alpha-\beta) \\
\delta(\beta-\alpha) & 0
\end{array}\right] \\
& |A B-B A|=(\alpha-\beta)^2 \gamma \delta \neq 0
\end{aligned}
\)
\(\therefore A B-B A\) is always an invertible matrix.
Hence, statement – 1 is true.
But \(A B-B A\) can be identity matrix if \(\gamma=-\delta\) or \(\delta=-\gamma\)
So, statement – 2 is false.
Consider the following relation \(\mathrm{R}\) on the set of real square matrices of order 3 .
\(R=\left\{(A, B) \mid A=P^{-1} B P\right.\) for some invertible matrix \(\left.P\right\}\)
Statement- \(1: R\) is equivalence relation.
Statement-2 : For any two invertible \(3 \times 3\) matrices \(M\) and \(N,(M N)^{-1}=N^{-1} M^{-1}\)
For reflexive
\(
(A, A) \in R
\)
\(A=P^{-1} A P\) is true, For \(P=\mathrm{I}\), which is an invertible matrix. \(\therefore \quad R\) is reflexive.
For symmetry
As \((A, B) \in \mathrm{R}\) for matrix \(P\)
\(
\begin{aligned}
& A=P^{-1} B P \\
& \Rightarrow \quad P A P^{-1}=B \\
& \Rightarrow \quad B=P A P^{-1} \\
& \Rightarrow \quad B=\left(P^{-1}\right)^{-1} \mathrm{~A}\left(\mathrm{P}^{-1}\right)
\end{aligned}
\)
\(\therefore \quad(\mathrm{B}, \mathrm{A}) \in \mathrm{R}[latex] for matrix [latex]P^{-1}\)
\(\therefore \quad \mathrm{R}\) is symmetric.
For transitivity
\(
\begin{array}{r}
A=P^{-1} B P \\
\text { and } B=P^{-1} C P
\end{array}
\)
\(
\begin{aligned}
& \Rightarrow A=P^{-1}\left(P^{-1} C P\right) P \\
& \Rightarrow A=\left(P^{-1}\right)^2 C P^2 \\
& \Rightarrow A=\left(P^2\right)^{-1} C\left(P^2\right)
\end{aligned}
\)
\(\therefore \quad(A, C) \in R\) for matrix \(P^2\)
\(\therefore R\) is transitive.
So \(R\) is equivalence
Let \(A\) be a \(2 \times 2\) matrix
Statement -1: \(\operatorname{adj}(\operatorname{adj} A)=A\)
Statement -2 : \(|\operatorname{adj} A|=|A|\)
(a) We know that \(|\operatorname{adj}(\operatorname{adj} A)|=|A|^{n-2} A\).
\(
=|A|^0 A=A
\)
Also \(|\operatorname{adj} A|=|A|^{\mathrm{n}-1}=|A|^{2-1}=|A|\)
\(\therefore\) Both the statements are true but statement- 2 is not a correct explanation for statement- 1.
Let \(A\) be a square matrix all of whose entries are integers. Then which one of the following is true?
(c) \(\because\) All entries of square matrix \(A\) are integers, therefore all cofactors should also be integers.
If \(\operatorname{det} A= \pm 1\) then \(A^{-1}\) exists. Also all entries of \(A^{-1}\) are integers.
If \(A^2-A+I=0\), then the inverse of \(A\) is
(d) Given \(A^2-A+I=0\)
\(
A^{-1} A^2-A^{-1} A+A^{-1} I=A^{-1} .0
\)
(Multiplying \(A^{-1}\) on both sides)
\(
\Rightarrow A-1+A^{-1}=0 \text { or } A^{-1}=I-A \text {. }
\)
Let \(A=\left(\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right)\). and \(B=\left(\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right)\). If \(B\) is the inverse of matrix \(A\), then \(\alpha\) is
(a)
\(
\begin{aligned}
& \text { Given that } 10 B=\left[\begin{array}{ccc}
4 & 2 & 2 \\
-5 & 0 & \alpha \\
1 & -2 & 3
\end{array}\right] \\
& \Rightarrow B=\frac{1}{10}\left[\begin{array}{ccc}
4 & 2 & 2 \\
-5 & 0 & \alpha \\
1 & -2 & 3
\end{array}\right]
\end{aligned}
\)
Also since, \(B=A^{-1} \Rightarrow A B=I\)
\(
\begin{gathered}
\Rightarrow \frac{1}{10}\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\left[\begin{array}{ccc}
4 & 2 & 2 \\
-5 & 0 & \alpha \\
1 & -2 & 3
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
\Rightarrow \frac{1}{10}\left[\begin{array}{ccc}
10 & 0 & 5-2 \\
0 & 10 & -5+\alpha \\
0 & 0 & 5+\alpha
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
\Rightarrow \frac{5-\alpha}{10}=0 \Rightarrow \alpha=5
\end{gathered}
\)
Let \(A=\left(\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right)\). The only correct statement about the matrix \(A\) is
(a) \(A=\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\)
clearly \(A \neq 0\). Also \(|A|=-1 \neq 0\)
\(\therefore A^{-1}\) exists, further \((-1) I=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right] \neq A\)
Also \(A^2=\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\)
\(
=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=I
\)
If \(S\) is the set of distinct values of ‘ \(b\) ‘ for which the following system of linear equations
\(
\begin{aligned}
& x+y+z=1 \\
& x+a y+z=1 \\
& a x+b y+z=0
\end{aligned}
\)
has no solution, then \(\mathrm{S}\) is :
(a)
\(
D=\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & a & 1 \\
a & b & 1
\end{array}\right|=0
\)
\(
\begin{aligned}
& \Rightarrow 1[a-b]-1[1-a]+1\left[b-a^2\right]=0 \Rightarrow(a-1)^2=0 \\
& \Rightarrow a=1
\end{aligned}
\)
For \(\mathrm{a}=1\), the First two equations are identical ie. \(x+y+z=1\)
To have no solution with \(x+b y+z=0\) \(\mathrm{b}=1\)
So \(b=\{1\} \Rightarrow\) It is singleton set.
The number of real values of \(\lambda\) for which the system of linear equations
\(
\begin{aligned}
& 2 x+4 y-\lambda z=0 \\
& 4 x+\lambda y+2 z=0 \\
& \lambda x+2 y+2 z=0
\end{aligned}
\)
has infinitely many solutions, is : \(\quad\)
Since the given system of linear equations has infinitely many solutions.
\(
\begin{aligned}
& \left|\begin{array}{ccc}
2 & 4 & -\lambda \\
4 & \lambda & 2 \\
\lambda & 2 & 2
\end{array}\right|=0 \\
& \lambda^3+4 \lambda-40=0
\end{aligned}
\)
\(\lambda\) has only 1 real root.
