Let $P_1=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], P_2=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$

$P_3=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 4\end{array}\right], P_4=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]$

$P_5=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right], P_6=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$. Then

$\text { and } \quad X=\sum_{k=1}^6 P_k\left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] P_k^T$

Then,

a) Ib $X\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\alpha\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$, then $\alpha=30$

b) X is symmetric matrix

c) X - 30 I is not an invertible matrix

$X\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] = (P_1+P_2+P_3+P_4+P_5+P_6)\left[\begin{array}{l}6 \\ 3 \\ 6\end{array}\right]$

$=2\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right] \left[\begin{array}{l}6 \\ 3 \\ 6\end{array}\right]$

$=2\left[\begin{array}{l}15 \\ 15 \\ 15\end{array}\right] =30\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$

$\Rightarrow X\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] =30\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] =30\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$

$(30)\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=0$

$\Rightarrow X = 30$

b) Need to show X is a symmetric matrix

$X^T=\left(\sum_{k=1}^6 P_k B P_k^T\right)^T$

$=\sum_{k=1}^6 P_k B^T P_k^T$

$=\sum_{k=1}^6 P_k B P_k^T=X$

$\Rightarrow$ X is a symmetric matrix

c) (X - 30 I) is not an invertible matrix

from part we have

$X\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] =30\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$

$\Rightarrow (X-30I)\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] =\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ is non-trivial solution of (X - 30I)y = 0

$\Rightarrow$ X - 30 I is not invertible because if X - 30 I is invertible then (X - 30 I) y = 0

has only zero solution.

Let A be an $n \times n$ matrix, X be an $n \times 1$ vector, and b be an $n \times 1$ vector. Then system of $n$ linear equations in variable

$x=\left[\begin{array}{c}x_1 \\ x_2 \ \vdots \\ x_n\end{array}\right]$

can be written as Ax = b _____ (1)

Any $x$ which satisfies (1) is $\wedge$ aid to be solution of system of Linear equations.

A system of linear equations can have

i) Unique solution

ii) Infinitely many solutions

iii) No solution

Example

i) $x_1+2 x_2=4$

$2 x_1+x_2 =2–(1)$

$x_1=0, x_2=2$ is a unique solution of (1)

ii) $x_1+ x_2=2$

$2x_1+2x_2 = 4$—(2)

It is easy to we that second equation can be obtained by multiplying first equation by 2.

$(\alpha, 2-\alpha)$ where $\alpha \in R$ is solution of (2).

That is, (2) has infinitely many solutions.

iii) $x_1+ x_2=2$

$2 x_1+2x_2 =3$–(3)

This system has no solution.

The question is – How do we decide about the solution of a system of linear equation

Ax = b

$a \rightarrow n \times n$

$x \rightarrow n \times 1$

$b \rightarrow n \times 1$

1. If rank(A) = rank (A | b) = n.

Then system of equations has unique solution.

In this case $\operatorname{det} A \neq 0$

2. If rank (A) = rank (A | b)

= m < n

Then, system of equations has infinitely many solutions.

3) If rank (A) $\neq$ rank (A | b)

Then the system of equations has no solution.

# Rank (A) can be obtained by reducing matrix A, using elementary row operations, to Echelon form.

The number of non-zeso rows of matrix in Echelon form gives the rank of a matrix.

$i) \left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 0 & -1\end{array}\right]$ rank = 3

$ii) \left[\begin{array}{lll}1 & 3 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0\end{array}\right]$ rank = 2

because the number of non-zero rows is 2.

Let $\alpha, \lambda, \mu \in \mathbb{R}$. Consider the system of linear equations

$\alpha x+2 y=\lambda$

$3 x-2 y=\mu$

For what values of $\alpha, \lambda, \mu$ the system of linear equation’s have

a) Unique solution

b) In finitely many solutions

c) No solution

$A=\left[\begin{array}{cc}\alpha & 2 \\ 3 & -2\end{array}\right], b=\left[\begin{array}{l}\lambda \\ \mu\end{array}\right]$

a) For unique solution

det A $\neq$ 0

rank (A | b) = rank (A) = 2

det A = $-2\alpha - b \neq$ 0

$\Rightarrow \alpha \neq -3$

$\Rightarrow$ For $\alpha \neq-3$ and $\lambda, \mu \in \mathbb{R}$ systems will have unique solution

$b) A=\left[\begin{array}{cc}\alpha & 2 \\ 3 & -2\end{array}\right], b=\left[\begin{array}{l}\lambda \\ \mu\end{array}\right]$

