Question

If we have a diagonal matrix \[A=diag\left( a,b,c \right)=\left[ \begin{matrix}
   a & 0 & 0 \\
   0 & b & 0 \\
   0 & 0 & c \\
\end{matrix} \right]\] such that \[abc\ne 0\] then \[{{A}^{-1}}=diag\left( \dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c} \right)=\left[ \begin{matrix}
   \dfrac{1}{a} & 0 & 0 \\
   0 & \dfrac{1}{b} & 0 \\
   0 & 0 & \dfrac{1}{c} \\
\end{matrix} \right].\]

Answer 1

Hint: We are given a matrix A and we are asked to find the inverse of A. We know that \[{{A}^{-1}}\] is given as \[{{A}^{-1}}=\dfrac{Adj\left( A \right)}{\left| A \right|}.\] So, we will first find the determinant of A. We will expand along row 1 and we will find the determinant of A. Then to find the Adj (A) we will find the cofactor for each position once we have the cofactor, we get Adj (A), then put the value in \[{{A}^{-1}}=\dfrac{Adj\left( A \right)}{\left| A \right|}\] to get the inverse of A.

Complete step-by-step solution:
We are given a matrix A as \[\left[ \begin{matrix}
   a & 0 & 0 \\
   0 & b & 0 \\
   0 & 0 & c \\
\end{matrix} \right].\] We are asked to find the inverse of this matrix A. We are clearly seen that our matrix A is a diagonal matrix. We know that the inverse of any matrix is given as
\[{{A}^{-1}}=\dfrac{Adj\left( A \right)}{\left| A \right|}\]
So, we will first find the determinant of A.
\[\left| A \right|=\left[ \begin{matrix}
   a & 0 & 0 \\
   0 & b & 0 \\
   0 & 0 & c \\
\end{matrix} \right]\]
Expanding along row 1, we get,
\[\Rightarrow a\left( bc-0 \right)+0+0\]
\[\Rightarrow abc\]
We are given that
\[\Rightarrow abc\ne 0\left[ \text{As given} \right]\]
Now, |A| = abc which is not zero. So, the inverse exists.
Now, we have to find the adjoint of A. In order to find the adjoint of A, we will find all the cofactors. So,
\[{{C}_{11}}={{\left( -1 \right)}^{1+1}}\left[ bc-0 \right]\]
\[\Rightarrow {{C}_{11}}={{\left( -1 \right)}^{2}}bc\]
\[\Rightarrow {{C}_{11}}=bc\]
Similarly,
\[{{C}_{12}}={{\left( -1 \right)}^{1+2}}\left[ 0-0 \right]\]
\[\Rightarrow {{C}_{12}}=0\]
\[{{C}_{13}}={{\left( -1 \right)}^{1+3}}\left[ 0-0 \right]\]
\[\Rightarrow {{C}_{13}}=0\]
\[{{C}_{21}}={{\left( -1 \right)}^{2+1}}\left[ 0-0 \right]\]
\[\Rightarrow {{C}_{21}}={{\left( -1 \right)}^{3}}\left( 0 \right)\]
\[\Rightarrow {{C}_{21}}=0\]
Similarly, we find,
\[{{C}_{22}}={{\left( -1 \right)}^{2+2}}\left( ac-0 \right)\]
\[\Rightarrow {{C}_{22}}={{\left( -1 \right)}^{4}}ac\]
\[\Rightarrow {{C}_{22}}=ac\]
Now,
\[{{C}_{23}}={{\left( -1 \right)}^{2+3}}\left( 0+0 \right)\]
\[\Rightarrow {{C}_{23}}=0\]
Now, we will find the cofactor along the row three. So, we get,
\[{{C}_{31}}={{\left( -1 \right)}^{3+1}}\left( 0-0 \right)\]
\[\Rightarrow {{C}_{31}}=0\]
\[{{C}_{32}}={{\left( -1 \right)}^{3+2}}\left[ 0-0 \right]\]
\[\Rightarrow {{C}_{32}}=0\]
And lastly,
\[{{C}_{33}}={{\left( -1 \right)}^{3+3}}\left[ ab-0 \right]\]
\[\Rightarrow {{C}_{33}}={{\left( -1 \right)}^{6}}\left( ab \right)\]
\[\Rightarrow {{C}_{33}}=ab\]
So, using this cofactor, we get our adjoint as,
\[\left( Adj\text{ }A \right)=\left[ \begin{matrix}
   bc & 0 & 0 \\
   0 & ac & 0 \\
   0 & 0 & ab \\
\end{matrix} \right]\]
So putting this adj A in the formula of \[{{A}^{-1}}.\] So, we get,
\[{{A}^{-1}}=\dfrac{Adj\left( A \right)}{\left| A \right|}\]
As, |A| = abc, so,
\[{{A}^{-1}}=\dfrac{1}{abc}\left[ \begin{matrix}
   bc & 0 & 0 \\
   0 & ac & 0 \\
   0 & 0 & ab \\
\end{matrix} \right]\]
Multiplying each term by \[\dfrac{1}{abc}\] we get,
\[{{A}^{-1}}=\left[ \begin{matrix}
   \dfrac{bc}{abc} & 0 & 0 \\
   0 & \dfrac{ac}{abc} & 0 \\
   0 & 0 & \dfrac{ab}{abc} \\
\end{matrix} \right]\]
\[\Rightarrow {{A}^{-1}}=\left[ \begin{matrix}
   \dfrac{1}{a} & 0 & 0 \\
   0 & \dfrac{1}{b} & 0 \\
   0 & 0 & \dfrac{1}{c} \\
\end{matrix} \right]\]
Hence, we get the inverse of A as \[\left[ \begin{matrix}
   \dfrac{1}{a} & 0 & 0 \\
   0 & \dfrac{1}{b} & 0 \\
   0 & 0 & \dfrac{1}{c} \\
\end{matrix} \right].\]

Note: Remember the determinant of any diagonal matrix is given by the product of its diagonal element as our matrix, \[A=\left[ \begin{matrix}
   a & 0 & 0 \\
   0 & b & 0 \\
   0 & 0 & c \\
\end{matrix} \right]\] in the diagonal matrix. So, we can easily find the value \[\left| A \right|=a\times b\times c=abc.\] We get from here that for a diagonal matrix, the inverse of the diagonal matrix is made by inversing the diagonal elements only.