Using properties of determinants- prove that 111-3 x 1-3 y 1111-3 z 1-93 x y z-x y-y z-z x-

Let #160; #8710;=111+3x1+3y1111+3z1Applying #160;R2 #8594;R2 #160; #160;R1, #160; #160;R3 #8594;R3 #160; #160;R1 #160; #8658; #8 ...

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Standard XII

Mathematics

Change of Variables

Using propert...

Question

Using properties of determinants, prove that $|\begin{matrix} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{matrix}| = 9 (3 x y z + x y + y z + z x) .$

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Solution

$Let ∆ = |\begin{matrix} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{matrix}| Applying R_{2} \to R_{2} - R_{1,} R_{3} \to R_{3} - R_{1} \Rightarrow ∆ = |\begin{matrix} 1 & 1 & 1 + 3 x \\ 3 y & 0 & - 3 x \\ 0 & 3 z & - 3 x \end{matrix}|$
Expanding along R₁,we get
$\begin{array}{rcl} ∆ & = & 1 (0 + 9 x z) - 1 (- 9 x y - 0) + (1 + 3 x) (9 y z - 0) \\ = & 9 x z + 9 x y + 9 y z + 27 x y z \\ = & 9 (3 x y z + x y + y z + z x) \end{array}$

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Usng properties of determinants , prove that $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 + 3 x 1 + 3 y & 1 & 1 1 & 1 + 3 z & 1 \end{matrix} ∣ ∣ ∣ ∣ = 9 (3 x y z + x y + y z + z x)$

Q. Using properties of determinants, prove that

∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 + 3 x 1 + 3 y & 1 & 1 1 & 1 + 3 z & 1 \end{matrix} ∣ ∣ ∣ ∣ = 9 (3 x y z + x y + y z + z x)

Using the properties of determinants, show that:

∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & y z y & y^{2} & z x z & z^{2} & x y \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (x y + y z + z x)

Q. Using properties of determinants, prove that

∣ ∣ ∣ ∣ ∣ \begin{matrix} (x + y)^{2} & z x & z y z x & (z + y)^{2} & x y z y & x y & (z + x)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 2 x y z (x + y + z)^{3}

Q. By using properties of determinants prove that

∣ ∣ ∣ ∣ ∣ \begin{matrix} (y + z)^{2} & x y & z x x y & (x + z)^{2} & y z x z & y z & (x + y)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

is divisible by

(x + y + z)^{n}

. Find

n

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Using properties of determinants, prove that 111+3x1+3y1111+3z1=93xyz+xy+yz+zx.

Let ∆=111+3x1+3y1111+3z1Applying R2→R2 - R1, R3→R3 - R1 ⇒∆=111+3x3y0-3x03z-3x Expanding along R1 ,we get ∆=1(0+9xz) - 1(-9xy-0)+(1+3x)(9yz-0)= 9xz+9xy+9yz+27xyz= 93xyz +xy +yz+zx

Using properties of determinants, prove that $|\begin{matrix} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{matrix}| = 9 (3 x y z + x y + y z + z x) .$