The system of linear equations
\(
\begin{array}{r}
\mathrm{x}+\lambda \mathrm{y}-\mathrm{z}=0 \\
\lambda \mathrm{x}-\mathrm{y}-\mathrm{z}=0 \\
\mathrm{x}+\mathrm{y}-\lambda \mathrm{z}=0
\end{array}
\)
has a non-trivial solution for:
(b) For trivial solution,
\(
\begin{aligned}
& \left|\begin{array}{ccc}
1 & \lambda & -1 \\
\lambda & -1 & -1 \\
1 & 1 & -\lambda
\end{array}\right|=0 \\
& \Rightarrow-\lambda(\lambda+1)(\lambda-1)=0 \\
& \Rightarrow \lambda=0,+1,-1
\end{aligned}
\)
The set of all values of \(\lambda\) for which the system of linear equations :
\(
\begin{aligned}
& 2 \mathrm{x}_1-2 \mathrm{x}_2+\mathrm{x}_3=\lambda \mathrm{x}_1 \\
& 2 \mathrm{x}_1-3 \mathrm{x}_2+2 \mathrm{x}_3=\lambda \mathrm{x}_2 \\
& -\mathrm{x}_1+2 \mathrm{x}_2=\lambda \mathrm{x}_3
\end{aligned}
\)
has a non-trivial solution,
(a)
\(
\begin{aligned}
& \left.\begin{array}{r}
2 \mathrm{x}_1-2 \mathrm{x}_2+\mathrm{x}_3=\lambda \mathrm{x}_1 \\
2 \mathrm{x}_1-3 \mathrm{x}_2+2 \mathrm{x}_3=\lambda \mathrm{x}_2 \\
-\mathrm{x}_1+2 \mathrm{x}_2=\lambda \mathrm{x}_3
\end{array}\right\} \\
& \Rightarrow \quad(2-\lambda) \mathrm{x}_1-2 \mathrm{x}_2+\mathrm{x}_3=0 \\
& 2 \mathrm{x}_1-(3+\lambda) \mathrm{x}_2+2 \mathrm{x}_3=0 \\
& -\mathrm{x}_1+2 \mathrm{x}_2-\lambda \mathrm{x}_3=0 \\
&
\end{aligned}
\)
For non-trivial solution,
\(
\Delta=0
\)
i.e. \(\left|\begin{array}{ccc}2-\lambda & -2 & 1 \\ 2 & -(3+\lambda) & 2 \\ -1 & 2 & -\lambda\end{array}\right|=0\)
\(
\begin{aligned}
& \Rightarrow(2-\lambda)[\lambda(3+\lambda)-4]+2[-2 \lambda+2]+1[4-(3+\lambda)]=0 \\
& \Rightarrow \lambda^3+\lambda^2-5 \lambda+3=0 \\
& \Rightarrow \lambda=1,1,3
\end{aligned}
\)
Hence \(\lambda\) has 2 values.
If \(a, b, c\) are non-zero real numbers and if the system of equations
\(
\begin{aligned}
& (a-1) x=y+z \\
& (b-1) y=z+x \\
& (c-1) z=x+y
\end{aligned}
\)
has a non-trivial solution, then \(a b+b c+c a\) equals:
Given system of equations can be written as
\(
\begin{aligned}
& (a-1) x-y-z=0 \\
& -x+(b-1) y-z=0
\end{aligned}
\)
\(
-x-y+(c-1) z=0
\)
For non-trivial solution, we have
\(
\begin{aligned}
& \left|\begin{array}{ccc}
a-1 & -1 & -1 \\
-1 & b-1 & -1 \\
-1 & -1 & c-1
\end{array}\right|=0 \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3 \\
& \left|\begin{array}{ccc}
a-1 & -1 & -1 \\
0 & b & -c \\
-1 & -1 & c-1
\end{array}\right|=0 \\
& \mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_3 \\
& \left|\begin{array}{ccc}
a-1 & 0 & -1 \\
0 & b+c & -c \\
-1 & -c & c-1
\end{array}\right|=0 \\
& \text { Apply } \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\
& \left|\begin{array}{ccc}
a-1 & 0 & -1 \\
0 & b+c & -c \\
-a & -c & c
\end{array}\right|=0 \\
& \Rightarrow(a-1)\left[b c+c^2-c^2\right]-1[a(b+c)]=0 \\
& \Rightarrow(a-1)[b c]-a b-a c=0 \\
& \Rightarrow a b c-b c-a b-a c=0 \\
& \Rightarrow a b+b c+c a=a b c \\
&
\end{aligned}
\)
The number of values of \(k\), for which the system of equations :
\(
\begin{aligned}
& (k+1) x+8 y=4 k \\
& k x+(k+3) y=3 k-1
\end{aligned}
\)
has no solution, is
(b) From the given system, we have
\(
\begin{aligned}
& \frac{k+1}{k}=\frac{8}{k+3} \neq \frac{4 k}{3 k-1} \\
& (\because \text { System has no solution) } \\
& \Rightarrow k^2+4 k+3=8 k \\
& \Rightarrow k=1,3
\end{aligned}
\)
If \(k=1\) then \(\frac{8}{1+3} \neq \frac{4.1}{2}\) which is false and if \(k=3\) then \(\frac{8}{6} \neq \frac{4.3}{9-1}\) which is true, therefore \(k=3\)
Hence for only one value of \(k\). System has no solution.
Consider the system of equations : \(x+a y=0, y+a z=0\) and \(z+a x=0\). Then the set of all real values of ‘ \(a\) ‘ for which the system has a unique solution is:
(b) Given system of equations is homogeneous which is
\(
\begin{aligned}
& x+a y=0 \\
& y+a z=0 \\
& z+a x=0
\end{aligned}
\)
It can be written in matrix form as
\(
\mathrm{A}=\left(\begin{array}{lll}
1 & a & 0 \\
0 & 1 & a \\
a & 0 & 1
\end{array}\right)
\)
Now, \(|\mathrm{A}|=\left[1-a\left(-a^2\right)\right]=1+a^3 \neq 0\)
So, system has only trivial solution.
Now, \(|\mathrm{A}|=0\) only when \(a=-1\)
So, system of equations has infinitely many solutions which is not possible because it is given that system has a unique solution.
Hence set of all real values of ‘ \(a\) ‘ is \(\mathrm{R}-\{-1\}\).
Statement-1: The system of linear equations
\(
\begin{aligned}
& x+(\sin \alpha) y+(\cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& x-(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}
\)
has a non-trivial solution for only one value of \(\alpha\) lying in the interval \(\left(0, \frac{\pi}{2}\right)\)
Statement-2: The equation in \(\alpha\)
\(
\left|\begin{array}{ccc}
\cos \alpha & \sin \alpha & \cos \alpha \\
\sin \alpha & \cos \alpha & \sin \alpha \\
\cos \alpha & -\sin \alpha & -\cos \alpha
\end{array}\right|=0
\)
has only one solution lying in the interval \(\left(0, \frac{\pi}{2}\right)\).
