Different way to determine rank of 3×3 -matrix with parameter by using row transformations A=(2−a1−aa1−a1−a02−2a1−a−(1−a)).

Step 1 You can simplify algorithm, borrowing techniques used for the computation of determinants: [2−a1−aa1−a1−a02−2a1−a−(1−a)]⇝[1−a1−a02−2a1−a−(1−a)2−a1−aa] Now, if a≠1 this matrix has the same rank as [1100−1−12−a1−aa]⇝[1100110−1a]⇝[11001100a+1]. You can now conclude (not forgetting to examine the case a=1).

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College
Algebra
Linear algebra
Matrix transformations

Lainey Goodwin

Answered question

2022-01-30

Different way to determine rank of

3 \times 3

-matrix with parameter by using row transformations

A = (\begin{matrix} 2 - a & 1 - a & a \\ 1 - a & 1 - a & 0 \\ 2 - 2 a & 1 - a & - (1 - a) \end{matrix}) .

Answer & Explanation

Dominique Green

Beginner2022-01-31Added 11 answers

Step 1 You can simplify algorithm, borrowing techniques used for the computation of determinants:

\begin{aligned} [\begin{array}{c} 2 - a & 1 - a & a \\ 1 - a & 1 - a & 0 \\ 2 - 2 a & 1 - a & - (1 - a) \end{array}] & ⇝ [\begin{array}{c} 1 - a & 1 - a & 0 \\ 2 - 2 a & 1 - a & - (1 - a) \\ 2 - a & 1 - a & a \end{array}] \end{aligned}

Now, if

a \neq 1

this matrix has the same rank as

\begin{array}{r} [\begin{array}{c} 1 & 1 & 0 \\ 0 & - 1 & - 1 \\ 2 - a & 1 - a & a \end{array}] ⇝ [\begin{array}{c} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & - 1 & a \end{array}] ⇝ [\begin{array}{c} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & a + 1 \end{array}] . \end{array}

You can now conclude (not forgetting to examine the case

a = 1

waijazar1

Beginner2022-02-01Added 13 answers

Gauss is very comfortable here; just switch column two and one. Consider then the case

a = 1

; here the rank is 1. Otherwise you will be done in two very easy steps without considering further cases.

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