Solve the system
$a+b=8$
$a+c=13$
$b+d=8$
$c-d=5$

Solution $\quad$ $\left[\begin{array}{cccc}1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1\end{array}\right]\left[\begin{array}{l}a \\ b \\ c \\ d\end{array}\right]=\left[\begin{array}{l}8 \\ 13 \\ 8 \\ 5\end{array}\right]$

$\quad$ $\left[\begin{array}{cccc|c} 1 & 1 & 0 & 0 & 8 \\ 1 & 0 & 1 & 0 & 13 \\ 0 & 1 & 0 & 1 & 8 \\ 0 & 0 & 1 & -1 & 5 \end{array}\right]$

$\begin{aligned} & R_2 \rightarrow R_2-R_1\left[\begin{array}{cccc|c}1 & 1 & 0 & 0 & 8 \\ 0 & -1 & 1 & 0 & 5 \\ 0 & 1 & 0 & 1 & 8 \\ 0 & 0 & 1 & -1 & 5\end{array}\right] \\ & R_2 \rightarrow-R_2\left[\begin{array}{cccc|c}1 & 1 & 0 & 0 & 8 \\ 0 & 1 & -1 & 0 & -5 \\ 0 & 1 & 0 & 1 & 8 \\ 0 & 0 & 1 & -1 & 5\end{array}\right] \\ & \begin{array}{l}R_1 \rightarrow R_1-R_2 \\ R_3 \rightarrow R_3-R_2\end{array}\left[\begin{array}{cccc|c}1 & 0 & 1 & 0 & 13 \\ 0 & 1 & -1 & 0 & -5 \\ 0 & 0 & 1 & 1 & 13 \\ 0 & 0 & 1 & -1 & 5\end{array}\right] \\ & \end{aligned}$

$\begin{aligned} & \begin{array}{l}R_1 \rightarrow R_1-R_3 \\ R_2 \rightarrow R_2+R_3 \\ R_4 \rightarrow R_4-R_3\end{array}\left[\begin{array}{cccc|c}1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 1 & 8 \\ 0 & 0 & 1 & 1 & 13 \\ 0 & 0 & 0 & -2 & -8\end{array}\right] \\ & R_4 \rightarrow \frac{-1}{2} R_4\left[\begin{array}{cccc|c}1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 1 & 8 \\ 0 & 0 & 1 & 1 & 13 \\ 0 & 0 & 0 & 1 & 4\end{array}\right] \\ & \begin{array}{l}R_1 \rightarrow R_1+R_4 \\ R_2 \rightarrow R_2-R_4 \\ R_3 \rightarrow R_3-R_4\end{array}\left[\begin{array}{llll|l}1 & 0 & 0 & 0 & 4 \\ 0 & 1 & 0 & 0 & 4 \\ 0 & 0 & 1 & 0 & 9 \\ 0 & 0 & 0 & 1 & 4\end{array}\right] \\ & \end{aligned}$
$a=4, b=4, c=9, d=4$

Solve the system

$\begin{gathered} x-3 y+2 z=0 \\ 2 x-5 y-2 z=0 \\ 4 x-11 y+2 z=0 \end{gathered}$

Solution
$\begin{aligned} & \left[\begin{array}{ccc}1 & -3 & 2 \\ 2 & -5 & -2 \\ 4 & -11 & 2\end{array}\right] \\ & \begin{array}{l}R_2 \rightarrow R_2-2 R_1 \\ R_3 \longrightarrow R_3-4 R_1\end{array}\left[\begin{array}{ccc}1 & -3 & 2 \\ 0 & 1 & -6 \\ 0 & 1 & -6\end{array}\right] \\ & \end{aligned}$

$\begin{aligned} & R_1 \rightarrow R_1+3 R_2 \\ & R_3 \longrightarrow R_3-R_2 \end{aligned}\left[\begin{array}{ccc} 1 & 0 & -16 \\ 0 & 1 & -6 \\ 0 & 0 & 0 \end{array}\right]$

