<PRE><B>Question 64626</B>
I&#39;m using Gauss-Jordan elimination method. You could follow these steps to produce your inverse matrix.

The augmented matrix is shown below, in sets of 4 equations, or rows.
The row operations, following a set of 4 eqns, apply to that set of eqns, with the result being in the following set of 4 eqns.

-2  6 -3  3 | -5 ---(1)
 4 -3  2 -1 | -3 ---(2)
-3 -4 -3  4 |-15 ---(3)
 5  2 -2  3 |-20 ---(4)

(3)x2, (4)x2

-2  6 -3  3 | -5 ---(5)
 4 -3  2 -1 | -3 ---(6)
-6 -8 -6  8 |-30 ---(7)
10  4 -4  6 |-40 ---(8)

(6) + 2x(5), (7) - 3x(5), (8) + 5x(5)

-2  6 -3  3 | -5 ---(9)
 0  9 -4  5 |-13 ---(10)
 0-26  3 -1 |-15 ---(11)
 0 34-19  21|-65 ---(12)

(12) + (11), (11) + 3x(10)

-2  6 -3  3 | -5 ---(13)
 0  9 -4  5 |-13 ---(14)
 0  1 -9 14 |-54 ---(15)
 0  8-16 20 |-80 ---(16)

(14) - 9x(15), (16) - 8x(15)

-2  6 -3  3 | -5 ---(17)
 0  9 77-121|473 ---(18)
 0  1 -9 14 |-54 ---(19)
 0  0 56-92 |352 ---(20)

(18) - (20), (20)÷4

-2  6 -3  3 | -5 ---(21)
 0  0 21-29 |473 ---(22)
 0  1 -9 14 |-54 ---(23)
 0  0 14-23 | 88 ---(24)

(22) - (24)

-2  6 -3  3 | -5 ---(25)
 0  0  7 -6 | 33 ---(26)
 0  1 -9 14 |-54 ---(27)
 0  0 14-23 | 88 ---(28)

(28) - 2x(26)

-2  6 -3  3 | -5 ---(29)
 0  0  7 -6 | 33 ---(30)
 0  1 -9 14 |-54 ---(31)
 0  0  0-11 | 22 ---(32)

(32)÷ -11,

-2  6 -3  3 | -5 ---(33)
 0  0  7 -6 | 33 ---(34)
 0  1 -9 14 |-54 ---(35)
 0  0  0  1 |  2 ---(36)

(33) - 3x(36), (34) + 6x(36), (35) - 14x(36)

-2  6 -3  0 |  1 ---(37)
 0  0  7  0 | 21 ---(38)
 0  1 -9  0 |-26 ---(39)
 0  0  0  1 |  2 ---(40)

(38)÷7,

-2  6 -3  0 |  1 ---(41)
 0  0  1  0 |  3 ---(42)
 0  1 -9  0 |-26 ---(43)
 0  0  0  1 |  2 ---(44)

(41) + 3x(42), (43) + 9x(42)

-2  6  0  0 | 10 ---(45)
 0  0  1  0 |  3 ---(46)
 0  1  0  0 |  1 ---(47)
 0  0  0  1 |  2 ---(48)

(45) - 6x(47)

-2  0  0  0 |  4 ---(49)
 0  0  1  0 |  3 ---(50)
 0  1  0  0 |  1 ---(51)
 0  0  0  1 |  2 ---(52)

(49)÷ -2

 1  0  0  0 | -2 ---(53)
 0  0  1  0 |  3 ---(54)
 0  1  0  0 |  1 ---(55)
 0  0  0  1 |  2 ---(56)

Rearrange,

 1  0  0  0 | -2 ---(57)
 0  1  0  0 |  1 ---(58)
 0  0  1  0 |  3 ---(59)
 0  0  0  1 |  2 ---(60)

The solution can now be read off as (-2, 1, 3, -2)
==================================================
The inverse matrix is,
|-29 -12 -9 3|
|33 22 -11 -11|
|181 256 19 -121|
|147 196 35 -77|
with a coefficient of (1/154).</PRE>