Linear Algebra help plox? (1 Viewer)

hayabusaboston

Well-Known Member
I just registered for a maths diploma and they accepted me only now after two full weeks so im trying to catch up on content

How u do dis?

2. Determine the values (if any) of t in R for which the following linear system has:
(a) no solutions,
(b) infinitely many solutions,
(c) a unique solution.
In the case (b) of infinitely many solutions, find all solutions.

tx-5y+3z=-8t
-tx+8y+3tz=11t
-x+2y+3z=1

He-Mann

Vexed?
Gaussian elimination.

hayabusaboston

Well-Known Member
Gaussian elimination.
can u help ploxx?

Im confused how to get to RRE with t's and numbers, + the rules for one, infinite and no solutions. would be helpful if someone did the question step by step so I understand for subsequent questions.

He-Mann

Vexed?
Do row operations as usual, group up the rows containing t to make it easier to analyse.:

\bg_white \begin{align*} (A|b) = \left(\begin{array}{ccc|c} t & -5 & 3 & -8t \\ -t & 8 & 3t & 11t \\ -1 & 2 & 3 & 1\end{array}\right) \stackrel{R1 \leftrightarrow R3}\longrightarrow & \left(\begin{array}{ccc|c} -1 & 2 & 3 & 1 \\ -t & 8 & 3t & 11t \\ t & -5 & 3 & -8t \end{array}\right) \\ \stackrel{R3 \to R2+R3}\longrightarrow & \left(\begin{array}{ccc|c} -1 & 2 & 3 & 1 \\ -t & 8 & 3t & 11t \\ 0 & 3 & 3 +3t & 3t \end{array}\right) \\ \stackrel{R2 \to R2 - tR1}\longrightarrow & \left(\begin{array}{ccc|c} -1 & 2 & 3 & 1 \\ 0 & 8-2t & 0 & 8t \\ 0 & 3 & 3 +3t & 3t \end{array}\right) \\ \stackrel{R2 \leftrightarrow R3}\longrightarrow & \left(\begin{array}{ccc|c} -1 & 2 & 3 & 1 \\ 0 & 3 & 3 +3t & 3t \\ 0 & 8-2t & 0 & 8t \end{array}\right) \\ \stackrel{R3 \to R3 - (8-2t)}\longrightarrow & \left(\begin{array}{ccc|c} -1 & 2 & 3 & 1 \\ 0 & 3 & 3 +3t & 3t \\ 0 & 0 & 2t - 8 & 10t - 8 \end{array}\right)\end{align*}

There's probably a faster sequence of row operations to reach RREF, but that's one way.

Start with (a), what values of t produces a contradiction? That's when you cannot solve the system.

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He-Mann

Vexed?
Suppose you had a row like this: (0 0 0 | 2317). Would it make sense?

hayabusaboston

Well-Known Member
Do row operations as usual, group up the rows containing t to make it easier to analyse.:

\bg_white \begin{align*} (A|b) = \left(\begin{array}{ccc|c} t & -5 & 3 & -8t \\ -t & 8 & 3t & 11t \\ -1 & 2 & 3 & 1\end{array}\right) \stackrel{R1 \leftrightarrow R3}\longrightarrow & \left(\begin{array}{ccc|c} -1 & 2 & 3 & 1 \\ -t & 8 & 3t & 11t \\ t & -5 & 3 & -8t \end{array}\right) \\ \stackrel{R3 \to R2+R3}\longrightarrow & \left(\begin{array}{ccc|c} -1 & 2 & 3 & 1 \\ -t & 8 & 3t & 11t \\ 0 & 3 & 3 +3t & 3t \end{array}\right) \\ \stackrel{R2 \to R2 - tR1}\longrightarrow & \left(\begin{array}{ccc|c} -1 & 2 & 3 & 1 \\ 0 & 8-2t & 0 & 8t \\ 0 & 3 & 3 +3t & 3t \end{array}\right) \\ \stackrel{R2 \leftrightarrow R3}\longrightarrow & \left(\begin{array}{ccc|c} -1 & 2 & 3 & 1 \\ 0 & 3 & 3 +3t & 3t \\ 0 & 8-2t & 0 & 8t \end{array}\right) \\ \stackrel{R3 \to R3 - (8-2t)}\longrightarrow & \left(\begin{array}{ccc|c} -1 & 2 & 3 & 1 \\ 0 & 3 & 3 +3t & 3t \\ 0 & 0 & 2t - 8 & 10t - 8 \end{array}\right)\end{align*}

There's probably a faster sequence of row operations to reach RREF, but that's one way.

Start with (a), what values of t produces a contradiction? That's when you cannot solve the system.
t=4 is that only solution that satisfies a)?

And t not equal to 4 for c)?

hayabusaboston

Well-Known Member
Suppose you had a row like this: (0 0 0 | 2317). Would it make sense?
Yea I know the no solution part i just was wondering about how to distinguish 1 vs infinite solutions. Something about not having an identity matrix isnt it? then infinite solutions? idk

He-Mann

Vexed?
Yea I know the no solution part i just was wondering about how to distinguish 1 vs infinite solutions. Something about not having an identity matrix isnt it? then infinite solutions? idk
If you want to get infinite solutions, then you need to prevent at least one row from giving any information. Intuition: for RREF, each row should give information about one of x, y, or z. But when you stop one row from finding either x, y, or z, then you won't be able to solve other rows. e.g., solving for y and z in y+z = 2 when you don't have any information on y or z. Thus, you get infinite solutions for y and z.

i.e., produce a zero row (0 0 0 | 0).

___

Your solutions are correct for (a) and (c).

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He-Mann

Vexed?
It seems that there is no t such that the system yields infinite solutions.

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