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Joint Entrance Examination
Mathematics
Matrices and Determinants
Question 298
Question 298 of 472
Medium
MCQ
If
A
=
(
1
2
x
3
−
1
2
)
A=\begin{pmatrix}1&2&x\\3&-1&2\end{pmatrix}
A
=
(
1
3
2
−
1
x
2
)
and
B
=
(
y
x
1
)
B=\begin{pmatrix}y\\x\\1\end{pmatrix}
B
=
y
x
1
be such that
A
B
=
(
6
8
)
AB=\begin{pmatrix}6\\8\end{pmatrix}
A
B
=
(
6
8
)
, then:
A
y
=
2
x
y=2x
y
=
2
x
B
y
=
−
2
x
y=-2x
y
=
−
2
x
C
y
=
x
y=x
y
=
x
D
y
=
−
x
y=-x
y
=
−
x
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Solutions · 1
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Official solution
A
B
=
(
y
+
2
x
+
x
3
y
−
x
+
2
)
=
(
y
+
3
x
3
y
−
x
+
2
)
=
(
6
8
)
AB=\begin{pmatrix}y+2x+x\\3y-x+2\end{pmatrix}=\begin{pmatrix}y+3x\\3y-x+2\end{pmatrix}=\begin{pmatrix}6\\8\end{pmatrix}
A
B
=
(
y
+
2
x
+
x
3
y
−
x
+
2
)
=
(
y
+
3
x
3
y
−
x
+
2
)
=
(
6
8
)
. So
y
+
3
x
=
6
y+3x=6
y
+
3
x
=
6
and
3
y
−
x
=
6
3y-x=6
3
y
−
x
=
6
. Solving:
x
=
6
5
,
y
=
12
5
x=\dfrac65,\ y=\dfrac{12}5
x
=
5
6
,
y
=
5
12
, giving
y
=
2
x
y=2x
y
=
2
x
.
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