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Mathematics Question on Determinants

The inverse of the matrix A=[200 030 004 ]A =\begin{bmatrix} {2}&{0} &{0}\\\ {0}&{3}& {0} \\\ {0}&{0}&{4}\\\ \end{bmatrix} is

A

[200 030 004 ]\begin{bmatrix} {2}&{0} &{0}\\\ {0}&{3}& {0} \\\ {0}&{0}&{4}\\\ \end{bmatrix}

B

[1200 0130 0014 ]\begin{bmatrix} {\frac {1}{2}}&{0} &{0}\\\ {0}&{\frac {1}{3}}& {0} \\\ {0}&{0}&{\frac {1}{4}}\\\ \end{bmatrix}

C

124[200 030 004 ] \frac {1}{24}\begin{bmatrix} {2}&{0} &{0}\\\ {0}&{3}& {0} \\\ {0}&{0}&{4}\\\ \end{bmatrix}

D

124[100 010 001]\frac{1}{24}\left[\begin{array}{lll}1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1\end{array}\right]

Answer

[1200 0130 0014 ]\begin{bmatrix} {\frac {1}{2}}&{0} &{0}\\\ {0}&{\frac {1}{3}}& {0} \\\ {0}&{0}&{\frac {1}{4}}\\\ \end{bmatrix}

Explanation

Solution

Given, A=[200 030 004 ]A =\begin{bmatrix} {2}&{0} &{0}\\\ {0}&{3}& {0} \\\ {0}&{0}&{4}\\\ \end{bmatrix}
Now, ∣A∣=2(12−0)=24|A|=2(12-0)=24
Cofactors of AA are
C11=(12−0)=12,C12=0,C13=0C_{11}=(12-0)=12, C_{12}=0, C_{13}=0
C21=0,C22=+8,C23=0,C31=0C_{21}=0, C_{22}=+8, C_{23}=0, C_{31}=0
C32=0,C33=6C_{32}=0, C_{33}=6
∴adj(A)=[C11C12C13 C21C22C23 C31C32C33]T=[1200 080 006]T\therefore adj(A)= \begin{bmatrix}C_{11} & C_{12} & C_{13} \\\ C_{21} & C_{22} & C_{23} \\\ C_{31} & C_{32} & C_{33}\end{bmatrix}^{T}= \begin{bmatrix}12 & 0 & 0 \\\ 0 & 8 & 0 \\\ 0 & 0 & 6\end{bmatrix}^{T}
=[1200 080 006]= \begin{bmatrix}12 & 0 & 0 \\\ 0 & 8 & 0 \\\ 0 & 0 & 6\end{bmatrix}
∴A−1=1∣A∣adj(A)\therefore A^{-1}=\frac{1}{|A|} adj (A)
=124[1200 060 006]=[1200 0130 0014]=\frac{1}{24} \begin{bmatrix}12 & 0 & 0 \\\ 0 & 6 & 0 \\\ 0 & 0 & 6\end{bmatrix}= \begin{bmatrix}\frac{1}{2} & 0 & 0 \\\ 0 & \frac{1}{3} & 0 \\\ 0 & 0 & \frac{1}{4}\end{bmatrix}