Question

If D = diag $\left(d_1,d_2,...,d_n\right)$ where $d_1\ne 0$, for i = 1, 2, ..., n, then $D^{-1}$ is equal to

• A. D${}^T$
• B. D
• C. Adj (D)
• D. diag $\left(d_1^{-1},d_2^{-1},...d_n^{-1}\right)$

Answer 1

Answer: (D)

D = diag $\left(d_1,d_2,...d_n\right)$
$D=\left|\begin{array}{cccccc}{d}_1& 0& 0& -& -& 0\\ 0& d_2& 0& -& -& 0\\ 0& 0& d_3& -& -& 0\\ -& -& -& -& -& -\\ -& -& -& -& -& -\\ 0& 0& 0& 0& -& 0\end{array}\right|$
$\left|D\right|=\left|\begin{array}{cccccc}{d}_1& 0& 0& -& -& 0\\ 0& d_2& 0& -& -& 0\\ 0& 0& d_3& -& -& 0\\ -& -& -& -& -& -\\ -& -& -& -& -& -\\ 0& 0& 0& 0& -& 0\end{array}\right|$
$=d_1{d}_2{d}_3.......d_n$
$adj\left(D\right)=\left[\begin{array}{cccccc}{d}_2{d}_3{d}_4...dn& 0& 0& -& -& 0\\ 0& d_1{d}_3{d}_4...dn& 0& -& -& 0\\ 0& 0& d_1{d}_2{d}_4...dn& -& -& 0\\ -& -& -& -& -& -\\ -& -& -& -& -& -\\ 0& 0& 0& 0& -& d_1{d}_2{d}_3...d_{n-1}\end{array}\right]$
$adj\left(D\right)=\frac{1}{d_1{d}_2{d}_3....d_n}$
$\left[\begin{array}{cccccc}{d}_2{d}_3{d}_4...dn& 0& 0& -& -& 0\\ 0& d_1{d}_3{d}_4...dn& 0& -& -& 0\\ 0& 0& d_1{d}_2{d}_4...dn& -& -& 0\\ -& -& -& -& -& -\\ -& -& -& -& -& -\\ 0& 0& 0& 0& -& d_1{d}_2{d}_3...d_{n-1}\end{array}\right]$
$=\left[\begin{array}{cccccc}\frac{1}{d_1}& 0& 0& -& -& 0\\ 0& \frac{1}{d_2}& 0& -& -& 0\\ 0& 0& \frac{1}{d_3}& -& -& 0\\ -& -& -& -& -& -\\ -& -& -& -& -& -\\ 0& 0& 0& 0& -& \frac{1}{d_n}\end{array}\right]$
$=\left[\begin{array}{cccccc}{d}_1^{-1}& 0& 0& -& -& 0\\ 0& d^{-1}& 0& -& -& 0\\ 0& 0& d_3^{-1}& -& -& 0\\ -& -& -& -& -& -\\ -& -& -& -& -& -\\ 0& 0& 0& 0& -& d_n^{-1}\end{array}\right]$
$=diag\left(d_1^{-1},d_2^{-1},....,d_n^{-1}\right)$