- 3
- -7
- 5
- -2
Solution:
Expanding along $\text{R}_1,$ we get
$\triangle=2(-1)-5(-1)=2+5$
$=3$
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If A is a singular matrix, then A (adj A) is a
($A$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the green region below the line $\mathrm{L}_{\mathrm{h}}$
($B$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the red region below the line $\mathrm{L}_h$
($C$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the red region below the line $L_h$
($D$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the green region below the line $\mathrm{L}_{\mathrm{h}}$