 Alyce Wilkinson

2021-08-11

Let M,N and K be matrices of type $3×2,2×3$ and $3×3$ respectively, such that det (MN) $=2$ and $det\left(K\right)=6$. Then which of the following is the value of the det ${\left[{\left(-3MN\right)}^{T}{K}^{-1}\right]}^{-1}$?
a) $-{3}^{2}.2$ b)$\frac{-1}{{3}^{3}.2}$ c) $\frac{2}{{3}^{2}}$ d)$\frac{-{3}^{2}}{4}$ e) $\frac{-1}{{3}^{2}}$ 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 $|k\cdot A|\mathrm{%}\left\{n\right\}={k}^{n}\cdot {|A|}^{n}$.
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 $|A\cdot B|=|A|\cdot |B|$.
Put the values of the determinants in equation (1) and solve.
${det\left[{\left(-3MN\right)}^{T}{K}^{-1}\right]}^{-1}={\left(-3\right)}^{-1}{\left(det\left(MN\right)\right)}^{T}-1{det\left(K\right)}^{-1}-1=-13\left(det\left(MN\right)T\right)-1det\left(K\right)\dots \dots \dots \dots \dots \dots \dots ..\left(1\right)=-13\left(2\right)-1\left(6\right)=-13126=-1$

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