\(
\text { (c) } \Delta_1=\left|\begin{array}{ccc}
1 & \sin \alpha & \cos \alpha \\
1 & \cos \alpha & \sin \alpha \\
1 & -\sin \alpha & \cos \alpha
\end{array}\right|
\)
\(
=\left|\begin{array}{ccc}
0 & \sin \alpha-\cos \alpha & \cos \alpha-\sin \alpha \\
0 & \cos \alpha+\sin \alpha & \sin \alpha-\cos \alpha \\
1 & -\sin \alpha & \cos \alpha
\end{array}\right|
\)
\(
\begin{aligned}
& =(\sin \alpha-\cos \alpha)^2-\left(\cos ^2 \alpha-\sin ^2 \alpha\right) \\
& =\sin ^2 \alpha+\cos ^2 \alpha-2 \sin \alpha \cdot \cos \alpha-\cos ^2 \alpha+\sin ^2 \alpha \\
& =2 \sin ^2 \alpha-2 \sin \alpha \cdot \cos \alpha \\
& =2 \sin \alpha(\sin \alpha-\cos \alpha)
\end{aligned}
\)
Now, \(\sin \alpha-\cos \alpha=0\) for only
\(
\begin{aligned}
& \alpha=\frac{\pi}{4} \text { in }\left(0, \frac{\pi}{2}\right) \\
& \Delta_1=2(\sin \alpha) \times 0=0
\end{aligned}
\)
since value of \(\sin \alpha\) is finite for \(\alpha \in\left(0, \frac{\pi}{2}\right)\)
Hence non-trivivial solution for only one value of \(\alpha\) in \(\left(0, \frac{\pi}{2}\right)\)
\(
\begin{aligned}
& \left|\begin{array}{ccc}
\cos \alpha & \sin \alpha & \cos \alpha \\
\sin \alpha & \cos \alpha & \sin \alpha \\
\cos \alpha & -\sin \alpha & -\cos \alpha
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
0 & \cos \alpha & \sin \alpha \\
2 \cos \alpha & -\sin \alpha & -\cos \alpha
\end{array}\right|=0 \\
& \Rightarrow 2 \cos \alpha\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\
& \therefore \cos \alpha=0 \text { or } \sin ^2 \alpha-\cos ^2 \alpha=0
\end{aligned}
\)
But \(\cos \alpha=0\) not possible for any value of \(\alpha \in\left(0, \frac{\pi}{2}\right)\) \(\therefore \sin ^2 \alpha-\cos ^2 \alpha=0 \Rightarrow \sin \alpha=-\cos \alpha\), which is also not possible for any value of \(\alpha \in\left(0, \frac{\pi}{2}\right)\) Hence, there is no solution.
If the system of linear equations:
\(
\begin{aligned}
& x_1+2 x_2+3 x_3=6 \\
& x_1+3 x_2+5 x_3=9 \\
& 2 x_1+5 x_2+a x_3=b
\end{aligned}
\)
is consistent and has infinite number of solutions, then :
(d) Given system of equations can be written in matrix form as \(\mathrm{AX}=\mathrm{B}\) where
\(
A=\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 3 & 5 \\
2 & 5 & a
\end{array}\right) \text { and } B=\left(\begin{array}{l}
6 \\
9 \\
b
\end{array}\right)
\)
Since, system is consistent and has infinitely many solutions
\(
\therefore(\operatorname{adj} . A) B=0
\)
\(
\begin{aligned}
& \Rightarrow\left(\begin{array}{ccc}
3 a-25 & 15-2 a & 1 \\
10-a & a-6 & -2 \\
-1 & -1 & 1
\end{array}\right)\left(\begin{array}{l}
6 \\
9 \\
b
\end{array}\right)=\left(\begin{array}{l}
0 \\
0 \\
0
\end{array}\right) \\
& \Rightarrow-6-9+b=0 \Rightarrow b=15 \\
& \text { and } 6(10-a)+9(a-6)-2(b)=0 \\
& \Rightarrow 60-6 a+9 a-54-30=0 \\
& \Rightarrow 3 a=24 \Rightarrow a=8 \\
& \text { Hence, } a=8, b=15 .
\end{aligned}
\)
Statement 1: If the system of equations \(x+k y+3 z=0\), \(3 x+k y-2 z=0,2 x+3 y-4 z=0\) has a non-trivial solution, then the value of \(k\) is \(\frac{31}{2}\).
Statement 2: A system of three homogeneous equations in three variables has a non-trivial solution if the determinant of the coefficient matrix is zero.
Given system of equations is
\(
\begin{aligned}
& x+k y+3 z=0 \\
& 3 x+k y-2 z=0 \\
& 2 x+3 y-4 z=0
\end{aligned}
\)
Since, system has non-trivial solution
\(
\therefore\left|\begin{array}{ccc}
1 & k & 3 \\
3 & k & -2 \\
2 & 3 & -4
\end{array}\right|=0
\)
\(
\begin{aligned}
& \Rightarrow 1(-4 k+6)-k(-12+4)+3(9-2 k)=0 \\
& \Rightarrow \quad 4 k+33-6 k=0 \Rightarrow k=\frac{33}{2}
\end{aligned}
\)
Hence, statement – 1 is false. Statement- 2 is the property. It is a true statement.
If the system of equations
\(
\begin{aligned}
& x+y+z=6 \\
& x+2 y+3 z=10 \\
& x+2 y+\lambda z=0
\end{aligned}
\)
has a unique solution, then \(\lambda\) is not equal to
(d) Given system of equations is
\(
\begin{aligned}
& x+y+z=6 \\
& x+2 y+3 z=10 \\
& x+2 y+\lambda z=0
\end{aligned}
\)
It has unique solution.
\(
\begin{aligned}
& \therefore\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 2 & \lambda
\end{array}\right| \neq 0 \\
& \Rightarrow 1(2 \lambda-6)-1(\lambda-3)+1(2-2) \neq 0 \\
& \Rightarrow 2 \lambda-6-\lambda+3 \neq 0 \Rightarrow \lambda-3 \neq 0 \Rightarrow \lambda \neq 3
\end{aligned}
\)
If the trivial solution is the only solution of the system of equations
\(
\begin{aligned}
& x-k y+z=0 \\
& k x+3 y-k z=0 \\
& 3 x+y-z=0
\end{aligned}
\)
then the set of all values of \(\mathrm{k}\) is:
(a)
\(
\begin{aligned}
& x-k y+z=0 \\
& k x+3 y-k z=0 \\
& 3 x+y-z=0
\end{aligned}
\)
The given system of equations will have non trivial solution, if
\(
\left|\begin{array}{ccc}
1 & -k & 1 \\
k & 3 & -k \\
3 & 1 & -1
\end{array}\right|=0
\)
\(
\begin{aligned}
& \Rightarrow 1(-3+k)+k(-k+3 k)+1(k-9)=0 \\
& \Rightarrow k-3+2 k^2+k-9=0 \\
& \Rightarrow k^2+k-6=0 \\
& \Rightarrow k=-3, k=2
\end{aligned}
\)
So the equation will have only trivial solution, when \(k \in \mathrm{R}-\{2,-3\}\)
The number of values of \(k\) for which the linear equations \(4 x+k y+2 z=0, k x+4 y+z=0\) and \(2 x+2 y+z=0\) possess a non-zero solution is
(a)
\(
\begin{aligned}
& \Delta=0 \\
& \Rightarrow\left|\begin{array}{lll}
4 & k & 2 \\
k & 4 & 1 \\
2 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow 4(4-2)-k(k-2)+2(2 k-8)=0 \\
& \Rightarrow 8-k^2+2 k+4 k-16=0 \\
& k^2-6 k+8=0 \\
& \Rightarrow(k-4)(k-2)=0 \Rightarrow k=4,2
\end{aligned}
\)
Consider the system of linear equations;
\(
\begin{aligned}
& x_1+2 x_2+x_3=3 \\
& 2 x_1+3 x_2+x_3=3 \\
& 3 x_1+5 x_2+2 x_3=1
\end{aligned}
\)
The system has
\(
\begin{aligned}
& \text { (c) } D=\left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 3 & 1 \\
3 & 5 & 2
\end{array}\right|=0 \\
& D_1=\left|\begin{array}{lll}
3 & 2 & 1 \\
3 & 3 & 1 \\
1 & 5 & 2
\end{array}\right| \neq 0
\end{aligned}
\)
\(\Rightarrow\) Given system, does not have any solution.
