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Mathematics Question on Determinants

If A=[1−22 02−3 3−24 ]A = \begin{bmatrix} {1}&{-2} &{2}\\\ {0}&{2}& {-3} \\\ {3}&{-2}&{4}\\\ \end{bmatrix} ,then A.adj(A)A . adj(A) is equal to

A

[500 050 005 ]\begin{bmatrix} {5}&{0} &{0}\\\ {0}&{5}& {0} \\\ {0}&{0}&{5}\\\ \end{bmatrix}

B

[511 151 115 ]\begin{bmatrix} {5}&{1} &{1}\\\ {1}&{5}& {1} \\\ {1}&{1}&{5}\\\ \end{bmatrix}

C

[000 000 000 ]\begin{bmatrix} {0}&{0} &{0}\\\ {0}&{0}& {0} \\\ {0}&{0}&{0}\\\ \end{bmatrix}

D

[800 080 008 ]\begin{bmatrix} {8}&{0} &{0}\\\ {0}&{8}& {0} \\\ {0}&{0}&{8}\\\ \end{bmatrix}

Answer

[800 080 008 ]\begin{bmatrix} {8}&{0} &{0}\\\ {0}&{8}& {0} \\\ {0}&{0}&{8}\\\ \end{bmatrix}

Explanation

Solution

A=[1−22 02−3 3−24]A = \begin{bmatrix}1&-2&2\\\ 0&2&-3\\\ 3&-2&4\end{bmatrix}
C11=8−6=2C_{11} = 8-6 =2
C12=−(0+9)=−9,C31=+6−4=2C_{12} =- \left(0+9\right) = - 9, C_{31} = +6-4=2
C13=0−6=−6,C32=−(−3−0)=3,C21=−(−8+4)=4,C33=2−0=2C_{13} = 0 -6 = - 6, C_{32} =- \left(-3 -0\right) = 3, C_{21} = - \left(-8+4\right) = 4, C_{33} = 2-0 = 2
C22=4−6=−2,C23=−(−2+6)=−4C_{22} = 4-6 = - 2 , C_{23} = - \left(-2 +6\right) = - 4
adj(A)=[2−9−6 4−2−4 232]T=[242 −9−23 −6−42]\text{adj} \left(A\right) = \begin{bmatrix}2&-9&-6\\\ 4&-2&-4\\\ 2&3&2\end{bmatrix} ^{T} = \begin{bmatrix}2&4&2\\\ -9&-2&3\\\ -6&-4&2\end{bmatrix}
Now , Aadj(A)=[1−22 02−3 3−24][242 −9−23 −6−42]A \text{adj} \left(A\right) = \begin{bmatrix}1&-2&2\\\ 0&2&-3\\\ 3&-2&4\end{bmatrix}\begin{bmatrix}2&4&2\\\ -9&-2&3\\\ -6&-4&2\end{bmatrix}
=[800 080 008]= \begin{bmatrix}8&0&0\\\ 0&8&0\\\ 0&0&8\end{bmatrix}
Alternative : A−1=adj(A)∣A∣A^{-1} = \frac{\text{adj} \left(A\right)}{\left|A\right|}
⇒AA−1=A.[adj(A)]∣A∣\Rightarrow A A^{-1} = \frac{A . \left[\text{adj} \left(A\right)\right]}{\left|A\right|}
⇒I∣A∣=A.[adj(A)]\Rightarrow I\left|A\right| = A . \left[\text{adj} \left(A\right)\right]
A=8A = 8
∴\therefore A [adj (A )] = 8I = [800 080 008]\begin{bmatrix}8&0&0\\\ 0&8&0\\\ 0&0&8\end{bmatrix}