$z$ - independent Variable Let $z=\lambda$

$\begin{gathered} x-16z=0 \Rightarrow x=16 \lambda \\ y-6z=0 \Rightarrow y=6 \lambda \\ (16 \lambda, 6 \lambda, \lambda), \quad \lambda \in \mathbb{R} . \end{gathered}$

Solve the system

$\begin{aligned} & t-u+2 v-3 w=9 \\ & 4 t+11 v-10 w=46 \\ & 3 t-u+8 v-6 w=27 \end{aligned}$

$\begin{aligned} & \left[\begin{array}{cccc|c} 1 & -1 & 2 & -3 & 9 \\ 4 & 0 & 11 & -10 & 46 \\ 3 & -1 & 4 & -6 & 27 \end{array}\right] \\ & R_2 \rightarrow R_2-4 R_1 , \quad R_3 \rightarrow R_3-3R_1 \left[\begin{array}{cccc|c} 1 & -1 & 2 & -3 & 9 \\ 0 & 4 & 3 & 2 & 10 \\ 0 & 2 & 2 & 3 & 0 \end{array}\right] \end{aligned}$

$R_2 \rightarrow \frac{1}{4} R_2\left[\begin{array}{cccc|c}1 & -1 & 2 & -3 & 9 \\ 0 & 1 & 3 / 4 & 1 / 2 & 5 / 2 \\ 0 & 2 & 2 & 3 & 0\end{array}\right]$

$\begin{aligned} & R_1 \rightarrow R_1+R_2 \\ & R_3 \rightarrow R_3-2 R_2\end{aligned}\left[\begin{array}{cccc|c}1 & 0 & 11 / 4 & -5 / 2 & 23 / 2 \\ 0 & 1 & 3 / 4 & 1 / 2 & 5 / 2 \\ 0 & 0 & 1 / 2 & 2 & -5\end{array}\right]$

$\begin{aligned} & R_3 \rightarrow 2 R_3\left[\begin{array}{cccc|c}1 & 0 & 11 / 4 & -5 / 2 & 23 / 2 \\ 0 & 1 & 3 / 4 & 1 / 2 & 5 / 2 \\ 0 & 0 & 1 & 4 & -10\end{array}\right] \\ & R_1 \rightarrow R_1-\frac{11}{4} R_3 ;R_2 \rightarrow R_2-\frac{3}{4} R_3\left[\begin{array}{cccc|c}1 & 0 & 0 & -\frac{5}{2}-11 & \frac{23}{2}+\frac{55}{2} \\ 0 & 1 & 0 & \frac{1}{2}-3 & \frac{5}{2}-\frac{15}{2} \\ 0 & 0 & 1 & 4 & -10\end{array}\right] \\ & \left[\begin{array}{cccc|c}1 & 0 & 0 & -27 / 2 & 39 \\ 0 & 1 & 0 & -5 / 2 & -5 \\ 0 & 0 & 1 & 4 & -10\end{array}\right]\end{aligned}$

$w$ - idependent variable. Let $w=\lambda$.

$\begin{aligned} & t-\frac{27}{2} w=39 \Rightarrow t=\frac{27}{2} \lambda+39 \\ & u-\frac{5}{2} w=-5 \Rightarrow u=\frac{5}{2} \lambda-5 \\ & v+4 w=-10 \Rightarrow v=-4 \lambda-10 \\ & \left(\frac{27}{2} \lambda+39, \frac{5}{2} \lambda-5,-4 \lambda-10, \lambda\right) . \end{aligned}$

Solve the system

$\begin{aligned} & x+2 y+3 z=1 \\ & 2 x+y+3 z=2 \\ & 5 x+5 y+9 z=4 . \end{aligned}$ $ \begin{aligned} & \text { Solution. }\left[\begin{array}{lll|c} 1 & 2 & 3 & 1 \\
2 & 1 & 3 & 2 \\
5 & 5 & 9 & 4 \end{array}\right] \\
& R_2 \rightarrow R_2-2 R_1 ; R_3 \rightarrow R_3-5 R_1\left[\begin{array}{ccc|c} 1 & 2 & 3 & 1 \\ 0 & -3 & -3 & 0 \\
0 & -5 & -6 & -1 \end{array}\right] \end{aligned} $

