[SOLVE] Matrix problem

**lin.13579** · Aug 31st 2009, 07:48 PM

Let A =
$<br /> \left|\begin{array}{cccc}2 & 1 & 1 \\ 3 & 2 & 1 \\ 1 & 1 & 1 \end{array}\right|<br />$

(a) Calculate det(A).
(b) Use the Inversion Algorithm to compute A^(−1).
(c) Consider a new matrix B which is the same as A except the 2 in the upper left hand corner of A is replaced by a 1. What do you say about the inverse of this new matrix?

Is anyone can explain how to do (b) and (c) please!!! !!!!!!!

My answer:
(a) (2*2*1)+(1*1*1)+(1*3*1)-(1*2*1)-(2*1*1)-(1*3*1) = 1
(b) 
(c)

**lin.13579** · Sep 2nd 2009, 04:14 PM

Let A =
$<br /> \left|\begin{array}{cccc}2 & 1 & 1 \\ 3 & 2 & 1 \\ 1 & 1 & 1 \end{array}\right|<br />$

(a) Calculate det(A).
(b) Use the Inversion Algorithm to compute A^(−1).
(c) Consider a new matrix B which is the same as A except the 2 in the upper left hand corner of A is replaced by a 1. What do you say about the inverse of this new matrix?

Is anyone can teach me how to do (b) and (c) please!!! !!!!!!!

My answer:
(a) (2*2*1)+(1*1*1)+(1*3*1)-(1*2*1)-(2*1*1)-(1*3*1) = 1
(b) 
(c)

**DJ Hobo** · Sep 3rd 2009, 12:51 AM

Well, since this has been up for a few days, and nobody else is helping, I thought I'd pitch in and add my thoughts.. I can't help with part (b), since I'm unfamiliar with the "Inversion Algorithm". However, part (c) is pretty straight forward.

Use the same method for finding the determinant that you used in part (a), except this time, replace element $A_{11}$ (the two in the upper-left corner) with a 1. You'll find that the determinant is zero. A determinant of zero means that the matrix is "singular", and therefore has no inverse. I'm sure you understand (or can figure out) why that is.

**quah13579** · Sep 3rd 2009, 05:29 PM

Originally Posted by DJ Hobo

Well, since this has been up for a few days, and nobody else is helping, I thought I'd pitch in and add my thoughts.. I can't help with part (b), since I'm unfamiliar with the "Inversion Algorithm". However, part (c) is pretty straight forward.

Use the same method for finding the determinant that you used in part (a), except this time, replace element $A_{11}$ (the two in the upper-left corner) with a 1. You'll find that the determinant is zero. A determinant of zero means that the matrix is "singular", and therefore has no inverse. I'm sure you understand (or can figure out) why that is.

easy

**JeffN12345** · Sep 3rd 2009, 11:10 PM

the inverse of a three by three matrix $A=\left(\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{ array}\right)$
is the matrix $A^{-1}=\frac{1}{DetA}\left(\begin{array}{ccc}ei-fh,&fg-di,&dh-eg\\hc-bi,&ai-cg,&gb-ah\\bf-ec,&cd-af,&ae-bd\end{array}\right)$
note that the inverse of a A's element (i,j) is given by $A^{-1}_{i,j}=\frac{Det\left(\begin{array}{cc}A_{(i+1)m od3,(j+1)mod3}&A_{(i+1)mod3,(j+2)mod3}\\A_{(i+2)mo d3,(j+1)mod3}&A_{(i+2)mod3,(j+2)mod3}\end{array}\r ight)}{DetA}$
a very similar formula is possible for square matrices of any size

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