Question

If a, b and c are non-zero numbers then the inverse of the matrix \[\text{A=}\left| \begin{matrix}
   a & 0 & 0 \\
   0 & b & 0 \\
   0 & 0 & c \\
\end{matrix} \right|\] is equal to
\[\begin{align}
  & A.\left| \begin{matrix}
   {{a}^{-1}} & 0 & 0 \\
   0 & {{b}^{-1}} & 0 \\
   0 & 0 & {{c}^{-1}} \\
\end{matrix} \right| \\
 & B.\dfrac{1}{abc}\left| \begin{matrix}
   {{a}^{-1}} & 0 & 0 \\
   0 & {{b}^{-1}} & 0 \\
   0 & 0 & {{c}^{-1}} \\
\end{matrix} \right| \\
 & \text{C}\text{.}\dfrac{1}{abc}\left| \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right| \\
 & \text{D}\text{.}\dfrac{1}{abc}\left| \begin{matrix}
   a & 0 & 0 \\
   0 & b & 0 \\
   0 & 0 & c \\
\end{matrix} \right| \\
\end{align}\]

Answer 1

Hint: We will use the basic definition in terms of matrices to solve the question. A minor is the determinant of a square matrix formed by deleting one row and one column from a longer matrix. The cofactor is the value obtained on minor with using sign '+' or '-' depending upon element position and the transpose of a matrix is obtained by reversing rows and columns of each element of the original matrix. The adjoint of a matrix is the transpose of the cofactor matrix. Now, we can use the below formula to get the answer, \[{{\text{A}}^{-1}}=\dfrac{1}{\left| A \right|}\left[ \text{adjA} \right]\]

Complete step-by-step solution:
Given that matrix,
\[\text{A=}\left| \begin{matrix}
   a & 0 & 0 \\
   0 & b & 0 \\
   0 & 0 & c \\
\end{matrix} \right|\]
Where, a, b and c are non-zero.
To determine ${{\text{A}}^{-1}}$ we will use the formula:
\[{{\text{A}}^{-1}}=\dfrac{1}{\left| A \right|}\left[ \text{adjA} \right]\]
Where, adj A is the adjoint of A and adjoint of A can be determined using minors and cofactors of matrix.
Minor of a matrix-A minor is the determinant of the square matrix formed by deleting one row and one column from some larger square matrix.
Example if \[\text{M=}\left| \begin{matrix}
   1 & 2 & 3 \\
   4 & 5 & 6 \\
   7 & 8 & 9 \\
\end{matrix} \right|\]
Then, minor along (1, 2) position of M would be
\[\text{M(1,2)=}\left| \begin{matrix}
   4 & 6 \\
   7 & 9 \\
\end{matrix} \right|\]
This is obtained by deleting row 1 and column 2 from the matrix M.
Cofactor of a matrix: Cofactor is the number we get when we remove the column and row of any element of matrix. We also take sign '+' or '-' into count here.
Example: like in the matrix M taken before
\[\text{M=}\left| \begin{matrix}
   1 & 2 & 3 \\
   4 & 5 & 6 \\
   7 & 8 & 9 \\
\end{matrix} \right|\]
The cofactor of \[\text{M(1,2)=}\left| \begin{matrix}
   4 & 6 \\
   7 & 9 \\
\end{matrix} \right|=4\times 9-7\times 6=-6\]
Also in cofactor, we take account of sign '+' or '-' when position of element is (i, j) then sign comes according as ${{\left( -1 \right)}^{i+j}}$
Now, here position is (1, 2) then sign \[\Rightarrow {{\left( -1 \right)}^{1+2}}={{\left( -1 \right)}^{3}}=-1\] then cofactor is \[\left( -6 \right)\left( -1 \right)=+6\]
Now, coming to original matrix
\[\text{A=}\left| \begin{matrix}
   a & 0 & 0 \\
   0 & b & 0 \\
   0 & 0 & c \\
\end{matrix} \right|\]
Then, determinant \[\left| A \right|=a\left[ bc-0 \right]=abc\]
So \[\left| A \right|=abc\]
Now, we compute adj (A).
Using the theory and example explained of minor and cofactor above, we get:
\[\text{adj}\left( A \right)=\left| \begin{matrix}
   {{\left( -1 \right)}^{1+1}}bc & {{\left( -1 \right)}^{1+2}}0 & {{\left( -1 \right)}^{1+3}}0 \\
   {{\left( -1 \right)}^{2+1}}0 & {{\left( -1 \right)}^{2+2}}ac & {{\left( -1 \right)}^{2+3}}0 \\
   {{\left( -1 \right)}^{3+1}}0 & {{\left( -1 \right)}^{3+2}}0 & {{\left( -1 \right)}^{3+3}}ab \\
\end{matrix} \right|\]
Then, \[\text{adj(A)=}{{\left| \begin{matrix}
   bc & 0 & 0 \\
   0 & ac & 0 \\
   0 & 0 & ab \\
\end{matrix} \right|}^{T}}\]
Here, \[{{\left| {} \right|}^{T}}\to \text{ T denotes transpose of matrix}\]
Transpose of a matrix: The transpose of a matrix is an operator which flips a matrix over its diagonal and reverse its rows and columns. Denoted by ${{A}^{T}}$
Now, \[\begin{align}
  & \text{adj(A)=}{{\left| \begin{matrix}
   bc & 0 & 0 \\
   0 & ac & 0 \\
   0 & 0 & ab \\
\end{matrix} \right|}^{T}} \\
 & \text{adj(A)=}\left| \begin{matrix}
   bc & 0 & 0 \\
   0 & ac & 0 \\
   0 & 0 & ab \\
\end{matrix} \right| \\
\end{align}\]
Therefore, \[\begin{align}
  & {{\text{A}}^{-1}}=\dfrac{1}{\left| A \right|}\left[ \text{adjA} \right] \\
 & {{\text{A}}^{-1}}=\dfrac{1}{abc}\left| \begin{matrix}
   bc & 0 & 0 \\
   0 & ac & 0 \\
   0 & 0 & ab \\
\end{matrix} \right| \\
\end{align}\]
Multiplying $\dfrac{1}{abc}$ inside of matrix in its all rows, we get
\[\begin{align}
  & {{\text{A}}^{-1}}=\left| \begin{matrix}
   \dfrac{bc}{abc} & 0 & 0 \\
   0 & \dfrac{ac}{abc} & 0 \\
   0 & 0 & \dfrac{ab}{abc} \\
\end{matrix} \right| \\
 & {{\text{A}}^{-1}}=\left| \begin{matrix}
   \dfrac{1}{a} & 0 & 0 \\
   0 & \dfrac{1}{b} & 0 \\
   0 & 0 & \dfrac{1}{c} \\
\end{matrix} \right| \\
 & {{\text{A}}^{-1}}=\left| \begin{matrix}
   {{a}^{-1}} & 0 & 0 \\
   0 & {{b}^{-1}} & 0 \\
   0 & 0 & {{c}^{-1}} \\
\end{matrix} \right| \\
\end{align}\]
Which is the required result.
Hence, option A is correct.

Note: The possibility of error in this question can be at a point where \[\dfrac{1}{abc}\] is multiplied inside the matrix of adj (A). When we multiply \[\dfrac{1}{abc}\] inside, we need to multiply it in all rows because this is a matrix, not a determinant. Determinants have a property to be multiplied to any one row or column, not matrices. So, take this into consideration while solving.