$\begin{aligned} & R_2 \rightarrow \frac{-1}{3} R_2\left[\begin{array}{ccc|c} 1 & 2 & 3 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & -5 & -6 & -1 \end{array}\right] \\ & \begin{array}{l} R_1 \rightarrow R_1-2 R_2 \\ R_3 \rightarrow R_3+5 R_2 \end{array}\left[\begin{array}{ccc|c} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & -1 & -1 \end{array}\right] \\ & R_3 \rightarrow-R_3\left[\begin{array}{lll|l} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{array}\right] \\ & \begin{array}{l} R_1 \rightarrow R_1-R_3 \\ R_2 \rightarrow R_2-R_3 \end{array}\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \end{array}\right] & \end{aligned}$

$x=0, y=-1, z=1 .$

$\begin{aligned} x+i y & =0 \\ -i x+z & =0 \\ y-z & =0 \end{aligned}$

Solution $\left[\begin{array}{ccc}1 & i & 0 \\ -i & 0 & 1 \\ 0 & 1 & -1\end{array}\right]$ $R_2 \rightarrow R_2+i R_1\left[\begin{array}{ccc} 1 & i & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \end{array}\right]$

$\begin{aligned} & R_2 \rightarrow-R_2\left[\begin{array}{ccc} 1 & i & 0 \\ 0 & 1 & -1 \\ 0 & 1 & -1 \end{array}\right] \\ & R_1 \rightarrow R_1-i R_2 ; R_3 \rightarrow R_3-R_2\left[\begin{array}{ccc} 1 & 0 & i \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right] \end{aligned}$

$z$ - independent variable.

$\begin{aligned} & z=\lambda . \\ & x+i z=0 \Rightarrow x=-i \lambda \\ & y-z=0 \Rightarrow y=\lambda . \\ & \quad(-i \lambda, \lambda, \lambda), \quad\lambda \in \mathbb{1R} . \end{aligned}$

Determine the values of $\lambda$ & $\mu$ for which the system of equation

$\begin{aligned} & x+2 y+3 z=6 \\ & x+3 y+5 z=9 \\ & 2 x+5 y+\lambda z=\mu \end{aligned}$ has

i) no solution

ii) unique solution &

iii) infinite number of solutions.

Solution $\left[\begin{array}{lll|l}1 & 2 & 3 & 6 \\ 1 & 3 & 5 & 9 \\ 2 & 5 & \lambda & \mu\end{array}\right]$

$(2 \lambda-16) z=2 \mu-30$ (or) $(\lambda-8) z=\mu-15$.

• $\lambda=8 \quad$ & $\mu \in \mathbb{R}$ \ $\{15\}$, the system has no soultion.
• $\lambda \neq 8\quad$ & $\mu \in \mathbb{R}$, the system has a unique solution.
• $\lambda=8\quad$ & $\mu=15$, the system has infinite number of solutions.

If the system

$\begin{aligned} x+a y & =0 \\ az+y & =0 \\ ax+z & =0 \end{aligned}$

has infinite number of solutions, then find the value of $a$.

Solution $\left[\begin{array}{lll}1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1\end{array}\right]$

If $a=0$, then the system has a unique solution. Let us assume that $a \neq 0$.

$R_3 \rightarrow R_3-a R_1\left[\begin{array}{ccc} 1 & a & 0 \\ 0 & 1 & a \\ 0 & -a^2 & 1 \end{array}\right]$

$\begin{aligned} & \begin{array}{l}R_1 \rightarrow R_1-a R_2 \\ R_3 \longrightarrow R_3+a^2 R_2\end{array}\left[\begin{array}{ccc}1 & 0 & -a^2 \\ 0 & 1 & a \\ 0 & 0 & 1+a^3\end{array}\right] \\ & 1+a^3=0 \Rightarrow a^3=-1 \text { i.e., } a=-1 \text {. } \\ & \end{aligned}$

If $a=-1$, the coefficient matrix has got rank 2 .

