If an asset costs Rs. 50,000 with an estimated useful life of 6 years and a scrap value of Rs. 5000. Then by using a linear depreciation method, the…

Given $A^2 = A$A2=A and $I$I is the identity matrix of the same order as $A$A.

Need to find $(A - 2I)^2 - (2A + I)^2 + 11A$(A−2I)2−(2A+I)2+11A

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Expanding $(A - 2I)^2$(A−2I)2:

$$(A - 2I)^2 = A^2 - 2A \cdot I - 2I \cdot A + (2I)^2$$(A−2I)2=A2−2A⋅I−2I⋅A+(2I)2

$$= A^2 - 2A - 2A + 4I$$=A2−2A−2A+4I

$$= A^2 - 4A + 4I$$=A2−4A+4I

Using $A^2 = A$A2=A:

$$(A - 2I)^2 = A - 4A + 4I$$(A−2I)2=A−4A+4I

$$= -3A + 4I$$=−3A+4I

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Expanding $(2A + I)^2$(2A+I)2:

$$(2A + I)^2 = (2A)^2 + 2(2A) \cdot I + I^2$$(2A+I)2=(2A)2+2(2A)⋅I+I2

$$= 4A^2 + 4A + I$$=4A2+4A+I

Using $A^2 = A$A2=A:

$$(2A + I)^2 = 4A + 4A + I$$(2A+I)2=4A+4A+I

$$= 8A + I$$=8A+I

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Substituting into the expression:

$$(A - 2I)^2 - (2A + I)^2 + 11A = (-3A + 4I) - (8A + I) + 11A$$(A−2I)2−(2A+I)2+11A=(−3A+4I)−(8A+I)+11A

$$= -3A + 4I - 8A - I + 11A$$=−3A+4I−8A−I+11A

$$= (-3 - 8 + 11)A + (4 - 1)I$$=(−3−8+11)A+(4−1)I

$$= 0A + 3I$$=0A+3I

$$= 3I$$=3I

Therefore, the value is $3I$3I.