InterviewSolution
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InterviewSolution
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Find the value of λ, a non-zero scalar, if \(\lambda\begin{bmatrix}1 & 0 & 2 \\[0.3em]1 & 4 & 5 \\[0.3em]\end{bmatrix}+\)\(2\begin{bmatrix}1 & 2 & 3 \\[0.3em]-1 & -3 & 2 \\[0.3em]\end{bmatrix}=\)\(\begin{bmatrix}4 & 4 & 10 \\[0.3em]4 & 2 & 14 \\[0.3em]\end{bmatrix}.\) |
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Answer» = \(\lambda\begin{bmatrix}1 & 0 & 2 \\[0.3em]1 & 4 & 5 \\[0.3em]\end{bmatrix}+\)\(\begin{bmatrix}2 & 4 &6 \\[0.3em]-2 & -6 & 4 \\[0.3em]\end{bmatrix}=\)\(\begin{bmatrix}4 & 4 & 10 \\[0.3em]4 & 2 & 14 \\[0.3em]\end{bmatrix}\) We know that, corresponding entries of equal matrices are equal. = \(\begin{bmatrix}\lambda+2 & 0+4 & 2\lambda+6 \\[0.3em]3\lambda-2 & 4\lambda-6 & 5\lambda+4 \\[0.3em]\end{bmatrix}=\)\(\begin{bmatrix}4 & 4 & 10 \\[0.3em]4 & 2 & 14 \\[0.3em]\end{bmatrix}\) = λ + 2 = 4 = λ = 4 - 2 = λ = 2 Hence, λ = 2 = λ[110425]+λ[102145]+[2−24−664]=[246−2−64]=[44421014][44104214]We know that, corresponding entries of equal matrices are equal. = [λ+23λ−20+44λ−62λ+65λ+4] =[λ+20+42λ+63λ−24λ−65λ+4] =[44421014][44104214] = λ + 2 = 4 = λ = 4 - 2 = λ = 2 Hence, λ = 2 |
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