\(\Rightarrow\) No solution
Let \(a, b, c\) be any real numbers. Suppose that there are real numbers \(x, y, z\) not all zero such that \(x=c y+b z, y=a z+c x\), and \(z=b x+a y\). Then \(a^2+b^2+c^2+2 a b c\) is equal to
(d)
The given equations are
\(
\begin{aligned}
& -x+c y+b z=0 \\
& c x-y+a z=0 \\
& b x+a y-z=0
\end{aligned}
\)
\(\because x, y, z\) are not all zero
\(\therefore\) The above system should not have unique (zero) solution
\(
\begin{aligned}
& \Rightarrow \Delta=0 \Rightarrow\left|\begin{array}{ccc}
-1 & c & b \\
c & -1 & a \\
b & a & -1
\end{array}\right|=0 \\
& \Rightarrow-1\left(1-a^2\right)-c(-c-a b)+b(a c+b)=0 \\
& \Rightarrow-1+a^2+b^2+c^2+2 a b c=0 \\
& \Rightarrow a^2+b^2+c^2+2 a b c=1
\end{aligned}
\)
The system of equations
\(
\begin{aligned}
& \alpha x+y+z=\alpha-1 \\
& x+\alpha y+z=\alpha-1 \\
& x+y+\alpha z=\alpha-1
\end{aligned}
\)
has infinite solutions, if \(\alpha\) is
(a)
\(
\begin{aligned}
&\begin{aligned}
& \alpha \mathrm{x}+\mathrm{y}+\mathrm{z}=\alpha-1 \\
& \mathrm{x}+\alpha \mathrm{y}+\mathrm{z}=\alpha-1 \\
& \mathrm{x}+\mathrm{y}+\mathrm{z} \alpha=\alpha-1
\end{aligned}\\
&\Delta=\left|\begin{array}{ccc}
\alpha & 1 & 1 \\
1 & \alpha & 1 \\
1 & 1 & \alpha
\end{array}\right|
\end{aligned}
\)
\(
\begin{aligned}
& =\alpha\left(\alpha^2-1\right)-1(\alpha-1)+1(1-\alpha) \\
& =\alpha(\alpha-1)(\alpha+1)-1(\alpha-1)-1(\alpha-1)
\end{aligned}
\)
For infinite solutions, \(\Delta=0\)
\(
\begin{aligned}
& \Rightarrow \quad(\alpha-1)\left[\alpha^2+\alpha-1-1\right]=0 \\
& \Rightarrow \quad(\alpha-1)\left[\alpha^2+\alpha-2\right]=0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow(\alpha-1)\left[\alpha^2+2 \alpha-\alpha-2\right]=0 \\
& \Rightarrow(\alpha-1)[\alpha(\alpha+2)-1(\alpha+2)]=0
\end{aligned}
\)
\(
\begin{aligned}
& (\alpha-1)=0, \alpha+2=0 \\
& \Rightarrow \alpha=-2,1 ;
\end{aligned}
\)
But \(\alpha \neq 1\).
\(
\therefore \alpha=-2
\)
If the system of linear equations
\(
x+2 a y+a z=0 ; x+3 b y+b z=0 \text {; }
\)
\(x+4 c y+c z=0\) has a non – zero solution, then \(\mathrm{a}, \mathrm{b}, \mathrm{c}\).
(d) For homogeneous system of equations to have non zero solution, \(\Delta=0\)
\(\left|\begin{array}{lll}1 & 2 a & a \\ 1 & 3 b & b \\ 1 & 4 c & c\end{array}\right|=0 \quad \quad C_2 \rightarrow C_2-2 C_3\)
\(\left|\begin{array}{ccc}1 & 0 & a \\ 1 & b & b \\ 1 & 2 c & c\end{array}\right|=0 \quad \quad R_3 \rightarrow R_3-R_2, R_2 \rightarrow R_2-R_1\)
On simplification, \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\therefore a, b, c\) are in Harmonic Progression.
For any \(y \in \mathbb{R}\), let \(\cot ^{-1}(y) \in(0, \pi)\) and \(\tan ^{-1}(y) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Then the sum of all the solutions of the equation \(\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\cot ^{-1}\left(\frac{9-y^2}{6 y}\right)=\frac{2 \pi}{3}\) for \(0<|y|<3\), is equal to
(c)
\(
\text { Case-I : } \mathrm{y} \in(-3,0)
\)
\(
\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\pi+\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3}
\)
\(
2 \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=-\frac{\pi}{3}
\)
\(
y^2-6 \sqrt{3} y-9=0 \Rightarrow y=3 \sqrt{3}-6(\because y \in(-3,0))
\)
\(
\text { Case-II : } y \in(0,3)
\)
\(
\begin{aligned}
& 2 \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3} \Rightarrow \sqrt{3} y^2+6 y-9 \sqrt{3}=0 \\
& y=\sqrt{3} \text { or } y=-3 \sqrt{3} \text { (rejected) } \\
& \text { sum }=\sqrt{3}+3 \sqrt{3}-6=4 \sqrt{3}-6
\end{aligned}
\)
Let \(\mathrm{M}=\left(\mathrm{a}_{\mathrm{ij}}\right), \mathrm{i}, \mathrm{j} \in\{1,2,3\}\), be the \(3 \times 3\) matrix such that \(\mathrm{a}_{\mathrm{ij}}=1\) if \(\mathrm{j}+1\) is divisible by \(\mathrm{i}\), otherwise \(a_{i j}=0\). Then which of the following statements is (are) true ?
