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For1i,j3,Letaij=π/2π/2cos(ix)cos(jx)dxand let A=[aij]3×3then

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a
Ais singular matrix
b
AX=Bhas a unique solution for every 3×3matrix B
c
A is a skew-symmetric matrix
d
A2=I

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detailed solution

Correct option is B

For 1≤i≤3,aii=∫-π/2π/2cos2(ix)dx=2∫0π/2cos2(ix)dx=∫0π/2[1+cos(2ix)]dx=x+sin(2ix)2i0π/2=π2+0=π2 For 1≤i,j≤3,i≠jaij=2∫0π/2cos(ix)cos(jx)dx=∫0π/2{cos[(i+j)x]+cos[(i-j)x]}dx=sin[(i+j)x]i+j+sin[(i-j)x]i-j0π/2=sin[(i+j)π/2]i+j+sin[(i-j)π/2]i-j∴a12=-13+1=23=a21a13=0=a31a23=15+1=65=a32 Thus, A=π/22/302/3π/26/506/5π/2|A|=π28-43π72≠0,A-1 exists. ∴ AX=Bhas a unique solution, for every matrix B


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