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  • Let <img alt="P= \begin{bmatrix} 3 &-1 &-2 \\ 2&0 &\alpha \\ 3&-5 &0 \end{bmatrix}" src="https://entrancecorner.codecogs.com/gif.latex?P%3D%20%5Cbegin%7Bbmatrix%7D%203%20%26-1%20%26-2%20%5C%5C%202%260%20%26%5Calpha%20%5C%5C%203%

Let P= \begin{bmatrix} 3 &-1 &-2 \\ 2&0 &\alpha \\ 3&-5 &0 \end{bmatrix}, where \alpha \in R. Suppose Q=\left [ q_{ij} \right ] is a matrix sattisfying PQ=kI_3 for some non-zero k\in R. If q_{23}=-\frac{k}{8} and \left | Q \right |=\frac{k^2}{2} then \alpha^{2}+k^2 is equal to_________.
Option: 1 17
Option: 2 16
Option: 3 15
Option: 4 14

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best_answer

 

\text { As } \mathrm{PQ}=\mathrm{kI} \quad \Rightarrow \quad \mathrm{Q}=\mathrm{kP}^{-1} \mathrm{I}

\text { now } \mathrm{Q}=\frac{\mathrm{k}}{|\mathrm{P}|}(\operatorname{adjP}) \mathrm{I}\\\\ \Rightarrow \quad \mathrm{Q}=\frac{\mathrm{k}}{(20+12 \alpha)}\left[\begin{array}{llc} - & - & - \\ - & - & (-3 \alpha-4) \\ - & - & - \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]

\because \mathrm{q}_{23}=\frac{-\mathrm{k}}{8} \Rightarrow \frac{\mathrm{k}}{(20+12 \alpha)}(-3 \alpha-4)=\frac{-\mathrm{k}}{8} \Rightarrow 2(3 \alpha+4)=5+3 \alpha

3 \alpha=-3 \Rightarrow \alpha=-1

\text { also }|\mathrm{Q}|=\frac{\mathrm{k}^{3}|\mathrm{I}|}{|\mathrm{P}|} \Rightarrow \frac{\mathrm{k}^{2}}{2}=\frac{\mathrm{k}^{3}}{(20+12 \alpha)}

(20+12 \alpha)=2 \mathrm{k} \quad \Rightarrow \quad 8=2 \mathrm{k} \quad \Rightarrow \quad \mathrm{k}=4

Posted by

Suraj Bhandari

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