If | beginmatrix x y z -x y z x -y z endmatrix |=kxyz, then k =

Question

If | beginmatrix x y z -x y z x -y z endmatrix |=kxyz, then k = (A)  1  (B)  3  (C)  4  (D)  2

Tardigrade Admin · Accepted Answer

| beginmatrix x y z -x y z x -y z endmatrix | =| beginmatrix x y z -x y z 0 0 2z endmatrix | (R3→ R3+R2) Expanding along R3, we get =2z(xy+xy)=4xyz On comparing with kxyz, we get k=4

Q. If $∣ ∣ x - x x y y - y z z z ∣ ∣ = k x yz,$ then $k$ =

$1$

$3$

$4$

$2$

$∣ ∣ x - x x y y - y z z z ∣ ∣$
$= ∣ ∣ x - x 0 y y 0 z z 2 z ∣ ∣$
$(R_{3} \to R_{3} + R_{2})$
Expanding along $R_{3},$
we get $= 2 z (x y + x y) = 4 x yz$
On comparing with kxyz, we get $k = 4$

Q. If ∣∣​x−xx​yy−y​zzz​∣∣​=kxyz, then k =

1

3

4

2

∣∣​x−xx​yy−y​zzz​∣∣​ =∣∣​x−x0​yy0​zz2z​∣∣​ (R3​→R3​+R2​) Expanding along R3​, we get =2z(xy+xy)=4xyz On comparing with kxyz, we get k=4

Q. If $∣ ∣ x - x x y y - y z z z ∣ ∣ = k x yz,$ then $k$ =

$1$

$3$

$4$

$2$

$∣ ∣ x - x x y y - y z z z ∣ ∣$
$= ∣ ∣ x - x 0 y y 0 z z 2 z ∣ ∣$
$(R_{3} \to R_{3} + R_{2})$
Expanding along $R_{3},$
we get $= 2 z (x y + x y) = 4 x yz$
On comparing with kxyz, we get $k = 4$