\(
\mathbf{M}=\left[\begin{array}{lll}
\mathrm{a}_{11} & \mathrm{a}_{12} & \mathrm{a}_{13} \\
\mathrm{a}_{21} & \mathrm{a}_{22} & \mathrm{a}_{23} \\
\mathrm{a}_{31} & \mathrm{a}_{32} & \mathrm{a}_{33}
\end{array}\right]=\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]
\)
\(
|\mathrm{M}|=-1+1=0 \Rightarrow \mathrm{M} \text { is singular so non-invertible }
\)
\(
\text { (B) } M\left[\begin{array}{l}
a_1 \\
a_2 \\
a_3
\end{array}\right]=\left[\begin{array}{l}
-a_1 \\
-a_2 \\
-a_3
\end{array}\right] \Rightarrow\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]\left[\begin{array}{l}
a_1 \\
a_2 \\
a_3
\end{array}\right]=\left[\begin{array}{l}
-a_1 \\
-a_2 \\
-a_3
\end{array}\right]
\)
\(
\left.\begin{array}{l}
\mathrm{a}_1+\mathrm{a}_2+\mathrm{a}_3=-\mathrm{a}_1 \\
\mathrm{a}_1+\mathrm{a}_3=-\mathrm{a}_2 \\
\mathrm{a}_2=-\mathrm{a}_3
\end{array}\right\} \Rightarrow \mathrm{a}_1=0 \text { and } \mathrm{a}_2+\mathrm{a}_3=0 \text { infinite solutions exists [B] is correct. }
\)
Option (D)
\(
\mathbf{M}-2 \mathbf{I}=\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]-2\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-1 & 1 & 1 \\
1 & -2 & 1 \\
0 & 1 & -2
\end{array}\right]
\)
\(
|M-2 I|=0 \Rightarrow[D] \text { is wrong }
\)
Option (C) :
\(
M X=0 \Rightarrow\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
\)
\(
\begin{aligned}
& x+y+z=0 \\
& x+z=0 \\
& y=0
\end{aligned}
\)
\(\therefore\) Infinite solution
[C] is correct
Let \(\mathrm{R}=\left\{\left(\begin{array}{lll}\mathrm{a} & 3 & \mathrm{~b} \\ \mathrm{c} & 2 & \mathrm{~d} \\ 0 & 5 & 0\end{array}\right): \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in\{0,3,5,7,11,13,17,19\}\right\}\). Then the number of invertible matrices in \(\mathrm{R}\) is
Let us calculate when \(|R|=0\)
Case-I: \(\mathrm{ad}=\mathrm{bc}=0\)
Now ad \(=0\)
\(\Rightarrow\) Total – (When none of a \& d is 0 )
\(=8^2-1=15\) ways
Similarly \(\mathrm{bc}=0 \Rightarrow 15\) ways
\(\therefore 15 \times 15=225\) ways of \(\mathrm{ad}=\mathrm{bc}=0\)
Case-II: \(\mathrm{ad}=\mathrm{bc} \neq 0\)
either \(\mathrm{a}=\mathrm{d}=\mathrm{b}=\mathrm{c} \quad\) OR \(\quad \mathrm{a} \neq \mathrm{d}, \mathrm{b} \neq \mathrm{d}\) but \(a d=\mathrm{bc}\)
\(
{ }^7 \mathrm{C}_1=7 \text { ways } \quad \quad \quad{ }^7 \mathrm{C}_2 \times 2 \times 2=84 \text { ways }
\)
Total 91 ways
\(
\begin{aligned}
& \therefore|R|=0 \text { in } 225+91=316 \text { ways } \\
& |R| \neq 0 \text { in } 8^4-316=3780
\end{aligned}
\)
Let for \(A=\left[\begin{array}{lll}1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2\end{array}\right],|A|=2\). If \(|2 \operatorname{adj}(2 \operatorname{adj}(2 A))|=32^n\), then \(3 n+\alpha\) is equal to
\(
\begin{aligned}
& |\mathrm{A}|=2 \\
& \operatorname{adj}(\mathrm{kA})=\mathrm{k}^{\mathrm{m}-1} \operatorname{adj} \mathrm{A} \quad\{\mathrm{m}=\text { order of matrix }\}
\end{aligned}
\)
\(
\begin{aligned}
& \operatorname{adj}(2 A)=2^2 \operatorname{adj} A=4 \operatorname{adj}(A) \\
& \operatorname{adj}(2 \operatorname{adj}(2 A))=\operatorname{adj}(8 \operatorname{adj} A) \\
& =8^2 \operatorname{adj} \operatorname{adj}(A)
\end{aligned}
\)
\(|2 \operatorname{adj} 2 \operatorname{adj}(2 \mathrm{~A})|=\left|2^7 \operatorname{adj} \operatorname{adj}(\mathrm{A})\right|\)
\(
\begin{aligned}
& =\left(2^7\right)^3|\mathrm{~A}|^{2^2} \\
& =2^{21}|\mathrm{~A}|^4 \\
& =2^{21} \cdot 2^4 \\
& \Rightarrow 2^{25}=(32)^{\mathrm{n}} \\
& \Rightarrow 2^{25}=2^{5 \mathrm{n}}
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow & \mathrm{n}=5 \\
& |\mathrm{~A}|=2 \\
& (6-1)-2(2 \alpha-1)+3(\alpha-3)=2 \\
\Rightarrow & 5-4 \alpha+2+3 \alpha-9 \\
\Rightarrow & \alpha=-4 \\
& 3 \mathrm{n}+\alpha=11
\end{aligned}
\)
If the system of equations
\(
\begin{aligned}
& 2 x+y-z=5 \\
& 2 x-5 y+\lambda z=\mu \\
& x+2 y-5 z=7
\end{aligned}
\)
has infinitely many solutions, then \((\lambda+\mu)^2+(\lambda-\mu)^2\) is equal to
(b)
\(
\begin{aligned}
& \Delta=0 \\
& \Rightarrow\left|\begin{array}{ccc}
2 & 1 & -1 \\
2 & -5 & \lambda \\
1 & 2 & -5
\end{array}\right|=0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow 2(25-2 \lambda)-1(-10-\lambda)-1(4+5)=0 \\
& \Rightarrow 51-3 \mathrm{x}=0 \\
& \Rightarrow \lambda=17 \\
& \Delta_{\mathrm{x}}=0
\end{aligned}
\)
\(
\begin{aligned}
& \left|\begin{array}{ccc}
5 & 1 & -1 \\
\mu & -5 & 17 \\
7 & 2 & -5
\end{array}\right|=0 \\
& \Rightarrow 5(25-34)-1(-5 \mu-119)-1(2 \mu+35)=0 \\
& \Rightarrow-45+5 \mu+119-2 \mu-35=0 \\
& \Rightarrow 39+3 \mu=0 \Rightarrow \mu=-13 \\
& (\lambda+\mu)^2+(\lambda-\mu)^2=4^2+(30)^2 \\
& =916
\end{aligned}
\)
The number of symmetric matrices of order 3, with all the entries from the set \(\{0,1,2,3,4,5,6,7,8,9\}\), is
\(
A=\left[\begin{array}{lll}
a & b & c \\
b & d & e \\
c & e & f
\end{array}\right], a, b, c, d, e, f \in\{0,1,2, \ldots .9\}, \text { Number of matrices }=10^6
\)
For the system of linear equations
\(
\begin{aligned}
& 2 x+4 y+2 a z=b \\
& x+2 y+3 z=4 \\
& 2 x-5 y+2 z=8
\end{aligned}
\)
which of the following is NOT correct?
