MHT CET 2021 20th September Evening Shift | MHT CET Year Wise Previous Years Questions

If $$\bar{r}=-4 \hat{i}-6 \hat{j}-2 \hat{k}$$ is a linear combination of the vectors $$\bar{a}=-\hat{i}+4 \hat{j}+3 \hat{k}$$ and $$\bar{b}=-8 \hat{i}-\hat{j}+3 \hat{k}$$, then

$$\overline{\mathrm{r}}=\frac{-4}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}$$

$$\overline{\mathrm{r}}=\frac{4}{3} \overline{\mathrm{a}}+\frac{2}{3} \bar{b}$$

$$\overline{\mathrm{r}}=\frac{-1}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}$$

$$\overline{\mathrm{r}}=\frac{1}{3} \overline{\mathrm{a}}-\frac{1}{3} \overline{\mathrm{b}}$$

MHT CET 2021 20th September Evening Shift

MCQ (Single Correct Answer)

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If $$A^{-1}=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$$ and $$B^{-1}=\left[\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right]$$, then $$(A B)^{-1}=$$

$$\left[\begin{array}{cc}2 & 7 \\ 3 & -1\end{array}\right]$$

$$\left[\begin{array}{cc}2 & -7 \\ -3 & 11\end{array}\right]$$

$$\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]$$

$$\left[\begin{array}{cc}2 & 3 \\ 7 & -11\end{array}\right]$$

MHT CET 2021 20th September Evening Shift

MCQ (Single Correct Answer)

-0

Two unbiased dice are thrown. Then the probability that neither a doublet nor a total of 10 will appear is

$$\frac{1}{12}$$

$$\frac{1}{36}$$

$$\frac{2}{9}$$

$$\frac{7}{9}$$