The number of all possible matrices of order 3 with each entry either 0 or 1 is:

The triangle has vertices at $(-3, 0)$(−3,0), $(3, 0)$(3,0), and $(0, k)$(0,k).

Points $(-3, 0)$(−3,0) and $(3, 0)$(3,0) both lie on the x-axis.

Point $(0, k)$(0,k) lies on the y-axis.

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The base of the triangle is the distance from $(-3, 0)$(−3,0) to $(3, 0)$(3,0):

Base $= 3 - (-3)$=3−(−3)

Base $= 6$=6 units

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The height is the perpendicular distance from $(0, k)$(0,k) to the x-axis.

Height $= |k|$=∣k∣ units

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Using the area formula:

$$\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}$$Area=21​×base×height

$$9 = \dfrac{1}{2} \times 6 \times |k|$$9=21​×6×∣k∣

$$9 = 3|k|$$9=3∣k∣

$$|k| = 3$$∣k∣=3

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Therefore, $k = 3$k=3 or $k = -3$k=−3

Both values give the same area since the formula uses absolute value.

The answer is $k = \pm 3$k=±3.