Linear Algebra

Question 1
Let y be a non-zero vector of size 2022 x 1. Which of the following statement(s) is/are TRUE?
A
yy^T is a symmetric matrix.
B
y^Ty is an eigenvalue of yy^T
C
yy^T has a rank of 2022.
D
yy^T is invertible.
GATE CE 2022 SET-2   Engineering Mathematics
Question 1 Explanation: 
Let vector
\begin{aligned} y&=\begin{bmatrix} 4\\ 4\\ 4 \end{bmatrix}_{3 \times 1}\\ y^T&=\begin{bmatrix} 4& 4& 4 \end{bmatrix}_{1 \times 3}\\ yy^T&=\begin{bmatrix} 4\\ 4\\ 4 \end{bmatrix}\begin{bmatrix} 4& 4& 4 \end{bmatrix}\\ &=\begin{bmatrix} 16 & 16 & 16\\ 16 & 16 & 16\\ 16 & 16 & 16 \end{bmatrix}\\ y^Ty&=[4^2+4^2+4^2]_{1 \times 1}\\ &=[48]_{1 \times 1}\\ \rho (y)&=\rho (y^T)=\rho (yy^T)=\rho (y^Ty)=1 \end{aligned}
From above information
yy^T is asymmetric.
y^Ty is an eigen value of yy^T .
yy^T has rank 1
det(yy^T) =0 so, yy^T is not invertible.
Question 2
P and Q are two square matrices of the same order. Which of the following statement(s) is/are correct?
A
If P and Q are invertible, then [PQ]^{-1}=Q^{-1}P^{-1}
B
If P and Q are invertible, then [QP]^{-1}=P^{-1}Q^{-1}
C
If P and Q are invertible, then [PQ]^{-1}=Q^{-1}P^{-1}
D
If P and Q are not invertible, then [PQ]^{-1}=P^{-1}Q^{-1}
GATE CE 2022 SET-2   Engineering Mathematics
Question 2 Explanation: 
If P and Q are invertible then (PQ)^{-1} = Q^{-1}P^{-1} is correct. Let,
\begin{aligned} PQ&=C \\ P^{-1}PQ&=P^{-1}C \\ Q&=P^{-1} C\\ Q^{-1}Q&=Q^{-1}P^{-1}C \\ I&=Q^{-1}P^{-1}C \\ IC^{-1}&=Q^{-1}P^{-1}CC^{-1} \\ C^{-1}&= Q^{-1}P^{-1}\\ (PQ)^{-1}&= Q^{-1}P^{-1} \end{aligned}
Hence, proved. Similarly, we can prove if P, Q are invertible then (QP)^{-1}= P^{-1}Q^{-1}
Question 3
The matrix M is defined as
M=\begin{bmatrix} 1 & 3\\ 4 & 2 \end{bmatrix}
and has eigenvalues 5 and -2. The matrix Q is formed as
Q=M^3-4M^2-2M
Which of the following is/are the eigenvalue(s) of matrix Q ?
A
15
B
25
C
-20
D
-30
GATE CE 2022 SET-1   Engineering Mathematics
Question 3 Explanation: 
Eigen values of M are 5, -2. So, eigen value of Q=M^3-4M^2-2M are
5^3-4 \times 5^2-2 \times 5=15
(-2)^3-4 \times (-2)^2-2 \times (-2)=-20
Question 4
The smallest eigenvalue and the corresponding eigenvector of the matrix \left[\begin{array}{cc} 2 & -2 \\ -1 & 6 \end{array}\right], respectively, are
A
1.55 and \left\{\begin{array}{l} 2.00 \\ 0.45 \end{array}\right\}
B
2.00 and \left\{\begin{array}{l} 1.00 \\ 1.00 \end{array}\right\}
C
1.55 and \left\{\begin{array}{l} -2.55 \\ -0.45 \end{array}\right\}
D
1.55 and \left\{\begin{array}{c} 2.00 \\ -0.45 \end{array}\right\}
GATE CE 2021 SET-2   Engineering Mathematics
Question 4 Explanation: 
\begin{aligned} A&=\left[\begin{array}{cc} 2 & -2 \\ -1 & 6 \end{array}\right] \Rightarrow|A-\lambda I|=0 \\ \Rightarrow \qquad \lambda&=(4+\sqrt{6}) \text { and }(4-\sqrt{6})\\ A X&=\lambda X\\ (A-\lambda I) X&=0 \end{aligned}
{\left[\begin{array}{cc} 2-(4-\sqrt{6}) & -2 \\ -1 & 6-(4-\sqrt{6}) \end{array}\right] \left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right] =\left[\begin{array}{l} 0 \\ 0 \end{array}\right]}
\begin{aligned} x_{1}&=\left(\frac{2}{-2+\sqrt{6}}\right) x_{2}\\ \text { Let, } \qquad x_{2}&=K \text { then } x_{1}=\left(\frac{2}{-2+\sqrt{6}}\right) \text { K }\\ \Rightarrow \qquad\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]&=\left[\begin{array}{c} \frac{2}{-2+\sqrt{6}} k \\ k \end{array}\right] \approx\left[\begin{array}{c} 2 \\ -2+\sqrt{6} \end{array}\right]=\left[\begin{array}{l} 2.00 \\ 0.45 \end{array}\right] \end{aligned}
Question 5
If A is a square matrix then orthogonality property mandates
A
A A^{T}=I
B
A A^{T}=0
C
A A^{T}=A^{-1}
D
A A^{T}=A^{2}
GATE CE 2021 SET-2   Engineering Mathematics
Question 5 Explanation: 
\text { If, } \qquad \qquad A A^{\top}=I \quad \text { or } A^{-1}=A^{T}
The matrix is orthogonal.
