Answer :

**To prove:** is divisible by (x + y + z)

**To find:** the quotient

Apply C_{1}→ C_{1} – C_{2} and C_{2}→ C_{2} – C_{3}

Taking common (x + y + z) from C_{1} and C_{2}

Apply R_{1}→ R_{1} + R_{2} + R_{3}

Expanding the determinant

= (x + y + z)^{2} (xy – z^{2} + yz – x^{2} + zx – y^{2}){(z – y)(y – x) – (x – z)^{2}}

= (x + y + z)^{2} (xy – z^{2} + yz – x^{2} + zx – y^{2}){(zy – y^{2} + xy – xz) – (x^{2} + z^{2} – 2xz)}

= (x + y + z)^{2} (xy – z^{2} + yz – x^{2} + zx – y^{2})(zy – y^{2} + xy – xz – x^{2} – z^{2} + 2xz)

= (x + y + z)^{2} (zy + xy + xz – x^{2} – y^{2} – z^{2})^{2}

Hence the given determinant is divisible by (x + y + z) and quotient is

**OR**

**Given:** System of equations:

8x + 4y + 3z = 19, 2x + y + z = 5, x + 2y + 2z = 7

**To find:** Solution of the system of the equations i.e. values of x, y and z which satisfy these equations

We know,

A = I.A where I is an identity matrix

Applying R_{1} R_{3}

Applying R_{2} → R_{2} – 2R_{1} and R_{3} → R_{3} – 8R_{1}

Applying R_{1} → R_{1} – 2R_{2} and R_{3} → R_{3} + 12R_{2}

Applying R_{3} → -R_{3} and R_{2} → R_{2} – R_{3}

To solve these equation and get values of x, y and z, we have:

AX = B where,

AX = B

⇒ X = A^{-1} B

Hence, solutions of the equations are **x = 1, y = 2, z = 1**

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