Question
If A is a singular matrix, then adj A is.
- non−singular
- singular
- symmetric
- not defined
Solution:
Given ∣A∣ = 0
We know ∣adjA∣ = ∣A∣ n - 1
∴ ∣adjA∣ = 0
Hence, adj A is singular
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Statement $1$ : The function $f$ has a local extremum at $x = 0$
Statement $2$ : The function $f$ is continuous and differentiable on $\left( { - \infty ,\infty } \right)$ and $f'(0) = 0$