# Given a matrix $A$ of rank $n$, show that $\det(\operatorname{adj}(A))=\det(A)^{n-1}$ and $\operatorname{adj}(\operatorname{adj}(A))=(\det A)^{n-2}A$

• Prove that if $A$ is regular then $\operatorname{adj}(\operatorname{adj}(A)) = (\det A)^{n-2} A$

#### Solutions Collecting From Web of "Given a matrix $A$ of rank $n$, show that $\det(\operatorname{adj}(A))=\det(A)^{n-1}$ and $\operatorname{adj}(\operatorname{adj}(A))=(\det A)^{n-2}A$"
For (1), if $rank(A) = n$, $adj(A)A = det(A)I$; taking determinants of both sides easily yields the result; if $rank(A) < n$, $det(A) = 0 = det(adj(A))$.
For (2), we use (1). If $rank A = n$, then
\begin{align} adj(adj(A))adj(A) &= det(adj(A))I\\ &= det(adj(A)) (1/det(A))adj(A)A\\ &= det(adj(A))^{n-2} adj(A)A adj(adj(A)) = det(adj(A))^{n-2}A, \end{align}
and, if $rank(A) < 0$, both sides are again zero.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://bootmath.com/given-a-matrix-a-of-rank-n-show-that-detoperatornameadjadetan-1-and-operatornameadjoperatornameadjadet-an-2a.html", "fetch_time": 1531841973000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-30/segments/1531676589752.56/warc/CC-MAIN-20180717144908-20180717164908-00597.warc.gz", "warc_record_offset": 51336954, "warc_record_length": 5110, "token_count": 328, "char_count": 849, "score": 4.09375, "int_score": 4, "crawl": "CC-MAIN-2018-30", "snapshot_type": "latest", "language": "en", "language_score": 0.48430728912353516, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"}