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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The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}-\text{y}}+\text{x}^2\text{e}^{-\text{y}}$ is:
$\text{y}=\text{e}^{\text{x}-\text{y}}-\text{x}^2\text{e}^{-\text{y}}+\text{c}$
$\text{e}^{\text{y}}-\text{e}^{\text{x}}=\frac{\text{x}^3}{3}+\text{c}$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}=\frac{\text{x}^3}{3}+\text{c}$
$\text{e}^{\text{x}}-\text{e}^{\text{y}}=\frac{\text{x}^3}{3}+\text{c}$