Question

$Find space the space matrix space x space so space that space x space space open square brackets table row 1 2 3 row 4 5 6 end table close square brackets equals space open square brackets table row cell negative 7 end cell cell negative 8 end cell cell negative 9 end cell row 2 4 6 end table close square brackets$

Solution

$space We space are space given space that x space open square brackets table row 1 2 3 row 4 5 6 end table close square brackets equals open square brackets table row cell negative 7 end cell cell negative 8 end cell cell negative 9 end cell row 2 4 6 end table close square brackets Where space x space is space straight a space matrix.$
Let x be of m X n type.
Since number of columns of x = number of rows of
$open square brackets table row 1 2 3 row 4 5 6 end table close square brackets therefore space straight n equals 2 open square brackets table row 1 2 3 row 4 5 6 end table close square brackets space is space space of space 2 space straight x space 3 space type x space open square brackets table row 1 2 3 row 4 5 6 end table close square brackets space space must space be space of space type space straight m space straight X space 3. therefore space straight m equals 2 therefore space straight X space is space straight a space square space matrix space of space type space 2 straight x 2. Let space space space space space space space space space straight X equals open square brackets table row straight a straight b row straight c straight d end table close square brackets Now space straight X space open square brackets table row 1 2 3 row 4 5 6 end table close square brackets equals open square brackets table row cell negative 7 end cell cell negative 8 end cell cell negative 9 end cell row cell negative 2 end cell 4 6 end table close square brackets rightwards double arrow open square brackets table row straight a straight b row straight c straight d end table close square brackets space open square brackets table row 1 2 3 row 4 5 6 end table close square brackets equals open square brackets table row cell negative 7 end cell cell negative 8 end cell cell negative 9 end cell row cell negative 2 end cell 4 6 end table close square brackets rightwards double arrow open square brackets table row cell straight a plus 4 straight b end cell cell 2 straight a plus 5 straight b end cell cell 3 straight a plus 6 straight b end cell row cell straight c plus 4 straight d end cell cell 2 straight c plus 5 straight d end cell cell 3 straight c plus 6 straight d end cell end table close square brackets equals open square brackets table row cell negative 7 end cell cell negative 8 end cell cell negative 9 end cell row cell negative 2 end cell 4 6 end table close square brackets$

By the definition of equality of matrices,
a + 4b = –7    ...(1)
2a + 5 b =–8    ..(2)
3a + 6 b = –9    –(3)
c + 4 d = 2 ......(4)
2c + 5 d = 4        (5)
3 c + 6 d = 6    ....(6)
Multiplying (1) by 2 and (2) by 1, we get,
2a + 8 b = –14    ...(7)
2a + 5 b = –8     .(8)
Subtracting (8) from (7), we get,
3b = –6 or b = – 2
∴ from (1), a - 8 = – 7 or a = 1
Now a = 1, b = -2 also satisfy (3)
∴ we have a= 1, b = – 2
Multiplying (4) by 2 and (5) by 1, we get
2 c + 8 d = 4    (9)
2;c + 5 d = 4    ...(10)
Subtracting (10) from (9), we get
3d = 0 or d = 0
∴from (4), c + 0 = 2 or c = 2
Now c = 2, d = 0 satisfy (6)

∴ we have c = 2, d = 0

$therefore space straight X equals open square brackets table row 1 cell negative 2 end cell row 2 0 end table close square brackets$

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Matrices

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