# Linear Algebra

 Question 1
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 1 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 2
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 2 Explanation:
$\text { If, } \qquad \qquad A A^{\top}=I \quad \text { or } A^{-1}=A^{T}$
The matrix is orthogonal.
 Question 3
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 3 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 4
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 4 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 5
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 5 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 6
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 6 Explanation:
|P|= 70 and Trace (P) = 15
So, only option, (1, 2, 5, 7) satisfies.
 Question 7
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 7 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
$=\begin{bmatrix} 1 &3 & 2 &\vdots &1 \\ 0 &-1 & -2 &\vdots &-1 \\ 0 &0 & 1 &\vdots &3 \\ 0 &0 & 0 &\vdots &0 \\ \end{bmatrix}$
$\Rightarrow \; \begin{bmatrix} 1 &3 & 2\\ 0&-1 &-2 \\ 0&0 &1 \\ 0 &0 &0 \end{bmatrix}\begin{bmatrix} x_1\\ X_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 1\\ -1\\ 3\\ 0 \end{bmatrix}\Rightarrow \; x_3=3$
 Question 8
The inverse of the matrix $\begin{bmatrix} 2 & 3 & 4\\ 4& 3 & 1\\ 1 & 2 & 4 \end{bmatrix}$ is
 A $\begin{bmatrix} 10 & -4 & -9\\ -15& 4 & 14\\ 5 & -1 & -6 \end{bmatrix}$ B $\begin{bmatrix} -10 & 4 & 9\\ 15& -4 & -14\\ -5 & 1 & 6 \end{bmatrix}$ C $\begin{bmatrix} -2 & \frac{4}{5} & \frac{9}{5}\\ 3& -\frac{4}{5} & -\frac{14}{5}\\ -1 & \frac{1}{5} & \frac{6}{5} \end{bmatrix}$ D $\begin{bmatrix} 2 & -\frac{4}{5} & -\frac{9}{5}\\ -3& \frac{4}{5} & \frac{14}{5}\\ 1 & -\frac{1}{5} & -\frac{6}{5} \end{bmatrix}$
GATE CE 2019 SET-2   Engineering Mathematics
Question 8 Explanation:
det (A) = 2(12 - 2) - 3 (16 - 1) + 4 (8 -3)
= 2(10) - 3 (15) + 4(5)
det (A) = - 5
Minor of 2 = 12 - 2 = 10
Minor of 3 = 16 - 1 = 15
Minor of 4 = 8- 3 = 5
Minor of 4 = 12 - 8 = 4
Minor of 3 = 8 - 4 = 4
Minor of 1 = 4 - 3 = 1
Minor of 1 = 3 - 12 = -9
Minor of 2 = 2 - 16 = -14
Minor of 4 = 6 - 12 = -6
Cofactors of $A=\begin{bmatrix} 10 &-15 &5 \\ -4& 4 & -1\\ -9& 14 &-6 \end{bmatrix}$
\begin{aligned} \text{adj} \;A &= (\text{cofactors of A})^T \\ \text{adj} \;A &=\begin{bmatrix} 10 &-4 &-9 \\ -15& 4 &14 \\ 5& -1 &-6 \end{bmatrix} \\ \therefore \;A^{-1}=\frac{\text{adj} \;A}{|A|} &=\frac{-1}{5}\begin{bmatrix} 10 &-4 &-9 \\ -15& 4 &14 \\ 5& -1 &-6 \end{bmatrix}\\ A^{-1}&=\begin{bmatrix} -2 &\frac{4}{5} &\frac{9}{5} \\ 3& \frac{-4}{5} & \frac{-14}{5}\\ -1& \frac{1}{5} &\frac{6}{5} \end{bmatrix} \end{aligned}
 Question 9
The rank of the following matrix is
$\begin{pmatrix} 1 &1 &0 &-2 \\ 2 & 0 &2 &2 \\ 4 &1 &3 &1 \\ \end{pmatrix}$
 A 1 B 2 C 3 D 4
GATE CE 2018 SET-2   Engineering Mathematics
Question 9 Explanation:
\begin{aligned} A&=\begin{pmatrix} 1 & 1 & 0 &-2 \\ 2 & 0 & 2 &2 \\ 4 & 1 & 3 & 1 \end{pmatrix} \\ R_{2}&\rightarrow R_{2}-2R_{1},\: R_{3}\rightarrow R_{3}-4R_{1} \\ & \begin{pmatrix} 1 &1 &0 & -2\\ 0 & -2 & 2 & 6\\ 0 & -3 & 3 & 9 \end{pmatrix} \\ R_{3} & \rightarrow R_{3}-\frac{3}{2}R_{2} \\ & \begin{pmatrix} 1 &1 &0 & -2\\ 0 & -2 & 2 & 6\\ 0 & 0 & 0 & 0 \end{pmatrix} \end{aligned}
Number of non zero rows $=2$
Rank of $A=2$
 Question 10
The matrix $\begin{pmatrix} 2 &-4 \\ 4 &-2 \end{pmatrix}$ has
 A real eigenvalues and eigenvectors B real eigenvalues but complex eigenvectors C complex eigenvalues but real eigenvectors D complex eigenvalues and eigenvectors
GATE CE 2018 SET-2   Engineering Mathematics
Question 10 Explanation:
\begin{aligned} A&=\begin{bmatrix} 2 & -4\\ 4 & -2 \end{bmatrix} \\ \left | A-\lambda I \right |&=0 \\ \begin{vmatrix} 2-\lambda & -4\\ 4 & -2-\lambda \end{vmatrix}&=0 \\ -4-2\lambda +2\lambda +\lambda ^{2}+16&=0 \\ \lambda ^{2}+12&=0 \\ \lambda &=\pm 2\sqrt{3}i\text{ (Complex eigen values)} \end{aligned}
$(1) \;\; \lambda =2\sqrt{3}i$
Consider $\left ( A-\lambda I \right )X=0$
$\begin{bmatrix} 2-2\sqrt{3}i &-4 \\ 4 & -2-2\sqrt{3}i \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$
\begin{aligned} 2-2\sqrt{3}ix_{1}&=4x_{2} \\ \frac{x_{1}}{4}&=\frac{x_{2}}{2-2\sqrt{3}i} \\ \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}&=\begin{bmatrix} 4\\ 2-2\sqrt{3}i \end{bmatrix}\end{aligned}
$(2)\;\; \lambda =-2\sqrt{3}i$
Consider $\left ( A-\lambda I \right )X=0$
$\begin{bmatrix} 2+2\sqrt{3}i & -4\\ 4 & -2+2\sqrt{3}i \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$
\begin{aligned} 2+2\sqrt{3}ix_{1}&=4x_{2} \\ \frac{x_{1}}{4}&=\frac{x_{2}}{2+2\sqrt{3}i} \\ \begin{bmatrix} x_{1}\\ x-{2} \end{bmatrix}&=\begin{bmatrix} 4\\ 2+2\sqrt{3}i \end{bmatrix} \end{aligned}
$\therefore$ Complex eigen values and complex eigen vectors.
There are 10 questions to complete.