Determinants Question 3
Question 3 - 2024 (27 Jan Shift 1)
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 $B _1, B _2, B _3$ are column matrices, and $A _1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$,
$AB _2=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right], AB _3=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$
If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^{3}+\beta^{3}$ is equal to
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Answer (28)
Solution
$A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right] \quad B=\left[B _1, B _2, B _3\right]$
$B _1=\left[\begin{array}{l}x _1 \\ y _1 \\ z _1\end{array}\right], \quad B _2=\left[\begin{array}{l}x _2 \\ y _2 \\ z _2\end{array}\right], \quad B _3=\left[\begin{array}{l}x _3 \\ y _3 \\ z _3\end{array}\right]$
$AB _1=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x _1 \\ y _1 \\ z _1\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$
$x _1=1, y _1=-1, z _1=-1$
$AB _2=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x _2 \\ y _2 \\ z _2\end{array}\right]=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right]$
$x _2=2, y _2=1, z _2=-2$
$AB _3=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x _3 \\ y _3 \\ z _3\end{array}\right]=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$
$x _3=2, y _3=0, z _3=-1$
$B=\left[\begin{array}{ccc}1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1\end{array}\right]$
$\alpha=|B|=3$
$\beta=1$
$\alpha^{3}+\beta^{3}=27+1=28$
