Question

If \[A = \left( {\begin{array}{*{20}{c}}
  x&0&0 \\
  0&y&0 \\
  0&0&z
\end{array}} \right)\] is a nonsingular matrix then find ${A^{ - 1}}$ by elementary row transformations. Hence, find the inverse of $\left( {\begin{array}{*{20}{c}}
  2&0&0 \\
  0&1&0 \\
  0&0&{ - 1}
\end{array}} \right)$.

Answer 1

Hint: For elementary row transformation we use $A = AI$ , to provide us the required solution which provides us certain operation to the matrix which can provide the inverse as in this question we can determine the matrix by applying the operation . then we compare it with $I = A{A^{ - 1}}$ to get the required solution.
Now in this we perform steps
1. Swap rows
2. Multiply or divide each elements in a row by a constant
3. Replace a row by adding or subtracting a multiple of another row to it.
We must do it to the whole row

Complete step-by-step answer:
Given:
\[A = \left( {\begin{array}{*{20}{c}}
  x&0&0 \\
  0&y&0 \\
  0&0&z
\end{array}} \right)\]
So we know that $A = AI$
$\Rightarrow$\[\left( {\begin{array}{*{20}{c}}
  x&0&0 \\
  0&y&0 \\
  0&0&z
\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right)\]
Now in this we have to convert \[A = \left( {\begin{array}{*{20}{c}}
  x&0&0 \\
  0&y&0 \\
  0&0&z
\end{array}} \right)\] in identity matrix $\because A{A^{ - 1}} = I$
So we perform row operation on it
$(1)$ divide first row to $x$ that is ${R_1} \to \dfrac{{{R_1}}}{x}$
$\Rightarrow$\[\left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&y&0 \\
  0&0&z
\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}
  {\dfrac{1}{x}}&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right)\]
2. divide first row to $y$ that is ${R_2} \to \dfrac{{{R_2}}}{y}$
$\Rightarrow$\[\left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&z
\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}
  {\dfrac{1}{x}}&0&0 \\
  0&{\dfrac{1}{y}}&0 \\
  0&0&1
\end{array}} \right)\]
3. divide first row to $z$ that is ${R_3} \to \dfrac{{{R_3}}}{z}$
$\Rightarrow$\[\left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}
  {\dfrac{1}{x}}&0&0 \\
  0&{\dfrac{1}{y}}&0 \\
  0&0&{\dfrac{1}{z}}
\end{array}} \right)\]
Now we compare it with
$I = A{A^{ - 1}}$
We get
$\Rightarrow$\[{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}
  {\dfrac{1}{x}}&0&0 \\
  0&{\dfrac{1}{y}}&0 \\
  0&0&{\dfrac{1}{z}}
\end{array}} \right)\]
Now we have to find the inverse of $\left( {\begin{array}{*{20}{c}}
  2&0&0 \\
  0&1&0 \\
  0&0&{ - 1}
\end{array}} \right)$
As we see that inverse of \[A = \left( {\begin{array}{*{20}{c}}
  x&0&0 \\
  0&y&0 \\
  0&0&z
\end{array}} \right)\] is \[\left( {\begin{array}{*{20}{c}}
  {\dfrac{1}{x}}&0&0 \\
  0&{\dfrac{1}{y}}&0 \\
  0&0&{\dfrac{1}{z}}
\end{array}} \right)\]
Now from here we can say $x = 2$, $y = 1$ and $z = - 1$ put these values in the inverse of matrix $A$
So inverse is \[\left( {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&0&0 \\
  0&{\dfrac{1}{1}}&0 \\
  0&0&{\dfrac{1}{{ - 1}}}
\end{array}} \right)\]
So final answer inverse of $\left( {\begin{array}{*{20}{c}}
  2&0&0 \\
  0&1&0 \\
  0&0&{ - 1}
\end{array}} \right)$ is \[\left( {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&0&0 \\
  0&1&0 \\
  0&0&{ - 1}
\end{array}} \right)\] .

Note: Inverse of any matrix can be found by using
1. First we have to find determinant of given Matrix
2. Second we have to find adjoint of given matrix by using $Adj(A) = {[cof({a_{ij}})]^T}$
After this we use this simple formula to find inverse
${A^{ - 1}} = \dfrac{1}{{\det (A)}}adj(A)$ .
An inverse can be found only of that matrix which is a square matrix that means order of row and order of column is the same.
And the determinant of the matrix must not be zero. This is instead of the real number not being zero to have an inverse, the determinant must not be zero to have an inverse.