$\therefore$ The system has infinite number of solutions if $a=-1$.

A square matrix $A$ is said to be orthogonal if $A A^{T}=I$.

If $A=\left[\begin{array}{ccc}0 & 2 \beta & \gamma \\ \alpha & \beta & -\gamma\\ \alpha & -\beta & \gamma\end{array}\right]$ is an orthoganal matrix then find the values of $\alpha, \beta$ & $\gamma$.

Solution

$\begin{aligned} & \quad A A^{\top}=I \\ & {\left[\begin{array}{ccc} 0 & 2 \beta & \gamma\\ \alpha & \beta & -\gamma\\ \alpha & -\beta & \gamma \end{array}\right]\left[\begin{array}{ccc} 0 & \alpha & \alpha \\ 2 \beta & \beta & -\beta \\ \gamma& -\gamma& \gamma \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \end{aligned}$

$\left[\begin{array}{ccc} 4 \beta^2+\gamma^2 & 2 \beta^2-\gamma^2 & -2 \beta^2+\gamma^2 \\ 2 \beta^2-\gamma^2 & \alpha^2+\beta^2+\gamma^2 & \alpha^2-\beta^2-\gamma^2 \\ -2 \beta^2+\gamma^2 & \alpha^2-\beta^2-\gamma^2 & \alpha^2+\beta^2+\gamma^2 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$

$\begin{aligned} & 4 \beta^2+\gamma^2=1 \\ & 2 \beta^2-\gamma^2=0 \\ & \alpha^2+\beta^2+\gamma^2=1 \\ & \alpha^2-\beta^2-\gamma^2=0 \\ & {\left[\begin{array}{ccc|c} 0 & 4 & 1 & 1 \\ 0 & 2 & -1 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & -1 & -1 & 0 \end{array}\right]} \end{aligned}$

$\begin{aligned} & R_1 \longleftrightarrow R_3 , R_2 \longleftrightarrow R_4\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & -1 & -1 & 0 \\ 0 & 4 & 1 & 1 \\ 0 & 2 & -1 & 0 \end{array}\right] \\ & R_2 \rightarrow R_2-R_1\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & -2 & -2 & -1 \\ 0 & 4 & 1 & 1 \\ 0 & 2 & -1 & 0 \end{array}\right] \\ & R_2 \rightarrow \frac{1}{-2} R_2\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 / 2 \\ 0 & 4 & 1 & 1 \\ 0 & 2 & -1 & 0 \end{array}\right] \\ & \end{aligned}$

$\begin{aligned} & \begin{array}{l}R_1 \rightarrow R_1-R_2 \\ R_3 \rightarrow R_3-4 R_2 \\ R_4 \rightarrow R_4-2 R_2\end{array}\left[\begin{array}{ccc|c}1 & 0 & 0 & 1 / 2 \\ 0 & 1 & 1 & 1 / 2 \\ 0 & 0 & -3 & -1 \\ 0 & 0 & -3 & -1\end{array}\right] \\ & R_3 \rightarrow \frac{1}{-3} R_3\left[\begin{array}{ccc|c}1 & 0 & 0 & 1 / 2 \\ 0 & 1 & 1 & 1 / 2 \\ 0 & 0 & 1 & 1 / 3 \\ 0 & 0 & -3 & -1\end{array}\right] \\ & \begin{array}{l}R_2 \rightarrow R_2-R_3 \\ R_4 \rightarrow R_4+3 R_3\end{array}\left[\begin{array}{lll|l}1 & 0 & 0 & 1 / 2 \\ 0 & 1 & 0 & 1 / 6 \\ 0 & 0 & 1 & 1 / 3 \\ 0 & 0 & 0 & 0\end{array}\right] \\ & \end{aligned}$

$\begin{array}{ll} \alpha^2=1 / 2, & \beta^2=1 / 6, \quad \gamma^2=1 / 3 \\ \alpha= \pm 1 / \sqrt{2} & \beta= \pm 1 / \sqrt{6} \quad \gamma= \pm 1 / \sqrt{3} \end{array}$