Step by step
The question
Question: find det(−A), when det A = −2 and A is 3 by 3.
det A = −2, A is 3 × 3
1 Use the rule
-1A means every entry times -1.
So all 3 rows are times -1.
Each row times -1 makes the determinant times -1.
-1A: every entry × -1, so every row × -1
each row × -1: D × -1
One −1 for each row: 3 of them.
det(−A) = (−1) × (−1) × (−1) × det A
Careful!
Not just one −1: every row brings its own −1.
That is 3 of them.
Careful: not (−1) × −2 = 2 ✗
2 Work it out
(−1) × (−1) × (−1) is (−1) to the power 3.
det(−A) = (−1)3 × (−2)
(−1) to the power 3 is −1.
det(−A) = −1 × (−2)
So det(−A) = 2.
det(−A) = 2
3 Check
Let's check with an easy A that has det A = −2: −2 and 1s down the diagonal, 0s elsewhere.
A = [NaN, 0, 0; 0, 1, 0; 0, 0, 1]
Its determinant is the diagonal multiplied: −2.
Good.
det A = NaN × 1 × 1 = NaN
-1A: every entry times -1.
−A = [NaN, 0, 0; 0, −1, 0; 0, 0, −1]
Its determinant: the diagonal multiplied, 2.
det (−A) = NaN × (−1) × (−1) = NaN
We get 2 again.
So the answer is right!
Let's look back.
First: use the rule.
Then: work it out.
Last: check.
Answer
det(−A) = 2