MATRICES - LINEAR EQUATIONS
----------------------------------------------

WANT FREE COACHING IN MATHS IN BANGALORE ?
CALL 9480193388 OR EMAIL freemaths10212@gmail.com

For details see post on 26-Feb-2015 at 6:00 am.

-----------------------------

2a)….
A=
1	1	-1	9
0	8	6	-6
-2	4	-6	40
NR3=R3+2R1
1	1	-1	9
0	8	6	-6
0	6	-8	58
NR1=R1-R2/8….NR2=R2/8….NR3=R3-6R2/8
1	0	-1.75	9.75
0	1	0.75	-0.75
0	0	-12.5	62.5
NR1=R1-1.75R3/12.5….NR2=R2+0.75R3/12.5…NR3=R3/-12.5
1	0	0	1
0	1	0	3
0	0	1	-5
HENCE THE SOLUTON IS
X=	1
Y=	3
Z=	-5
ANSWER
2b)….
A=
2	-2	4	0	0
-3	3	-6	5	15
1	-1	2	0	0
NR1=R1/2….NR2=R2+3R1/2….NR3=R3-R1/2
1	-1	2	0	0
0	0	0	5	15
0	0	0	0	0
NR2=R2/5
1	-1	2	0	0
0	0	0	1	3
0	0	0	0	0
LAST ROW IS ALL ZEROS ..SO L.D. & CONSISTENT EQNS.
X-Y+2Z=0……X=Y-2Z
T=3
WE HAVE 2 L.I. EQNS. IN 4 VARIABLES SAY X,Y,Z,T
HENCE 4-2 = 2 FREE VARIABLES…TAKING Y & Z AS FREE VARIABLES
THE SOLUTON SET IS IN GENERAL
X=	Y-2Z
Y=	Y
Z=	Z
T=	3
WHERE Y & Z CAN BE ANY REAL NUMBERS ….
SAY ….
X=	1		-2		-1
Y=	1	…..OR …..	0	…..OR …..	1	….ETC…
Z=	0		1		1
T=	3		3		3
ANSWER
3…....
A=
4	-2	6
-2	1	-3
RANK SHALL SATISFY 2 CONDITIONS …
1. THERE SHALL BE AT LEAST ONE NON ZERO DETERMINANT
THAT CAN BE OBTAINED FROM THE MATRIX BY SELECTING SOME ROWS/COLUMNS
2.THE ORDER OF SUCH DETERMINANT SHALL BE THE MAXIMUM POSSIBLE ORDER.
LET US CHECK ……NR1=R1+2R2
0	0	0
-2	1	-3
WE CAN HAVE A NON ZERO DETERMINANT OF ORDER 1 ONLY
SO RANK = 1
4…..
A=				A^2=
-10	-25	1		0	0	-36
4	10	1		0	0	13
0	0	-1		0	0	1
RANK OF A = 2 SINCE WE HAVE |Am|=-35 AS NON ZERO WHERE
Am=
-25	1
10	1
WHERE AS RANK A^2 = 1 , SINCE WE HAVE NO II ORDER NON ZERO DETERMINANT
NOW CONSIDER …
B=				B^2=
1	2	3		24	33	42
4	5	6		54	75	96
5	7	9		78	108	138
WE CAN SEE HERE THAT RANK OF B = 2 = RANK OF B^2
SO WE HAVE AN EXAMPLE WHERE ..
RANK OF A = RANK OF B = 2
BUT RANK OF A^2 = 1 IS NOT EQUAL TO RANK OF B^2=2
6….
IN A NON SQUARE MATRIX OF SAY M ROWS AND N COLUMNS
WE HAVE 2 CASES
CASE 1 ….. M < N
THAT IS THERE ARE MORE COLUMNS THAN ROWS …
THAT IS THERE ARE MORE THAN M COLUMN VECTORS
IN RM SPACE OF DIMENSION M ..
BUT WE CAN AT MOST HAVE HAVE M LINEARLY INDEPENDENT
COLUMN VECTORS IN RM OF DIMENSION M.
SO THE N COLUMN VECTORS ARE L.D.
THAT IS THE COLUMNS ARE L.D.
CASE 2 ……M > N
THAT IS THERE ARE MORE   ROWS THAN COLUMNS .. …
USING THE PROPERTY THAT |A| = | A TRANSPOSE|,
BY TAKING TRANSPOSE OF THIS MATRIX AND APPLYING THE
SAME LOGIC AS ABOVE , WE CAN PROVE THAT THE ROWS
WILL BE L.D.