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Byju's Answer
Standard IX
Mathematics
(x±a)(x±b) = x²±(a+b)x+ab.
Write the val...
Question
Write the value of the determinant
x
+
y
y
+
z
z
+
x
z
x
y
-
3
-
3
-
3
.
Open in App
Solution
x
+
y
y
+
z
z
+
x
z
x
y
-
3
-
3
-
3
=
x
+
y
+
z
x
+
y
+
z
z
+
x
+
y
z
x
y
-
3
-
3
-
3
Applying
R
1
→
R
1
+
R
2
=
x
+
y
+
z
1
1
1
z
x
y
-
3
-
3
-
3
Taking
x
+
y
+
z
common
from
R
1
=
x
+
y
+
z
1
1
1
z
x
y
-
3
-
3
-
3
Applying
R
3
→
R
3
+
3
R
1
=
x
+
y
+
z
1
1
1
z
x
y
0
0
0
=
0
Expanding
along
the
last
row
Hence, the value of the determinant
x
+
y
y
+
z
z
+
x
z
x
y
-
3
-
3
-
3
is 0.
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4
Similar questions
Q.
Write the value of
△
=
∣
∣ ∣
∣
x
+
y
y
+
z
z
+
x
z
x
y
−
3
−
3
−
3
∣
∣ ∣
∣
.
Q.
Using the properties of determinants, show that:
∣
∣ ∣ ∣
∣
x
x
2
y
z
y
y
2
z
x
z
z
2
x
y
∣
∣ ∣ ∣
∣
=
(
x
−
y
)
(
y
−
z
)
(
z
−
x
)
(
x
y
+
y
z
+
z
x
)
Q.
If
∣
∣ ∣ ∣
∣
x
k
x
k
+
2
x
k
+
3
y
k
y
k
+
2
y
k
+
3
z
k
z
k
+
2
z
k
+
3
∣
∣ ∣ ∣
∣
=
(
x
−
y
)
(
y
−
z
)
(
z
−
x
)
(
1
x
+
1
y
+
1
z
)
then find the value of
−
k
.
Q.
Using properties of determinants, prove that
ω
∣
∣ ∣ ∣
∣
x
y
z
x
2
y
2
z
2
y
+
z
z
+
x
x
+
y
∣
∣ ∣ ∣
∣
=
(
x
−
y
)
(
y
−
z
)
(
z
−
x
)
(
x
+
y
+
z
)
Q.
Using properties of determinant, prove
that:
∣
∣ ∣ ∣
∣
x
x
2
1
+
p
x
3
y
y
2
1
+
p
y
3
z
z
2
1
+
p
z
3
∣
∣ ∣ ∣
∣
=
(
1
+
p
x
y
z
)
(
x
−
y
)
(
y
−
z
)
(
z
−
x
)
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(x±a)(x±b) = x²±(a+b)x+ab.
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