Question 6
The rank of the matrix \left[\begin{array}{cccc} 5 & 0 & -5 & 0 \\ 0 & 2 & 0 & 1 \\ -5 & 0 & 5 & 0 \\ 0 & 1 & 0 & 2 \end{array}\right] is
A
1
B
2
C
3
D
4
GATE CE 2021 SET-2   Engineering Mathematics
Question 6 Explanation: 
\begin{aligned} \left[\begin{array}{cccc} 5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ -5 & 0 & -1 & 0 \\ 0 & 1 & 0 & 2 \end{array}\right] & \stackrel{R_{1} \longleftrightarrow R_{1}+R_{3}}{\longrightarrow}\left[\begin{array}{llll} 5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2 \end{array}\right] \\ & \stackrel{R_{4} \longleftrightarrow R_{4}-\frac{1}{2} R_{2}}{\longrightarrow}\left[\begin{array}{llll} 5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{3}{2} \end{array}\right]\\ &R_{3} \longleftrightarrow R_{4}\left[\begin{array}{llll}5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & \frac{3}{2} \\ 0 & 0 & 0 & 0\end{array}\right] \end{aligned}
Rank(A) = 3
Question 7
If P=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] and Q=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] then Q^{T} P^{T} is
A
\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]
B
\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]
C
\left[\begin{array}{ll} 2 & 1 \\ 4 & 3 \end{array}\right]
D
\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right]
GATE CE 2021 SET-1   Engineering Mathematics
Question 7 Explanation: 
\begin{array}{l} \quad P Q=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right] \\ (P Q)^{\top}=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right] \end{array}
Now using Reversal law
Q^{\top} P^{\top}=(P Q) T=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right]
Question 8
The rank of matrix \left[\begin{array}{llll} 1 & 2 & 2 & 3 \\ 3 & 4 & 2 & 5 \\ 5 & 6 & 2 & 7 \\ 7 & 8 & 2 & 9 \end{array}\right] is
A
1
B
2
C
3
D
4
GATE CE 2021 SET-1   Engineering Mathematics
Question 8 Explanation: 
Using R_{2} \rightarrow R_{2} \rightarrow 3 R_{1}, R_{3} \rightarrow R_{3}-5 R_{1}, R_{4} \rightarrow R_{4}-7 R_{1}
A=\left[\begin{array}{cccc} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & -4 & -8 & -8 \\ 0 & -6 & -12 & -12 \end{array}\right]
Using R_{3} \rightarrow R_{3}-2 R_{2}, R_{4} \rightarrow R_{4}-3 R_{2}
A=\left[\begin{array}{cccc} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]
So, \rho(A)= No. of non-zero rows = 2.
Question 9
A 4x4 matrix [P] is given below
[P]=\begin{bmatrix} 0 &1 &3 &0 \\ -2& 3 & 0 & 4\\ 0& 0& 6 & 1\\ 0& 0& 1& 6 \end{bmatrix}
The eigen values of [P] are
A
0,3,6,6
B
1,2,3,4
C
3,4,5,7
D
1,2,5,7
GATE CE 2020 SET-2   Engineering Mathematics
Question 9 Explanation: 
|P|= 70 and Trace (P) = 15
So, only option, (1, 2, 5, 7) satisfies.
Question 10
Consider the system of equations
\begin{bmatrix} 1 & 3 &2 \\ 2& 2 & -3\\ 4 &4 &-6 \\ 2& 5 &2 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 1\\ 1\\ 2\\ 1 \end{bmatrix}
The value of x_3 (round off to the nearest integer), is _______.
A
1
B
2
C
3
D
4
GATE CE 2020 SET-1   Engineering Mathematics
Question 10 Explanation: 
[A:B]=\begin{bmatrix} 1 &3 & 2 &\vdots &1 \\ 2 &2 & -3 &\vdots &1 \\ 4 &4 & -6 &\vdots &2 \\ 3 &5 & 2 &\vdots &1 \\ \end{bmatrix}
Converting into an Echelin Form
=\begin{bmatrix} 1 &3 & 2 &\vdots &1 \\ 0 &-1 & -2 &\vdots &-1 \\ 0 &0 & 1 &\vdots &3 \\ 0 &0 & 0 &\vdots &0 \\ \end{bmatrix}
\Rightarrow \; \begin{bmatrix} 1 &3 & 2\\ 0&-1 &-2 \\ 0&0 &1 \\ 0 &0 &0 \end{bmatrix}\begin{bmatrix} x_1\\ X_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 1\\ -1\\ 3\\ 0 \end{bmatrix}\Rightarrow \; x_3=3
There are 10 questions to complete.

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