\(
\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
2 & 4 & 2 \mathrm{a} \\
1 & 2 & 3 \\
2 & -5 & 2
\end{array}\right|=18(3-\mathrm{a}) \\
\Delta_{\mathrm{x}} & =\left|\begin{array}{ccc}
\mathrm{b} & 4 & 2 \mathrm{a} \\
4 & 2 & 3 \\
8 & -5 & 2
\end{array}\right|=(64+19 \mathrm{~b}-72 \mathrm{a})
\end{aligned}
\)
For unique solution \(\Delta=0\)
\(
\Rightarrow \mathrm{a} \neq 3 \text { and } \mathrm{b} \in \mathrm{R}
\)
For infinitely many solution :
\(
\begin{aligned}
& \Delta=\Delta_x=\Delta_y=\Delta_z=0 \\
& \Rightarrow a=3 \quad \because \Delta=0 \\
& \text { and } \mathrm{b}=8 \quad \because \Delta_{\mathrm{x}}=0
\end{aligned}
\)
Let \(B=\left[\begin{array}{ccc}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{array}\right], \alpha>2\) be the adjoint of a matrix \(A\) and \(|A|=2\). then
\(\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] \mathrm{B}\left[\begin{array}{c}\alpha \\ -2 \alpha \\ \alpha\end{array}\right]\) is equal to
Given, \(B=\left[\begin{array}{ccc}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{array}\right]\)
\(
\begin{aligned}
& |\mathrm{B}|=4 \\
& 1(8-3 \alpha)-3(4-3 \alpha)+\alpha(\alpha-2 \alpha)=4 \\
& -\alpha^2+6 \alpha-8=0 \\
& \alpha=2,4
\end{aligned}
\)
Given \(\alpha>2\)
So, \(\alpha=2\) is rejected
\(
\left[\begin{array}{lll}
4 & -8 & 4
\end{array}\right]\left[\begin{array}{ccc}
1 & 3 & 4 \\
1 & 2 & 3 \\
4 & 4 & 4
\end{array}\right]\left[\begin{array}{c}
4 \\
-8 \\
4
\end{array}\right]=-16
\)
Let the determinant of a square matrix \(A\) of order \(m\) be \(m-n\), where \(m\) and \(n\) satisfy \(4 m+n=22\) and \(17 m+\) \(4 \mathrm{n}=93\). If \(\operatorname{det}(\mathrm{n} \operatorname{adj}(\operatorname{adj}(\mathrm{mA})))=3^{\mathrm{a}} 5^{\mathrm{b}} 6^{\mathrm{c}}\), then \(\mathrm{a}+\mathrm{b}+\mathrm{c}\) is equal to
\(
\begin{aligned}
& |\mathrm{A}|=\mathrm{m}-\mathrm{n} \\
& 4 \mathrm{~m}+\mathrm{n}=22 \\
& 17 \mathrm{~m}+4 \mathrm{n}=93 \\
& \mathrm{~m}=5, \mathrm{n}=2 \\
& |\mathrm{~A}|=3
\end{aligned}
\)
\(
\mid 2 \operatorname{adj}(\operatorname{adj} 5 A))\left.\left|=2^5\right| 5 A\right|^{16}
\)
\(
\begin{aligned}
& =2^5 \cdot 5^{80}|\mathrm{~A}|^{16} \\
& =2^5 \cdot 5^{80} \cdot 3^{16} \\
& =3^{11} \cdot 5^{80} \cdot 6^5
\end{aligned}
\)
\(
a+b+c=96
\)
Let \(\mathrm{P} b\) a square matrix such that \(\mathrm{P}^2=\mathrm{I}-\mathrm{P}\). For \(\alpha, \beta, \gamma, \delta \in \mathrm{N}\), if \(\mathrm{P}^\alpha+\mathrm{P}^\beta=\gamma \mathrm{I}-29 \mathrm{P}\) and \(\mathrm{P}^\alpha-\mathrm{P}^\beta=\delta \mathrm{I}-13 \mathrm{P}\), then \(\alpha+\beta+\gamma-\delta\) is equal to:
\(
\begin{aligned}
& \mathrm{P}^2=\mathrm{I}-\mathrm{P} \\
& \mathrm{P}^\alpha+\mathrm{P}^\beta=\gamma \mathrm{I}-29 \mathrm{P} \\
& \mathrm{P}^\alpha-\mathrm{P}^\beta=\delta \mathrm{I}-13 \mathrm{P} \\
& \mathrm{P}^4=(\mathrm{I}-\mathrm{P})^2=\mathrm{I}+\mathrm{P}^2-2 \mathrm{P} \\
& \mathrm{P}^4=\mathrm{I}+\mathrm{I}-\mathrm{P}-2 \mathrm{P}=2 \mathrm{I}-3 \mathrm{P} \\
& \mathrm{P}^8=\left(\mathrm{P}^4\right)^2=(2 \mathrm{I}-3 \mathrm{P})^2=4 \mathrm{I}+9 \mathrm{P}^2-12 \mathrm{P} \\
& =4 \mathrm{I}+9(\mathrm{I}-\mathrm{P})-12 \mathrm{P} \\
& \mathrm{P}^8=13 \mathrm{I}-21 \mathrm{P} \dots(1) \\
& \mathrm{P}^6=\mathrm{P}^4 \cdot \mathrm{P}^2 \quad=(2 \mathrm{I}-3 \mathrm{P})(\mathrm{I}-\mathrm{P}) \\
& =2 \mathrm{I}-5 \mathrm{P}+3 \mathrm{P}^2 \\
& =2 \mathrm{I}-5 \mathrm{P}+3(\mathrm{I}-\mathrm{P}) \\
& =5 \mathrm{I}-8 \mathrm{P} \dots(2) \\
&
\end{aligned}
\)
(1) \(+(2)\)
\(
\mathrm{P}^8+\mathrm{P}^6=18 \mathrm{I}-29 \mathrm{P}
\)
\(
\begin{aligned}
& (1)-(2) \\
& \mathrm{P}^8-\mathrm{P}^6=8 \mathrm{I}-13 \mathrm{P}
\end{aligned}
\)
Comparing with given equation
\(
\begin{gathered}
\alpha=8, \quad \beta=6 \\
\gamma=18 \\
\delta=8 \\
\alpha+\beta+\gamma-\delta=32-8=24
\end{gathered}
\)
If the system of equations
\(
\begin{aligned}
& x+y+a z=b \\
& 2 x+5 y+2 z=6 \\
& x+2 y+3 z=3
\end{aligned}
\)
has infinitely many solutions, then \(2 a+3 b\) is equal to :
\(
\begin{aligned}
& x+y+a z=b \\
& 2 x+5 y+2 z=6 \\
& x+2 y+3 z=3
\end{aligned}
\)
For \(\infty\) solution
\(
\Delta=0, \Delta_{\mathrm{x}}=0, \Delta_{\mathrm{y}}=0, \Delta_{\mathrm{z}}=0
\)
\(
\Delta=\left|\begin{array}{lll}
1 & 1 & \mathrm{a} \\
2 & 5 & 2 \\
1 & 2 & 3
\end{array}\right|=0 \Rightarrow 11-4-\mathrm{a}=0 \Rightarrow \mathrm{a}=7
\)
\(
\Delta_z=\left|\begin{array}{lll}
1 & 1 & b \\
2 & 5 & 6 \\
1 & 2 & 3
\end{array}\right|=0 \Rightarrow 3-0-\mathrm{b}=0 \Rightarrow \mathrm{b}=3
\)
\(
\text { Hence } 2 a+3 b=23
\)
If \(\mathrm{A}=\left[\begin{array}{cc}1 & 5 \\ \lambda & 10\end{array}\right], \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}\) and \(\alpha+\beta=-2\), then \(4 \alpha^2+\beta^2+\lambda^2\) is equal to :
