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  • If possible, using elementary row transformations, find the inverse of the following matrices [2 0 -1]

If possible, using elementary row transformations, find the inverse of the following matrices
\begin{bmatrix} 2 &0 &-1 \\5 &1 &0 \\0 &1 &3 \end{bmatrix}

Answers (1)

Let A =  \begin{bmatrix} 2 &0 &-1 \\5 &1 &0 \\0 &1 &3 \end{bmatrix}

To apply elementary row transformations, we write:

A = IA where I is the identity matrix

We proceed with solving our problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that

I = XA

And this X is called inverse of A = A^{-1}

Note: Never apply row and column transformations simultaneously over a matrix.

So we get,

\begin{aligned} &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-(5 / 2) \mathrm{R}_{1}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A\\ &\text { Applying } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2}\\ &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1 \end{array}\right] \mathrm{A}\\ \end{aligned}

\begin{aligned} &\text { Applying } R_{2} \rightarrow R_{2}-5 R_{3}\\ &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -15 & 6 & -5 \\ \frac{5}{2} & -1 & 1 \end{array}\right]\\ &\text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+2 \mathrm{R}_{3}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 6 & -2 & 2 \\ -15 & 6 & -5 \\ \frac{5}{2} & -1 & 1 \end{array}\right]\\ &\text { Applying } \mathrm{R}_{1} \rightarrow(1 / 2) \mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow 2 \mathrm{R}_{3}\\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ -5 & -2 & 2 \end{array}\right] \mathrm{A}\\ \end{aligned}

\begin{aligned} &\text { As we have Identity matrix in LHS, we get, }\\ &\therefore \mathrm{A}^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ -5 & -2 & 2 \end{array}\right] \end{aligned}

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