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I'm trying to show that A be a $ 3 x 3 $ upper triangular matrix with $det \ne 0 $. Show by explicit computation that $A^{-1}$ is also upper triangular. Simple showing is enough for me.

$$A= \begin{bmatrix}\color{blue}a & \color{blue}b & \color{blue}c \\0 & \color{blue}d & \color{blue}e \\ 0 & 0 &\color{blue}f\end{bmatrix}$$

Can someone explain and show it?

$$ A=\left(\begin{array}{rrr}% a&b&c\\% 0&d&e\\% 0&0&f\\% \end{array}\right)% $$ $$ x11=\left(\begin{array}{rrr}% d&e\\% 0&f\\% \end{array}\right)% =df, x12=-\left(\begin{array}{rrr}% 0&e\\% 0&f\\% \end{array}\right)% =0, x13=\left(\begin{array}{rrr}% 0&d\\% 0&0\\% \end{array}\right)% =0 $$ $$ x21=-\left(\begin{array}{rrr}% b&c\\% 0&f\\% \end{array}\right)% =-bf, X22=\left(\begin{array}{rrr}% a&c\\% 0&f\\% \end{array}\right)% =af, X23=-\left(\begin{array}{rrr}% a&b\\% 0&0\\% \end{array}\right)% =0 $$ $$ x31=\left(\begin{array}{rrr}% b&c\\% d&e\\% \end{array}\right)% =bc-cd, x32=-\left(\begin{array}{rrr}% a&c\\% 0&e\\% \end{array}\right)% =ac, x31=\left(\begin{array}{rrr}% a&b\\% 0&d\\% \end{array}\right)% =ad $$ $$ adjoint A = \left(\begin{array}{rrr}% df&0&0\\% -bf&af&0\\% bc-cd&-ac&ad\\% \end{array}\right)% $$ $$ det A = a\left(\begin{array}{rrr}% d&e\\% 0&f\\% \end{array}\right)% =adf $$ $$ Inverse-A =1/adf \left(\begin{array}{rrr}% df&0&0\\% -bf&af&0\\% bc-cd&-ac&ad\\% \end{array}\right)% $$ It came out lower triangular matrix. Is there any way to make it upper triangular matrix?

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  • $\begingroup$ Why did you remove the context? We can't help you very well if you don't share your thoughts and efforts. $\endgroup$ – Cameron Buie Nov 12 '15 at 1:26
  • $\begingroup$ See the answer given by WA. $\endgroup$ – lhf Nov 12 '15 at 1:38
  • $\begingroup$ Related and related. $\endgroup$ – Cameron Buie Nov 12 '15 at 1:48
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Instead of proceeding in terms of adjoints, it would be a great deal easier to proceed as follows.

First of all, note that each of $a,c,f$ is nonzero. Next, consider the augmented matrix $[A\mid I_3],$ where $I_3$ is the $3\times3$ identity matrix. Perform row operations on the augmented matrix until it is in the form $[I_3\mid B].$ The matrix $B$ is the inverse of $A,$ which you can (and should) verify by showing that $AB=I_3.$

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Your work is fine but you forgot to transpose the adjoint!

The inverse is essentially given by the adjugate matrix, which is the transpose of the cofactor matrix.

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