Question

Let $D = diag\left[ {{d_1},{d_2},{d_3}} \right]$ , where none of ${d_1},{d_2},{d_3}$ is 0, prove that ${D^{ - 1}} = diag\left[ {{d_1}^{ - 1},{d_2}^{ - 1},{d_3}^{ - 1}} \right].$

Answer 1

Here, we will prove that the inverse of a diagonal matrix is the inverses of the diagonals. We will find whether there exists an inverse by finding the determinant. We will find the minors, cofactors and transposes of the matrices and by using the inverse of the matrix formula to find the inverse of the diagonal matrix and prove the required. A matrix is defined as an array of numbers in the form of rows and columns.

Formula Used:
We will use the following formula:
1. Co-Factors of the matrix is given by Cofactor of $A{\rm{ = }}\left[ {{{\left( { - 1} \right)}^{i + j}}{a_{ij}}} \right]$
2. Transpose of a matrix is given by the formula ${A^T} = {\left[ {{a_{ij}}} \right]_{n \times m}}$ where $A = {\left[ {{a_{ij}}} \right]_{m \times n}}$
3. Adjoint of a Matrix is given by the formula $Adj\left( A \right) = {A^T}$
4. The inverse of a matrix is given by ${D^{ - 1}} = \dfrac{{Adj\left( D \right)}}{{\left| D \right|}}$

