# Linear Algebra

 Question 1
Cholesky decomposition is carried out on the following square matrix $[\mathrm{A}]$.

$[A]=\left[\begin{array}{cc} 8 & -5 \\ -5 & a_{22} \end{array}\right]$

Let $\mathrm{I}_{\mathrm{ij}}$ and $\mathrm{a}_{\mathrm{ij}}$ be the $(i, j)^{\text {th }}$ elements of matrices $[L]$ and $[A]$, respectively. If the element $I_{22}$ of the decomposed lower triangular matrix $[\mathrm{L}]$ is 1.968 , what is the value (rounded off to the nearest integer) of the element $a_{22}$ ?
 A 5 B 7 C 9 D 11
GATE CE 2023 SET-2   Engineering Mathematics
Question 1 Explanation:
We know, cholesky decomposition,
$A=LL^{T}$
Where, $L=$ lower tringular matrix
$\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[\begin{array}{cc}L_{11} & 0 \\ L_{21} & L_{22}\end{array}\right]\left[\begin{array}{cc}\mathrm{L}_{11} & L_{21} \\ 0 & L_{22}\end{array}\right]$

$\left[\begin{array}{cc}8 & -5 \\ -5 & \mathrm{a}_{22}\end{array}\right]=\left[\begin{array}{cc}\mathrm{L}_{11}^{2} & \mathrm{~L}_{11} \mathrm{~L}_{21} \\ \mathrm{~L}_{11} \mathrm{~L}_{21} & \mathrm{~L}_{21}^{2}+\mathrm{L}_{22}^{2}\end{array}\right]$
On comparison on both sides,
$\mathrm{L}_{11}=\sqrt{8}=2 \sqrt{2}$
$and,\;\; L_{11} L_{21}=-5$
$\mathrm{L}_{21}=\frac{-5}{2 \sqrt{2}}$
$and,\;\; a_{22}=L_{21}^{2}+L_{22}^{2}$
$=\left(\frac{-5}{2 \sqrt{2}}\right)^{2}+1.968^{2}$
$=6.998 \approx 7$
 Question 2
For the matrix
$[A]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{array}\right]$
which of the following statements is/are TRUE?
 A $[A]\{X\}=\{b\}$ has a unique solution B $[A]\{X\}=\{b\}$ does not have a unique solution C $[A]$ has three linearly independent eigenvectors D $[\mathrm{A}]$ is a positive definite matrix
GATE CE 2023 SET-2   Engineering Mathematics
Question 2 Explanation:
$|A|=\left|\begin{array}{ccc}1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1\end{array}\right|$
$=1(2-1)+1(-1)+0$
$=1-1=0$

So $A X=B$ does not have unique solution because $\rho(A) \lt 3$
$|A-\lambda|=0 \Rightarrow\left|\begin{array}{ccc}1-\lambda & -1 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & -1 & 1-\lambda\end{array}\right|=0$
$(1-\lambda)((2-\lambda)(1-\lambda)-1)+1(-1+\lambda)+0=0$
$(1-\lambda)\left(\lambda^{2}-3 \lambda+1\right)-1+\lambda=0$
$(1-\lambda)\left(\lambda^{2}-3 \lambda+1-1\right)=0$
$(\lambda-1)\left(\lambda^{2}-3 \lambda\right)=0$
$\lambda=0,1,3$

Matrix $A$ has three distinct Eigen values so have three linearly independent eigen vectors. so option (C) is correct.
Given matrix is symmetric matrix with real value entries. Hence $A$ is not a positive definite matrix. because
$|1|=1 \gt 0$
$\left|\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right|=2-1=1>0$
$\left|\begin{array}{ccc}1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1\end{array}\right|=0$ (which is not positive)
Hence option (D) is incorrect.

 Question 3
For the matrix
$[A]=\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 3 & 1 & 2 \end{array}\right]$
Which of the following statements is/are TRUE?
 A The eigenvalues of $[A]^T$ are same as the eigenvalues of $[A]$ B The eigenvalues of $[A]^{-1}$are the reciprocals of the eigenvalues of $[A]$ C The eigenvectors of $[A]^T$ are same as the eigenvectors of $[A]$ D The eigenvectors of $[A]^{-1}$ are same as the eigenvectors of $[A]$
GATE CE 2023 SET-1   Engineering Mathematics
Question 3 Explanation:
$A x=\lambda x \ldots$ (i)
$A^{T} x=\lambda x \ldots$ (ii)
$A$ and $A^T$ both have same eigen values and eigen vectors.
$A x=\lambda x \ldots(i)$
$\Rightarrow \quad A^{-1} A x=A^{-1}(\lambda x)=\lambda A^{-1} x$
$\Rightarrow \quad x=\lambda A^{-1} x$
$A^{-1} x=\frac{1}{\lambda} x$
So, eigen value and eigen vector of $A^{-1}$ is $\frac{1}{\lambda} x$ and $\mathrm{x}$.
 Question 4
The state equation of a second order system is

$\dot{x}(t)=A x(t), x(0)$ is the initial condition.

Suppose $\lambda_{1}$ and $\lambda_{2}$ are two distinct eigenvalues of $A$ and $v_{1}$ and $v_{2}$ are the corresponding eigenvectors. For constants $\alpha_{1}$ and $\alpha_{2}$, the solution, $x(t)$, of the state equation is
 A $\sum_{i=1}^{2} \alpha_{i} e^{\lambda_{i} t} v_{i}$ B $\sum_{i=1}^{2} \alpha_{i} e^{2 \lambda_{i} t} v_{i}$ C $\sum_{i=1}^{2} \alpha_{i} e^{3 \lambda_{i} t} v_{i}$ D $\sum_{i=1}^{2} \alpha_{i} e^{4 \lambda_{i} t} v_{i}$
GATE EC 2023   Engineering Mathematics
 Question 5
Let $x$ be an $n \times 1$ real column vector with length $l=\sqrt{x^{T} x}$. The trace of the matrix $P=x x^{T}$ is
 A $l^{2}$ B $\frac{l^{2}}{4}$ C $l$ D $\frac{l^{2}}{2}$
GATE EC 2023   Engineering Mathematics
Question 5 Explanation:
Given,
$l=\sqrt{x^{T} x}, P=\left(x x^{T}\right)_{n \times n}$

Let
\begin{aligned} (x)_{n \times 1} & =\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ \vdots \\ x_{n} \end{array}\right] \\ l & =\sqrt{x^{T} x}=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\ldots x_{n}^{2}} \\ P & =x x^{T} \\ &=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ \cdot \\ x_{n} \end{array}\right]\left[x_{1} x_{2} x_{3} \ldots x_{n}\right]\\ P&=\left[ \begin{array}{lllll} x_{1}^{2} & & & & \\ & x_{1}^{2} & & & \\ & & - & & \\ & & & - & \\ & & & & x_{n}^{2} \end{array} \right] \end{aligned}
Trace of $P=x_{1}^{2}+x_{2}^{2}+ . . .+ x_{n}^{2}=l^2$

There are 5 questions to complete.

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### 3 thoughts on “Linear Algebra”

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