CUET UG 2024 — Mathematics Algebra

If $A$A is a square matrix and $I$I is an identity matrix such that $A^{2}=A$A2=A, then $A(I-2 A)^{3}+2 A^{3}$A(I−2A)3+2A3 is equal to:

We are given that $A^2 = A$A2=A, which means $A$A is an idempotent matrix.

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First, let's find $A^3$A3 using the given property:

$$A^3 = A^2 \cdot A = A \cdot A = A$$A3=A2⋅A=A⋅A=A

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We'll first find $(I-2A)^2$(I−2A)2:

$$(I-2A)^2 = (I-2A)(I-2A) = I^2 - 2A \cdot I - 2I \cdot A + 4A^2$$(I−2A)2=(I−2A)(I−2A)=I2−2A⋅I−2I⋅A+4A2

$$= I - 2A - 2A + 4A^2$$=I−2A−2A+4A2

$$= I - 4A + 4A^2$$=I−4A+4A2

Since $A^2 = A$A2=A, we have:

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

Now we can calculate $(I-2A)^3$(I−2A)3:

$$(I-2A)^3 = (I-2A)^2(I-2A) = I(I-2A) = I - 2A$$(I−2A)3=(I−2A)2(I−2A)=I(I−2A)=I−2A

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Next, calculate $A(I-2A)^3$A(I−2A)3:

$$A(I-2A)^3 = A(I-2A) = A - 2A^2$$A(I−2A)3=A(I−2A)=A−2A2

Since $A^2 = A$A2=A, we have:

$$A(I-2A)^3 = A - 2A = -A$$A(I−2A)3=A−2A=−A

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Finally, calculate the complete expression $A(I-2A)^3+2A^3$A(I−2A)3+2A3:

$$A(I-2A)^3+2A^3 = -A + 2A^3 = -A + 2A = A$$A(I−2A)3+2A3=−A+2A3=−A+2A=A

Therefore, $A(I-2A)^3+2A^3 = A$A(I−2A)3+2A3=A