Let \(A=\begin{bmatrix}a_1\\a_2\end{bmatrix}\) and \(B=\begin{bmatrix}b_1\\b_2\end{bmatrix}\) be two \(2\times1\) matrices with real entries such that \(A=XB\), where
\(X=\frac{1}{\sqrt3}\begin{bmatrix}1&-1\\1&k\end{bmatrix}\), and \(k\in R\).
If \(a_1^2+a_2^2=\frac23(b_1^2+b_2^2)\) and \((k^2+1)b_2^2\ne-2b_1b_2\), then the value of k is __________.
\(X=\frac{1}{\sqrt3}\begin{bmatrix}1&-1\\1&k\end{bmatrix}\), and \(k\in R\).
If \(a_1^2+a_2^2=\frac23(b_1^2+b_2^2)\) and \((k^2+1)b_2^2\ne-2b_1b_2\), then the value of k is __________.
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Correct Answer: 1
Solution and Explanation
\(a_1^2+a_2^2=\frac13\left[(b_1-b_2)^2+(b_1+kb_2)^2\right]=\frac13\left[2b_1^2+2(k-1)b_1b_2+(1+k^2)b_2^2\right]\).
Set equal to \(\frac23(b_1^2+b_2^2)\): \(2(k-1)b_1b_2+(k^2+1)b_2^2-2b_2^2=0\), i.e. \(2(k-1)b_1b_2+(k^2-1)b_2^2=0\).
Factor: \((k-1)\left[2b_1b_2+(k+1)b_2^2\right]=0\).
The second factor is the case excluded by the given condition, so \(k=1\).
Answer: 1
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