\(
\begin{aligned}
& |A-x|=0 \Rightarrow\left|\begin{array}{cc}
1-x & 5 \\
\lambda & 10-x
\end{array}\right|=0 \Rightarrow x^2-11 x+10-5 \lambda=0 \\
& \Rightarrow(10-5 \lambda) A^{-1}=-A+11 I \\
& \therefore \alpha=\frac{-1}{10-5 \lambda} \text { and } \beta=\frac{+11}{10-5 \lambda}
\end{aligned}
\)
\(
\begin{gathered}
\alpha+\beta=-2 \Rightarrow \frac{10}{10-5 \lambda}=-2 \Rightarrow 10-5 \lambda=-5 \Rightarrow \lambda=3 \\
\therefore \alpha=\frac{1}{5} \quad \& \quad \beta=\frac{-11}{5} \\
\therefore 4 a^2+\beta^2+\lambda^2=\frac{4}{25}+\frac{121}{25}+3^2=14 \text { Ans. }
\end{gathered}
\)
Let \(\mathrm{S}\) be the set of all values of \(\theta \in[-\pi, \pi]\) for which the system of linear equations
\(
\begin{aligned}
& x+y+\sqrt{3} z=0 \\
& -x+(\tan \theta) y+\sqrt{7} z=0 \\
& x+y+(\tan \theta) z=0
\end{aligned}
\)
has non-trivial solution. Then \(\frac{120}{\pi} \sum_{\theta=S} \theta\) is equal to
For non trivial solutions
\(
\begin{aligned}
& \mathrm{D}=0 \\
& \left|\begin{array}{ccc}
1 & 1 & \sqrt{3} \\
-1 & \tan \theta & \sqrt{7} \\
1 & 1 & \tan \theta
\end{array}\right|=0 \\
& \tan ^2 \theta-(\sqrt{3}-1)-\sqrt{3}=0 \\
& \tan \theta=\sqrt{3},-1 \\
& \theta=\left\{\frac{\pi}{3}, \frac{-2 \pi}{3}, \frac{-\pi}{4}, \frac{3 \pi}{4}\right\} \\
& \frac{120}{\pi}(\Sigma \theta)=\frac{120}{\pi} \times \frac{\pi}{6}=20
\end{aligned}
\)
\(
\text { Let } \mathrm{P}=\left[\begin{array}{cc}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
-\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{array}\right], \mathrm{A}=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \text { and } Q=P A P^T \text {. If } P^T Q^{2007} P=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]
\),Â
\(
\text { then } 2 a+b-3 c-4 d \text { equal to }
\)
\(
\begin{aligned}
& \mathrm{Q}=\mathrm{PAP}^{\mathrm{T}} \\
& \mathrm{P}^{\mathrm{T}} \cdot \mathrm{Q}^{2007} \cdot \mathrm{P}=\mathrm{P}^{\mathrm{T}} \cdot \mathrm{Q} \cdot \mathrm{Q} \ldots \mathrm{Q} \cdot \mathrm{P}
\end{aligned}
\)
\(
=\mathrm{P}^{\mathrm{T}}\left(\mathrm{PAP}^{\mathrm{T}}\right)\left(\mathrm{P} \cdot \mathrm{AP}^{\mathrm{T}}\right) \ldots\left(\mathrm{PAP}^{\mathrm{T}}\right) \mathrm{P}
\)
\(
\Rightarrow\left(\mathrm{P}^{\mathrm{T}} \mathrm{P}\right) \mathrm{A}\left(\mathrm{P}^{\mathrm{T}} \mathrm{P}\right) \mathrm{A} \ldots \mathrm{A}\left(\mathrm{P}^{\mathrm{T}} \mathrm{P}\right)
\)
\(
\mathrm{P}^{\mathrm{T}} \cdot \mathrm{P}=\left[\begin{array}{cc}
\sqrt{3} / 2 & -1 / 2 \\
1 / 2 & \sqrt{3} / 2
\end{array}\right]\left[\begin{array}{cc}
-\sqrt{3} / 2 & 1 / 2 \\
-1 / 2 & \sqrt{3} / 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\mathrm{I}
\)
\(
\therefore \mathrm{P}^{\mathrm{T}} \cdot \mathrm{Q}^{2007} \cdot \mathrm{P}=\mathrm{A}^{2007}
\)
\(
A^2=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]
\)
\(
\therefore A^{2007}=\left[\begin{array}{cc}
1 & 2007 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]
\)
\(
\begin{aligned}
& a=1, b=2007, c=0, d=1 \\
& 2 a+b-3 c-4 d=2+2007-4=2005
\end{aligned}
\)
\(
\text { Let } A=\left[\begin{array}{ccc}
2 & 1 & 0 \\
1 & 2 & -1 \\
0 & -1 & 2
\end{array}\right]
\)
\(
\text { If }|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 \mathrm{~A}))|=(16)^{\mathrm{n}} \text {, then } \mathrm{n} \text { is equal to }
\)
\(
\begin{aligned}
& |A|=2[3]-1[2]=4 \\
& \therefore|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))| \\
& =|2 A|^{(n-1)^3} \Rightarrow|2 A|^8=16^n \\
& \Rightarrow\left(2^3|A|\right)^8=16^n \\
& \Rightarrow\left(2^3 \times 2^2\right)^8=16^n
\end{aligned}
\)
\(
\begin{aligned}
& =2^{40}=16^n \\
& =16^{10}=16^n \Rightarrow n=10
\end{aligned}
\)
If \(A=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{array}\right]\), then \(|\operatorname{adj}(\operatorname{adj}(2 A))|\) is equal to
\(
\begin{aligned}
& |A|=\frac{1}{5 ! 6 ! 7 !} 5 ! 6 ! 7 !\left|\begin{array}{lll}
1 & 6 & 42 \\
1 & 7 & 56 \\
1 & 8 & 72
\end{array}\right| \\
& R_3 \rightarrow R_3 \rightarrow R_2 \\
& R_2 \rightarrow R_2 \rightarrow R_1 \\
& |A|=\left|\begin{array}{lll}
1 & 8 & 42 \\
0 & 1 & 14 \\
0 & 1 & 16
\end{array}\right|=2 \\
& |\operatorname{adjadj}(2 A)|=|2 A|^{(n-1)^2} \\
& =|2 A|^4 \\
& =\left(2^3|A|\right)^4 \\
& =2^{12}|A|^4 \Rightarrow 2^{16}
\end{aligned}
\)
Let \(\mathrm{S}\) be the set of values of \(\lambda\), for which the system of equations
\(
\begin{aligned}
& 6 \lambda x-3 y+3 z=4 \lambda^2 \\
& 2 x+6 \lambda y+4 z=1
\end{aligned}
\)
\(
3 x+2 y+3 \lambda z=\lambda \text { has no solution. }
\)
Then \(12 \sum_{\mathrm{l} \in \mathrm{S}}|\lambda|\) is equal to _____.
