JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right], B=\left[B_1, B_2, B_3\right]\), where \(B_1\), \(\mathrm{B}_2, \mathrm{~B}_3\) are column matrices, and \(\mathrm{AB}_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\), \(\mathrm{AB}_2=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right], \mathrm{AB}_3=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]\) If \(\alpha=|B|\) and \(\beta\) is the sum of all the diagonal elements of \(B\), then \(\alpha^3+\beta^3\) is equal to
- A \(28\)
- B \(24\)
- C \(23\)
- D \(45\)
Answer & Solution
Correct Answer
(A) \(28\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right] \quad B=\left[B_1, B_2, B_3\right] \\ & B_1=\left[\begin{array}{l}x_1 \\ y_1 \\ z_1\end{array}\right], \quad B_2=\left[\begin{array}{l}x_2 \\ y_2 \\ z_2\end{array}\right], \quad…
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