If D=diag⁡d1,d2,d3,…,dn, where di≠0 for all i=1,2,…,n, then D−1 is equal to

Let D=d100…00d20…0⋮⋮⋮ 000…dn.Then, |D|=d1d2…dnNow, Cofactor of D11=d2 d3…dnCofactor of D22=d1 d3…dnetcand, Cofactor of Dij=0 when i≠j∴ D−1=1|D|adj⁡D=1d1d2…dnd2d3…dn00…0 0 d2d3…dn0…0 ⋮ ⋮⋮ 0 00…d1d2…dn−1 ∴ D−1=1d100…001d20…0⋮⋮⋮ 000…1dn=diag⁡d1−1d2−1…dn−1

If D=diag⁡d1,d2,d3,…,dn, where di≠0 for all i=1,2,…,n, then D−1 is equal to

Let D=d100…00d20…0⋮⋮⋮ 000…dn.Then, |D|=d1d2…dnNow, Cofactor of D11=d2 d3…dnCofactor of D22=d1 d3…dnetcand, Cofactor of Dij=0 when i≠j∴ D−1=1|D|adj⁡D=1d1d2…dnd2d3…dn00…0 0 d2d3…dn0…0 ⋮ ⋮⋮ 0 00…d1d2…dn−1 ∴ D−1=1d100…001d20…0⋮⋮⋮ 000…1dn=diag⁡d1−1d2−1…dn−1

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If $D = diag (d_{1}, d_{2}, d_{3}, \dots, d_{n}),$ where $d_{i} \neq 0$ for all $i = 1, 2, \dots, n,$ then $D^{- 1}$ is equal to

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$diag (d_{1}^{- 1} d_{2}^{- 1}, \dots, d_{n}^{- 1})$

none of these

$D$

answer is B.

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Detailed Solution

Let $D = [\begin{array}{ccccc} d_{1} & 0 & 0 & \dots & 0 \\ 0 & d_{2} & 0 & \dots & 0 \\ ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & \dots & d_{n} \end{array}] .$ Then, $| D | = d_{1} d_{2} \dots d_{n}$

Now, Cofactor of $D_{11} = d_{2} d_{3} \dots d_{n}$

Cofactor of $D_{22} = d_{1} d_{3} \dots d_{n} etc$

and, Cofactor of $D_{i j} = 0$ when $i \neq j$

$∴ D^{- 1} = \frac{1}{| D |} adj D$

$\begin{aligned} = \frac{1}{d_{1} d_{2} \dots d_{n}} [\begin{array}{ccccccc} d_{2} d_{3} \dots & d_{n} & 0 & 0 & \dots & 0 \\ 0 & d_{2} d_{3} \dots d_{n} & 0 & \dots & 0 \\ ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & \dots & d_{1} d_{2} \dots d_{n - 1} \end{array}] \\ ∴ D^{- 1} = [\begin{array}{ccccc} \frac{1}{d_{1}} & 0 & 0 & \dots & 0 \\ 0 & \frac{1}{d_{2}} & 0 & \dots & 0 \\ ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & \dots & \frac{1}{d_{n}} \end{array}] = diag (d_{1}^{- 1} d_{2}^{- 1} \dots d_{n}^{- 1}) \end{aligned}$