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NDA Math Matrices

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Question : 57 of 69

Marks: +1, -0

If

A=\begin{bmatrix}2 & 7 \\ 1 & 5\end{bmatrix}

A + 3A^{-1}

where I is the identity matrix of order 2.

3I
5I
7I
None of the above

Solution:

A=\begin{bmatrix}2 & 7 \\ 1 & 5\end{bmatrix}

Calculating the determinant value of A, we get

|A|=3

Now the inverse of the matrix of A will be

A^{-1}=\frac{1}{3}\begin{bmatrix}5 & -7 \\ -1 & 2\end{bmatrix}

=\begin{bmatrix}\frac{5}{3} & \frac{-7}{3} \\ \frac{-1}{3} & \frac{2}{3}\end{bmatrix}

Hence

A+3A^{-1}

=\begin{bmatrix}2 & 7 \\ 1 & 5\end{bmatrix}+3\begin{bmatrix}\frac{5}{3} & \frac{-7}{3} \\ \frac{-1}{3} & \frac{2}{3}\end{bmatrix}

=\begin{bmatrix}2 & 7 \\ 1 & 5\end{bmatrix}+\begin{bmatrix}5 & -7 \\ -1 & 2\end{bmatrix}

=\begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}

=7I

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