Question

Q= кө (1+lambda/2)

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Answer 1

Answer:

We have [λ2−2λ+11−λ2+3λλ−21−λ2]=Aλ2+Bλ+C

Putting λ=0, we get

C=[11−21]

Putting λ=1, we get

A+B+C=[03−10] ...(1)

Putting λ=−1, we get

A−B+C=[4−3−30]

Subtracting (2) from (1), we get

2B[03−10]−[4−3−30]=[−4620]

∴B=[−2310]

Explanation:

Answer 2

Answer:

Answer:

We have [λ2−2λ+11−λ2+3λλ−21−λ2]=Aλ2+Bλ+C

Putting λ=0, we get

C=[11−21]

Putting λ=1, we get

A+B+C=[03−10] ...(1)

Putting λ=−1, we get

A−B+C=[4−3−30]

Subtracting (2) from (1), we get

2B[03−10]−[4−3−30]=[−4620]

∴B=[−2310]

Explanation: