Using the property of determinants and without expanding- beginvmatrixxx a - x-ay - b - y-b lz - c - z-c vmatrix -0

Let A= x amp;a amp;x+a y amp;b amp;y+b z amp;c amp; z+c Applying C_3→ C_3 C_1, we get A= x amp; a amp;a y amp;b amp;b ...

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Standard XII

Mathematics

Applications of Cross Product

Using the pro...

Question

Using the property of determinants and without expanding.
$∣ ∣ ∣ ∣ \begin{matrix} x & a & x + a y & b & y + b z & c & z + c \end{matrix} ∣ ∣ ∣ ∣ = 0$

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Solution

Let $A = ∣ ∣ ∣ ∣ \begin{matrix} x & a & x + a y & b & y + b z & c & z + c \end{matrix} ∣ ∣ ∣ ∣$
Applying $C_{3} \to C_{3} - C_{1}$ , we get $A = ∣ ∣ ∣ ∣ \begin{matrix} x & a & a y & b & b z & c & c \end{matrix} ∣ ∣ ∣ ∣ = 0$
[Since, the two columns $(C_{2} = C_{3})$ of the determinants are identical]

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Q. Using the property of determinants and without expanding, prove that

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Using the property of determinants and without expanding. ∣∣ ∣∣xax+ayby+bzcz+c∣∣ ∣∣=0

Let A=∣∣ ∣∣xax+ayby+bzcz+c∣∣ ∣∣ Applying C3→C3−C1, we get A=∣∣ ∣∣xaaybbzcc∣∣ ∣∣=0 [Since, the two columns (C2=C3) of the determinants are identical]

Using the property of determinants and without expanding.
$∣ ∣ ∣ ∣ \begin{matrix} x & a & x + a y & b & y + b z & c & z + c \end{matrix} ∣ ∣ ∣ ∣ = 0$