# Let M,N and K be matrices of type 3 \times

Let M,N and K be matrices of type $$\displaystyle{3}\times{2},{2}\times{3}$$ and $$\displaystyle{3}\times{3}$$ respectively, such that det (MN) $$\displaystyle={2}$$ and $$\displaystyle{\det{{\left({K}\right)}}}={6}$$. Then which of the following is the value of the det $$\displaystyle{\left[{\left(-{3}{M}{N}\right)}^{{{T}}}{K}^{{-{1}}}\right]}^{{-{1}}}$$?
a) $$\displaystyle-{3}^{{{2}}}{.2}$$ b)$$\displaystyle{\frac{{-{1}}}{{{3}^{{{3}}}{.2}}}}$$ c) $$\displaystyle{\frac{{{2}}}{{{3}^{{{2}}}}}}$$ d)$$\displaystyle{\frac{{-{3}^{{{2}}}}}{{{4}}}}$$ e) $$\displaystyle{\frac{{-{1}}}{{{3}^{{{2}}}}}}$$

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Aamina Herring
Step 1
Determinant properties help to find the determinant of the given matrix.
Determinant property for the multiples helps to do the required, which is defined as $$\displaystyle{\left|{k}\cdot{A}\right|}\%{\left\lbrace{n}\right\rbrace}={k}^{{{n}}}\cdot{\left|{A}\right|}^{{{n}}}$$.
Here k is the real number whereas A is the square matrix.
The determinant of the matrix is equal to the determinant of its transpose.
Step 2
Apply the determinant properties for the multiples to find the required answer.
Apply the determinant properties for the product of matrices, which is defined as $$\displaystyle{\left|{A}\cdot{B}\right|}={\left|{A}\right|}\cdot{\left|{B}\right|}$$.
Put the values of the determinants in equation (1) and solve.
$$\displaystyle{{\det{{\left[{\left(-{3}{M}{N}\right)}^{{{T}}}{K}^{{-{1}}}\right]}}}^{{-{1}}}=}{\left(-{3}\right)}^{{-{1}}}{\left({\det{{\left({M}{N}\right)}}}\right)}^{{{T}}}-{1}{{\det{{\left({K}\right)}}}^{{-{1}}}-}{1}=-{13}{\left({\det{{\left({M}{N}\right)}}}{T}\right)}-{1}{\det{{\left({K}\right)}}}\ldots\ldots\ldots\ldots\ldots\ldots\ldots..{\left({1}\right)}=-{13}{\left({2}\right)}-{1}{\left({6}\right)}=-{13126}=-{1}$$