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IPMAT Indore 2022 - QA (SA) Q10 Explanation
December 26, 2024
Explanation:
(32)
Given A = [ 1 0 0 0 0 1 0 1 0 ]
A^2= [ 1 0 0 0 0 1 0 1 0 ][ 1 0 0 0 0 1 0 1 0 ] = [ 1 0 0 0 1 0 0 0 1 ] = I
∴ We can say that A^6= A^2. A^2. A^2 = I.I.I. = I
A^3 = A^2 .A = I.A = A
A^9 = A^3 .A^3 .A^3 = A.A.A = A^2.A = I.A = A
A^9 + A^6 + A^3 + A = A.A.A. = A^2.A = A
Now,
A^9 + A^6 + A^3 + A = A + I + A + A
= 3A + I
= 3 [ 1 0 0 0 0 1 0 1 0 ] + [ 1 0 0 0 1 0 0 1 0 ]
= [ 3 0 0 0 0 3 0 3 0 ] + [ 1 0 0 0 1 0 0 0 1 ]
A^9 + A^6 + A^3 + A = [ 4 0 0 0 1 3 0 3 1 ]
Determinant of above matrix = 4(1×1-3×3)
= 4 (1-9) = 4(-8)
= -32
Absolute value of -32 = |-32| = 32.