If \(R = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1}\\ 2&1&{ - 1}\\ 2&3&2 \end{array}} \right]\), the the top row of R^-1 is

CONCEPT:

Let A = [a_ij] be a square matrix of order n and C = [c_ij] is its co-factored matrix. Then, matrix C^T= [c_ji] is called the adjoint of matrix A and is denoted as: adj (A) = C^T= [c_ji], 1 ≤ i, j ≤ n.

Any non-singular matrix A = [a_ij] of order n is said to be invertible or has an inverse, if there exists another non-singular square matrix B of order n. The inverse of a square matrix say A of order n is given by \({A^{ - 1}} = \frac{1}{{\left| A \right|}} \cdot adj\;A\)

CALCULATION:

Given: \(R = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 1}\\ 2&1&{ - 1}\\ 2&3&2 \end{array}} \right]\)

First let's find the adjoint of the matrix R

\(⇒ adj\;\left( R \right) = \left[ {\begin{array}{*{20}{c}} 5&{ - 3}&1\\ { - 6}&0&{ - 1}\\ 4&{ - 3}&1 \end{array}} \right]\)

⇒ |R| = 1 × (2 + 3) - 0 × (4 + 2) - 1 × (6 - 2)

⇒ |R| = 1

∵ |R| ≠ 0 ⇒ R^-1 exists

As we know that, the  inverse of a square matrix say A of order n is given by \({A^{ - 1}} = \frac{1}{{\left| A \right|}} \cdot adj\;A\)

\(\Rightarrow {R^{ - 1}} = \;\left[ {\begin{array}{*{20}{c}} 5&{ - 3}&1\\ { - 6}&0&{ - 1}\\ 4&{ - 3}&1 \end{array}} \right]\)

So, the first row of R^-1 = [5  -3 1]

Hence, option B is the correct answer.