KCET · Maths · Matrices
If \(A=\left[\begin{array}{lll}2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]\), then \(|\operatorname{adj} A|\) is equal to
- A 0
- B 9
- C \(\frac{1}{9}\)
- D 81
Answer & Solution
Correct Answer
(D) 81
Step-by-step Solution
Detailed explanation
Given, \(A=\left[\begin{array}{lll}2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]\)
\(\mathrm{C}_{11}=4, \quad \mathrm{C}_{12}=1, \quad \mathrm{C}_{13}=-2\)
\(C_{21}=-2, \quad C_{22}=4, \quad C_{23}=1\)
\(\mathrm{C}_{31}=1, \quad \mathrm{C}_{32}=-2, \quad \mathrm{C}_{33}=4\)
\(\therefore \quad \operatorname{adj}(A)=\left[\begin{array}{rrr}4 & 1 & -2 \\ -2 & 4 & 1 \\ 1 & -2 & 4\end{array}\right]^{\mathrm{T}}\)
\(=\left[\begin{array}{rrr}4 & -2 & 1 \\ 1 & 4 & -2 \\ -2 & 1 & 4\end{array}\right]\)
\(\therefore \quad|\operatorname{adj} \mathrm{A}|=4(16+2)+2(4-4)+(1+8)\)
\(=72+0+9=81\)
\(\mathrm{C}_{11}=4, \quad \mathrm{C}_{12}=1, \quad \mathrm{C}_{13}=-2\)
\(C_{21}=-2, \quad C_{22}=4, \quad C_{23}=1\)
\(\mathrm{C}_{31}=1, \quad \mathrm{C}_{32}=-2, \quad \mathrm{C}_{33}=4\)
\(\therefore \quad \operatorname{adj}(A)=\left[\begin{array}{rrr}4 & 1 & -2 \\ -2 & 4 & 1 \\ 1 & -2 & 4\end{array}\right]^{\mathrm{T}}\)
\(=\left[\begin{array}{rrr}4 & -2 & 1 \\ 1 & 4 & -2 \\ -2 & 1 & 4\end{array}\right]\)
\(\therefore \quad|\operatorname{adj} \mathrm{A}|=4(16+2)+2(4-4)+(1+8)\)
\(=72+0+9=81\)
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