By using properties of deteminants- beginvmatrix 0 - a -b-a - 0 -cb - c - 0 vmatrix -0

Let A= 0 amp; a amp; b a amp; 0 amp; c b amp;c amp;0 =1/c 0 amp; ac amp; bc a amp; 0 amp; c b amp;c amp;0 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Scalar Multiplication of a Matrix

By using prop...

Question

By using properties of deteminants.

$∣ ∣ ∣ ∣ \begin{matrix} 0 & a & - b - a & 0 & - c b & c & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$

Open in App

Solution

Let $A = ∣ ∣ ∣ ∣ \begin{matrix} 0 & a & - b - a & 0 & - c b & c & 0 \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{c} ∣ ∣ ∣ ∣ \begin{matrix} 0 & a c & - b c - a & 0 & - c b & c & 0 \end{matrix} ∣ ∣ ∣ ∣$ (using $R_{1} \to c R_{1}$ )

$= \frac{1}{c} ∣ ∣ ∣ ∣ \begin{matrix} a b & a c & 0 - a & 0 & - c b & c & 0 \end{matrix} ∣ ∣ ∣ ∣$ (using $R_{1} \to R_{1} - b R_{2}$ )

$= \frac{a}{c} ∣ ∣ ∣ ∣ \begin{matrix} b & c & 0 - a & 0 & - c b & c & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$ [Since, the two row $R_{1}$ and $R_{2}$ are identical.]

Note: Suppose, if we multiply any row (or column) by any constant k, then we also divide that row (or column) by the same constant k and take common $\frac{1}{k}$ outside the determinant.

Suggest Corrections

Similar questions

Q. By using properties of determinants Find :

∣ ∣ ∣ ∣ \begin{matrix} 0 & a & - b - a & 0 & - c b & c & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0

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

Using the properties of determinants, find the value of

∣ ∣ ∣ ∣ \begin{matrix} 0 & a & - b - a & 0 & - c b & c & 0 \end{matrix} ∣ ∣ ∣ ∣

Q. Using properties of determinants, prove that

∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(a + b)^{2}}{c} & c & c a & \frac{(b + c)^{2}}{a} & a b & b & \frac{(c + a)^{2}}{b} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 2 (a + b + c)^{3}

OR

If

p \neq 0, q \neq 0

and

∣ ∣ ∣ ∣ \begin{matrix} p & q & p α + q q & r & p α + r p α + q & q α + r & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0

, then, using properties of determinants.
Prove that at least one of the following statements is true:

(a) p, q, r are in G.P.

(b)

α

is a root of the equation

p x^{2} + 2 q x + r = 0

Q. Using properties of deteminants, prove that

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{{(a + b)}^{2}}{c} & c & c a & \frac{{(b + c)}^{2}}{a} & a b & b & \frac{{(c + a)}^{2}}{b} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 2 {(a + b + c)}^{3}

Join BYJU'S Learning Program

By using properties of deteminants. ∣∣ ∣∣0a−b−a0−cbc0∣∣ ∣∣=0

By using properties of deteminants.

$∣ ∣ ∣ ∣ \begin{matrix} 0 & a & - b - a & 0 & - c b & c & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$