$[A | b]=\left[\begin{array}{cc}\alpha & 2 \\ 3 & -2\end{array}\right] \left[\begin{array}{l}\lambda \\ \mu\end{array}\right]$

$~\left[\begin{array}{cc}\alpha & 2 \\ 3+\alpha & 0\end{array}\right], \left[\begin{array}{l}\lambda \\ \lambda+\mu\end{array}\right]$

$R_2 \rightarrow R_2+R_1$

For infinitely many solution

rank (A | b) = rank (A) < 2

If $\alpha$ + 3 = 0 and $\lambda + \mu = 0$

ie $\alpha=-3 \quad 4 \quad \lambda=-\mu$

Then rank (A | b) = rank (A) = 1

$\Rightarrow$ system of equations will have infinitely many solutions

$[A \mid b] \sim\left[\begin{array}{cc|c}\alpha & 2 & 1 \\ 3+\alpha & 0 & \lambda+\mu\end{array}\right]$

If $\alpha=-3 ~ and ~ \lambda \neq-\mu$

$\Rightarrow$ rank (A | b) = 2

rank (A) = 1

i.e. rank (A) $\neq$ rank (A | b)

$\Rightarrow$ systam has no solution.

For a real number $\alpha$ of the system

$\left[\begin{array}{ccc}1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}1 \\ -1 \\ 1\end{array}\right]$

of linear equations has infinitely many solutions, then what will be the value of $1+\alpha+\alpha^2$ ?

Consider the augmented matrix

$\left[\begin{array}{lll}A & \mid & b\end{array}\right]=\left[\begin{array}{ccc|c}1 & \alpha & \alpha^2 & 1 \\ \alpha & 1 & \alpha & -1 \\ \alpha^2 & \alpha & 1 & 1\end{array}\right]$

$\sim\left[\begin{array}{ccc|c}1 & \alpha & \alpha^2 & 1 \\ 0 & 1-\alpha^2 & \alpha-\alpha^3 & -1-\alpha \\ 0 & \alpha-\alpha^3 & 1-\alpha^4 & 1-\alpha^2\end{array}\right]$

$\sim\left[\begin{array}{ccc|c}1 & \alpha & \alpha^2 & 1 \\ 0 & 1-\alpha^2 & \alpha-\alpha^3 & -1-\alpha \\ 0 & 0 & 1-\alpha^2 & 1+\alpha\end{array}\right]$

$R_2 \rightarrow R_2-\alpha R_1$

$R_3 \rightarrow R_3-\alpha^2 R_1$

$R_3 \rightarrow R_3-\alpha R_2$

$\left[\begin{array}{ll}A |b\end{array}\right]=\left[\begin{array}{ccc|c}1 & \alpha & \alpha^2 & 1 \\ 0 & 1-\alpha^2 & \alpha-\alpha^3 & -1-\alpha \\ 0 & 0 & 1-\alpha^2 & 1+\alpha\end{array}\right]$

Since the system has infinitely many solutions

$\Rightarrow$ rank (A | b) = rank (A) < 3

If 1 - $\alpha^2 = 0$ and 1 + $\alpha = 0$

$\Rightarrow \alpha = -1$

$\Rightarrow 1+\alpha+\alpha^2 = 1$

$\Rightarrow$ rank (A | b) = rank k(A) = 2 < 3

$\Rightarrow$ for $\alpha$ = -1 system has infinite solutions.

x - 2y + 3z = -1

-x + y - 2z = K

x - 3y + 4z = 1

For what values of K system of equations has no solution.

Soln. $[A \mid b]=\left[\begin{array}{ccc|c}1 & -2 & 3 & -1 \\ -1 & 1 & -2 & k \\ 1 & -3 & 4 & 1\end{array}\right]$

$R_2 \rightarrow R_2+R_1, R_3 \rightarrow R_3-R_1$

$\sim\left[\begin{array}{ccc|c}1 & -2 & 3 & -1 \\ 0 & -1 & 1 & k-1 \\ 0 & -1 & 1 & 2\end{array}\right] R_3 \rightarrow R_3-R_2$

$\sim\left[\begin{array}{ccc|c}1 & -2 & 3 & -1 \\ 0 & -1 & 1 & k-1 \\ 0 & 0 & 0 & 3-k\end{array}\right]$

rank (A) = 2, For system to have ho solution we need rank (A | b) = 3

$\Rightarrow k \neq 3$, rank (A | b) = 3

$\Rightarrow$ For $k \neq 3$ system will have no solution.