Question

a. Can [ i ] be a column of a unitary matrix?
[-i ]
Justify your answer
b. Check whether or not the following linear system is consistent:
3+t = x+y+z, 2+2t= x+2y-z, 4-t= x-y +4z

Answer 1

Part a

Yes, [i] can be a column at a unitary matrix as if we choose $1 \times 1$ matrix. I.e i=j, meaning that row = column.

So in this case we can say that [i] as a column of a unitary matrix.

Consqueantly

Yes, [-i] can be a column at a unitary matrix as if we choose $1 \times 1$ matrix. I.e -i=-j, meaning that row = column.

So in this case we can say that [-i] as a column of a unitary matrix.

Part b

$x+y+z =3+t \\ x+2y-z=2+2t\\ x-y +4z=4-t\\$

It can be written as

$x+y+z -t=3 \\ x+2y-z-2t=2\\ x-y +4z+t=4\\$

Now we can convert it into matrix [A:b] form and do row reduction form

$=\begin{bmatrix} 1 & 1 & 1 & -1| 3 \\ 1 & 2 & -1 & -2| 2\\ 1 & -1 & 4 & 1| 4 \end{bmatrix}\\$

$=\begin{bmatrix} 1 & 1 & 1 & -1| 3 \\ 0 & 1 & -2 & -1| -1\\ 0 & -2 & 3 & 2| 1 \end{bmatrix}$

$=\begin{bmatrix} 1 & 0 & 3 & 0| 4 \\ 0 & 1 & -2 & -1| -1\\ 0 & 0 & -1 & 0| -1 \end{bmatrix}$

$=\begin{bmatrix} 1 & 0 & 0 & 0| 1 \\ 0 & 1 & 0 & -1| 1\\ 0 & 0 & 1 & 0| 1 \end{bmatrix}$

x= 1, y=1,z= 1

So the solution is consistent, the value of t is not fixed as t is a free variable. So it has infinitely many solutions.