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

& 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{aligned}$$

Answer 1

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,

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{aligned} & \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{aligned}$$ Therefore, $$\begin{aligned} & {{\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{aligned}$$ Multiplying $\dfrac{1}{abc}$ inside of matrix in its all rows, we get $$\begin{aligned} & {{\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{aligned}$$ Which is the required result. **Hence, option A is correct.** **Note:** The possibility of error in this question can be at a point where $$\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.