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Question
If ` f(x)=|[a,-1,0],[ax,a,-1],[ax^2,ax,a]| ` , using properties of determinants find the value of f(2x) − f(x).
Solution
`f(x)=|[a,-1,0],[ax,a,-1],[ax^2,ax,a]|`
`=>f(x)=|[a,-1,0],[ax,a,-1],[ax^2,ax,a]|`
Applying C2→C2+C1, we get
`f(x)=a|[1,0,0],[x,x+a,-1],[x^2,x^2+ax,a]|`
`=>f(x)=a(a^2+ax+ax+x^2)`
`=>f(x)=a(a^2+2ax+x^2)`
Also,
`f(2x)=|[a,-1,0],[2ax,a,-1],[4ax^2,2ax,a]|`
`f(2x)=a|[1,-1,0],[2x,a,-1],[4x^2,2ax,a]|`
Applying C2→C2+C1, we get
`f(2x)=a|[1,0,0],[2x,2x+a,-1],[4x^2,4x^2+2ax,a]|`
`⇒f(2x)=a{a(2x+a)+4x^2+2ax}`
`⇒f(2x)=a(4x^2+a^2+4ax)`
`∴ f(2x)−f(x)=a(4x^2+a^2+4ax−a^2−2ax−x^2) `
`=ax(3x+2a)`
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