\(
\begin{aligned}
& \Delta=\left|\begin{array}{ccc}
6 \lambda & -3 & 3 \\
2 & 6 \lambda & 4 \\
3 & 2 & 3 \lambda
\end{array}\right|=0 \\
& 2 \lambda\left(9 \lambda^2-4\right)+(3 \lambda-6)+(2-9 \lambda)=0 \\
& 18 \lambda^3-14 \lambda-4=0 \\
& (\lambda-1)(3 \lambda+1)(3 \lambda+2)=0 \\
& \Rightarrow \lambda=1,-1 / 3,-2 / 3
\end{aligned}
\)
For each values of \(\lambda, \Delta_1=\left|\begin{array}{ccc}6 \lambda & -3 & 4 \lambda^2 \\ 2 & 6 \lambda & 1 \\ 3 & 2 & \lambda\end{array}\right| \neq 0\)
\(
12\left(1+\frac{1}{3}+\frac{2}{3}\right)=24
\)
If \(A\) is a \(3 \times 3\) matrix and \(|A|=2\), then \(\left|3 \operatorname{adj}\left(|3 A| A^2\right)\right|\) is equal to :
\(
\begin{aligned}
& \text { Given }|\mathrm{A}|=\mathbf{2} \\
& \text { Now, }\left|3 \operatorname{adj}\left(|3 \mathrm{~A}| \mathrm{A}^2\right)\right| \\
& |3 \mathrm{~A}|=3^3 \cdot|\mathrm{A}| \\
& =3^3 .(2)
\end{aligned}
\)
\(
\begin{aligned}
& \text { Adj. }\left(|3 \mathrm{~A}| \mathrm{A}^2\right)=\operatorname{adj}\left\{\left(3^3 \cdot 2\right) \mathrm{A}^2\right\} \\
& =\left(2 \cdot 3^3\right)^2(\operatorname{adj} \mathrm{A})^2 \\
& =2^2 \cdot 3^6 \cdot(\operatorname{adj} \mathrm{A})^2
\end{aligned}
\)
\(
\begin{aligned}
\left|3 \operatorname{adj}\left(|3 \mathrm{~A}| \mathrm{A}^2\right)\right| & =\left|2^2 \cdot 3 \cdot 3^6(\operatorname{adj} \mathrm{A})^2\right| \\
& =\left(2^2 \cdot 3^7\right)^3|\operatorname{adj} \mathrm{A}|^2 \\
& =2^6 \cdot 3^{21}\left(|\mathrm{~A}|^2\right)^2 \\
& =2^6 \cdot 3^{21}\left(2^2\right)^2 \\
& =2^{10} \cdot 3^{21} \\
& =2^{10} \cdot 3^{10} \cdot 3^{11}
\end{aligned}
\)
\(
\left|3 \operatorname{adj}\left(|3 A| A^2\right)\right|=6^{10} \cdot 3^{11}
\)
For the system of linear equations
\(
\begin{aligned}
& 2 x-y+3 z=5 \\
& 3 x+2 y-z=7 \\
& 4 x+5 y+\alpha z=\beta,
\end{aligned}
\)
which of the following is NOT correct?
Given
\(
\begin{aligned}
& 2 x-y+3 z=5 \\
& 3 x+2 y-z=7 \\
& 4 x+5 y+\alpha z=\beta
\end{aligned}
\)
\(
\Delta=\left|\begin{array}{ccc}
2 & -1 & 3 \\
3 & 2 & -1 \\
4 & 5 & \alpha
\end{array}\right|=7 \alpha+35
\)
\(
\Delta=7(\alpha+5)
\)
For unique solution \(\Delta \neq 0\)
\(
\alpha \neq-5
\)
For inconsistent & Infinite solution
\(
\begin{aligned}
& \Delta=0 \\
& \alpha+5=0 \Rightarrow \alpha=-5
\end{aligned}
\)
\(
\begin{aligned}
\Delta_1 & =\left|\begin{array}{ccc}
5 & -1 & 3 \\
7 & 2 & -1 \\
\beta & 5 & -5
\end{array}\right|=-5(\beta-9) \\
\Delta_2 & =\left|\begin{array}{ccc}
2 & 5 & 3 \\
3 & 7 & -1 \\
4 & \beta & -5
\end{array}\right|=11(\beta-9) \\
\Delta_3 & =\left|\begin{array}{ccc}
2 & -1 & 5 \\
3 & 2 & 7 \\
4 & 5 & \beta
\end{array}\right|
\end{aligned}
\)
For Inconsistent system : –
At least one \(\Delta_1, \Delta_2 \& \Delta_3\) is not zero \(\alpha=-5, \beta=8\) option (a) True Infinite solution:
\(
\Delta_1=\Delta_2=\Delta_3=0
\)
From here \(\beta-9=0 \Rightarrow \beta=9\)
\(\alpha=-5 \&\) option (d) True
\(
\beta=9
\)
Unique solution
\(
\alpha \neq-5, \beta=8 \rightarrow \text { option (c) True }
\)
Option (b) False
For Infinitely many solution \(\alpha\) must be -5 .
If the system of linear equations
\(
\begin{aligned}
& 7 x+11 y+\alpha z=13 \\
& 5 x+4 y+7 z=\beta \\
& 175 x+194 y+57 z=361
\end{aligned}
\)
has infinitely many solutions, then \(\alpha+B+2\) is equal to :
\(
\begin{aligned}
& 7 x+11 y+\alpha z=13 \\
& 5 x+4 y+7 z=\beta \\
& 175 x+194 y+57 z=361
\end{aligned}
\)
\(4 \mathrm{sc}\) condition of Infinite Many solution \(\Delta=0\) \& \(\Delta \mathrm{x}, \Delta \mathrm{y}, \Delta \mathrm{z}=0\) check.
After solving we get \(\alpha+13+2=4\)
Let \(A\) be a \(2 \times 2\) matrix with real entries such that \(A^{\prime}=\alpha A+I\), where \(a \in \mathbb{R}-\{-1,1\}\). If \(\operatorname{det}\left(A^2-A\right)=4\), then the sum of all possible values of \(\alpha\) is equal to :
\(
\begin{aligned}
& \mathrm{A}^{\mathrm{T}}=\alpha \mathrm{A}+\mathrm{I} \\
& \mathrm{A}=\alpha \mathrm{A}^{\mathrm{T}}+\mathrm{I} \\
& \mathrm{A}=\alpha(\alpha \mathrm{A}+\mathrm{I})+\mathrm{I} \\
& \mathrm{A}=\alpha^2 \mathrm{~A}+(\alpha+1) \mathrm{I} \\
& \mathrm{A}\left(1-\alpha^2\right)=(\alpha+1) \mathrm{I}
\end{aligned}
\)
\(
A=\frac{I}{1-\alpha} \dots(1)
\)
\(
|A|=\frac{1}{(1-\alpha)^2} \dots(2)
\)
\(
\left|\mathrm{A}^2-\mathrm{A}\right|=|\mathrm{A}||\mathrm{A}-1| \ldots \text { (3) }
\)
\(
\begin{aligned}
& A-I=\frac{I}{I-\alpha}-I=\frac{\alpha}{1-\alpha} I \\
& |A-I|=\left(\frac{\alpha}{1-\alpha}\right)^2 \ldots(4)
\end{aligned}
\)
\(
\begin{aligned}
& \text { Now }\left|A^2-A\right|=4 \\
& |A||A-I|=4 \\
& \Rightarrow \frac{1}{(1-\alpha)^2} \frac{\alpha^2}{\left(1-\alpha^2\right)}=4 \\
& \Rightarrow \frac{\alpha}{(1-\alpha)^2}= \pm 2 \\
& \Rightarrow 2(1-\alpha)^2= \pm \alpha \\
& \left(C_1\right) 2(1-\alpha)^2=\alpha
\end{aligned}
\)
\(
2 \alpha^2-5 \alpha+2=0<_{\alpha_2}^{\alpha_1}
\)
\(
\alpha_1+\alpha_2=\frac{5}{2}
\)
\(
\left(C_2\right) 2(1-\alpha)^2=-\alpha
\)
\(
\begin{aligned}
& 2 \alpha^2-3 \alpha+2=0 \\
& \alpha \notin \mathrm{R}
\end{aligned}
\)
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