Solve:

(i) x+y-2z =0 (ii)2x+3y+4z =0 (iii)3x+y+z =0 (iv) x+2y-3z = -4

2x+y-3z =0 x+y+z =0 x-4y+3z =02x+3y+2z =2

5x+4y-9z =0 2x-y+3z =0 2x+5y-2z =0 3x-3y-4z =11

(1) Given homogeneous equations are –

Let AX = 0 be a homogeneous system of *n* linear equations with *n* unknowns.

Where

Applying *c*_{2} → *c*_{2 }– *c*_{1} and *c*_{3} → *c*_{3 }+ 2 *c*_{1} , we get

= 1 (–1 + 1) = 0

i.e. Matrix A is singular.

Then, the system has infinitely many solutions.

Put *z* = *k* (any real number) and solve first two equations for *x* and *y*.

*x* + *y* = 2*k* ..... (4)

and 2*x* + *y* = 3*k* ..... (5)

or AX = B

So, A^{–1} will exist.

Now put *x* = *y* = *z* = *k* in (3), we get –

5*k* + 4*k* – 9*k* = 0

⇒ 0 = 0

which is true.

Hence, *x* = *k , *y = *k ,* *z* = *k *where *k *is any real number satisfy the given system of equations

**Q4 **Given system of equations is –

The given system can be written as –

AX = B

Since is non-singular and hence a unique solution given by

X = A^{–1} B will exist.

∴ X = A^{–1} B

⇒ *x* = 3, *y* = –2, *z* = 1

(2) and (3) queries can be solved in the same way. If you face any problem, please do get back to us.

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