Complete step-by-step answer:
We are given that $D = diag\left[ {{d_1},{d_2},{d_3}} \right]$ .
So, the given matrix is a diagonal matrix.
So, the matrix is D = \left[ {\begin{array}{*{20}{c}}{{d_1}}&0&0\\\0&{{d_2}}&0\\\0&0&{{d_3}}\end{array}} \right]
Now, we will find the determinant of the matrix$D$.
\Rightarrow \left| {\begin{array}{*{20}{c}}{{d_1}}&0&0\\\0&{{d_2}}&0\\\0&0&{{d_3}}\end{array}} \right| = {d_1}\left| {\begin{array}{*{20}{c}}{{d_2}}&0\\\0&{{d_3}}\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}0&0\\\0&{{d_3}}\end{array}} \right| + 0\left| {\begin{array}{*{20}{c}}0&{{d_2}}\\\0&0\end{array}} \right|
\Rightarrow \left| {\begin{array}{*{20}{c}}{{d_1}}&0&0\\\0&{{d_2}}&0\\\0&0&{{d_3}}\end{array}} \right| = {d_1}\left( {{d_2}{d_3} - 0} \right) - 0\left( {0 - 0} \right) + 0\left( {0 - 0} \right)
\Rightarrow \left| {\begin{array}{*{20}{c}}{{d_1}}&0&0\\\0&{{d_2}}&0\\\0&0&{{d_3}}\end{array}} \right| = {d_1}{d_2}{d_3}
We are given that none of ${d_1},{d_2},{d_3}$ is 0 i.e., ${d_1} = {d_2} = {d_3} \ne 0$
\Rightarrow \left| {\begin{array}{*{20}{c}}{{d_1}}&0&0\\\0&{{d_2}}&0\\\0&0&{{d_3}}\end{array}} \right| = {d_1}{d_2}{d_3} \ne 0
Since the determinant of the matrix is not zero, then there exist an inverse of the matrix.
Now, we will find the minors of the matrices
\Rightarrow \left[ {\begin{array}{*{20}{c}}{\left| {\begin{array}{*{20}{c}}{{d_2}}&0\\\0&{{d_3}}\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}0&0\\\0&{{d_3}}\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}0&{{d_2}}\\\0&0\end{array}} \right|}\\\\{\left| {\begin{array}{*{20}{c}}0&0\\\0&{{d_3}}\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}{{d_1}}&0\\\0&{{d_3}}\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}{{d_1}}&0\\\0&0\end{array}} \right|}\\\\{\left| {\begin{array}{*{20}{c}}0&0\\\\{{d_2}}&0\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}{{d_1}}&0\\\0&0\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}{{d_1}}&0\\\0&{{d_2}}\end{array}} \right|}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{d_2}{d_3} - 0}&{0 - 0}&{0 - 0}\\\\{0 - 0}&{{d_1}{d_3} - 0}&{0 - 0}\\\\{0 - 0}&{0 - 0}&{{d_1}{d_2} - 0}\end{array}} \right]
\Rightarrow \left[ {\begin{array}{*{20}{c}}{\left| {\begin{array}{*{20}{c}}{{d_2}}&0\\\0&{{d_3}}\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}0&0\\\0&{{d_3}}\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}0&{{d_2}}\\\0&0\end{array}} \right|}\\\\{\left| {\begin{array}{*{20}{c}}0&0\\\0&{{d_3}}\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}{{d_1}}&0\\\0&{{d_3}}\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}{{d_1}}&0\\\0&0\end{array}} \right|}\\\\{\left| {\begin{array}{*{20}{c}}0&0\\\\{{d_2}}&0\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}{{d_1}}&0\\\0&0\end{array}} \right|}&{\left| {\begin{array}{*{20}{c}}{{d_1}}&0\\\0&{{d_2}}\end{array}} \right|}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{d_2}{d_3}}&0&0\\\0&{{d_1}{d_3}}&0\\\0&0&{{d_1}{d_2}}\end{array}} \right]
Co-Factors of the matrix is given by $Cofactor{\rm{ }}of{\rm{ }}A{\rm{ = }}\left[ {{{\left( { - 1} \right)}^{i + j}}{a_{ij}}} \right]$
$\Rightarrow$ Cofactor of D = \left[ {\begin{array}{*{20}{c}}{{{\left( { - 1} \right)}^{1 + 1}}{d_2}{d_3}}&{{{\left( { - 1} \right)}^{1 + 2}}0}&{{{\left( { - 1} \right)}^{1 + 3}}0}\\\\{{{\left( { - 1} \right)}^{2 + 1}}0}&{{{\left( { - 1} \right)}^{2 + 2}}{d_1}{d_3}}&{{{\left( { - 1} \right)}^{2 + 3}}0}\\\\{{{\left( { - 1} \right)}^{3 + 1}}0}&{{{\left( { - 1} \right)}^{3 + 2}}0}&{{{\left( { - 1} \right)}^{3 + 3}}{d_1}{d_2}}\end{array}} \right]
$\Rightarrow$ Cofactor of D = \left[ {\begin{array}{*{20}{c}}{{{\left( { - 1} \right)}^2}{d_2}{d_3}}&{{{\left( { - 1} \right)}^3}0}&{{{\left( { - 1} \right)}^4}0}\\\\{{{\left( { - 1} \right)}^3}0}&{{{\left( { - 1} \right)}^4}{d_1}{d_3}}&{{{\left( { - 1} \right)}^5}0}\\\\{{{\left( { - 1} \right)}^4}0}&{{{\left( { - 1} \right)}^5}0}&{{{\left( { - 1} \right)}^6}{d_1}{d_2}}\end{array}} \right]
We know that ${\left( { - 1} \right)^n} = 1$ when $n$is an even integer; ${\left( { - 1} \right)^n} = - 1$ when $n$ is an odd integer.
$\Rightarrow$ Cofactor of D = \left[ {\begin{array}{*{20}{c}}{{d_2}{d_3}}&0&0\\\0&{{d_1}{d_3}}&0\\\0&0&{{d_1}{d_2}}\end{array}} \right]
Transpose of a matrix is given by the formula ${A^T} = {\left[ {{a_{ij}}} \right]_{n \times m}}$ where $A = {\left[ {{a_{ij}}} \right]_{m \times n}}$
$\Rightarrow$ Transpose of a matrix with cofactors = {\left[ {\begin{array}{*{20}{c}}{{d_2}{d_3}}&0&0\\\0&{{d_1}{d_3}}&0\\\0&0&{{d_1}{d_2}}\end{array}} \right]^T}
$\Rightarrow$ Transpose of a matrix with cofactors{D^T} = \left[ {\begin{array}{*{20}{c}}{{d_2}{d_3}}&0&0\\\0&{{d_1}{d_3}}&0\\\0&0&{{d_1}{d_2}}\end{array}} \right]
Now, we will find the Adjoint of the given matrix.
Adjoint of a Matrix is given by the formula $Adj\left( A \right) = {A^T}$
$\Rightarrow AdjD = {D^T}$
\Rightarrow AdjD = \left[ {\begin{array}{*{20}{c}}{{d_2}{d_3}}&0&0\\\0&{{d_1}{d_3}}&0\\\0&0&{{d_1}{d_2}}\end{array}} \right]
The inverse of a matrix is given by ${D^{ - 1}} = \dfrac{{Adj\left( D \right)}}{{\left| D \right|}}$
Thus, we get
\Rightarrow {D^{ - 1}} = \dfrac{1}{{{d_1}{d_2}{d_3}}}\left[ {\begin{array}{*{20}{c}}{{d_2}{d_3}}&0&0\\\0&{{d_1}{d_3}}&0\\\0&0&{{d_1}{d_2}}\end{array}} \right]
By multiplying the term, we get
\Rightarrow {D^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\dfrac{{{d_2}{d_3}}}{{{d_1}{d_2}{d_3}}}}&0&0\\\0&{\dfrac{{{d_1}{d_3}}}{{{d_1}{d_2}{d_3}}}}&0\\\0&0&{\dfrac{{{d_1}{d_2}}}{{{d_1}{d_2}{d_3}}}}\end{array}} \right]
By cancelling out the term, we get
\Rightarrow {D^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\dfrac{1}{{{d_1}}}}&0&0\\\0&{\dfrac{1}{{{d_2}}}}&0\\\0&0&{\dfrac{1}{{{d_3}}}}\end{array}} \right]
\Rightarrow {D^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{{d_1}^{ - 1}}&0&0\\\0&{{d_2}^{ - 1}}&0\\\0&0&{{d_3}^{ - 1}}\end{array}} \right]
$\Rightarrow {D^{ - 1}} = diag\left[ {{d_1}^{ - 1},{d_2}^{ - 1},{d_3}^{ - 1}} \right]$
Therefore, ${D^{ - 1}} = diag\left[ {{d_1}^{ - 1},{d_2}^{ - 1},{d_3}^{ - 1}} \right]$ is proved.

Note: We know that the adjoint of a square matrix is defined as the transpose of a matrix with cofactors. For every square matrix, we can associate a number which is called as the determinant of the matrix. The transpose of a matrix is defined as the interchanging of rows and columns. The determinant of a matrix is a value obtained after crossing out a row and column by multiplying the determinant of a square matrix.