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Mathematics Question on Matrices

If [31 41]X=[5−1 23]\begin{bmatrix}3&1\\\ 4&1\end{bmatrix} X = \begin{bmatrix}5&-1\\\ 2&3\end{bmatrix} then X is equal to:

A

[−34 14−13]\begin{bmatrix}-3&4\\\ 14&-13\end{bmatrix}

B

[3−4 −1413] \begin{bmatrix}3&-4\\\ -14&13\end{bmatrix}

C

[34 1413]\begin{bmatrix}3&4\\\ 14&13\end{bmatrix}

D

[−34 −1413]\begin{bmatrix}-3&4\\\ -14&13\end{bmatrix}

Answer

[−34 14−13]\begin{bmatrix}-3&4\\\ 14&-13\end{bmatrix}

Explanation

Solution

Given : [31 41]X=[5−1 23]\begin{bmatrix}3&1\\\ 4&1\end{bmatrix} X = \begin{bmatrix}5&-1\\\ 2&3\end{bmatrix} Thus X=[31 41]−1[5−1 23]X = \begin{bmatrix}3&1\\\ 4&1\end{bmatrix}^{-1} \begin{bmatrix}5&-1\\\ 2&3\end{bmatrix} Let A=[31 41]A = \begin{bmatrix}3&1\\\ 4&1\end{bmatrix} a11a_{11} = co-factor of a11=1a_{11} = 1 a12a_{12} = co-factor of a12=(−1)1+2.4=−4a_{12} = (-1)^{1+2}. 4 = - 4 a21a_{21} = co-factor of a21=(−1)2+1.1=−1a_{21} = (-1)^{2+1} . 1 = - 1 a22a_{22} = co-factor of a22=3a_{22} = 3 ∣A∣=3−4=−1| A | = 3 - 4 = -1 So A−1=[1−4 −13]−1=[−11 4−3] A^{-1} = \frac{\begin{bmatrix}1&-4\\\ -1&3\end{bmatrix}}{-1} = \begin{bmatrix}-1&1\\\ 4&-3\end{bmatrix} Thus X=[−11 4−13][5−1 23]X = \begin{bmatrix}-1&1\\\ 4&-13\end{bmatrix}\begin{bmatrix}5&-1\\\ 2&3\end{bmatrix} X=[−34 14−13]X = \begin{bmatrix}-3&4\\\ 14&-13\end{bmatrix}