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Joint Entrance Examination
Mathematics
Matrices and Determinants
Question 69
Question 69 of 472
Medium
MCQ
Let
A
A
A
be a
3
×
3
3\times 3
3
×
3
real matrix such that
A
(
(
∣
1
0
1
)
)
∣
=
2
(
(
∣
1
0
1
)
)
∣
,
A
(
(
∣
−
1
0
1
)
)
∣
=
4
(
(
∣
−
1
0
1
)
)
∣
,
A
(
(
∣
0
1
0
)
)
∣
=
2
(
(
∣
0
1
0
)
)
∣
A((|\begin{gathered} 1 \\ 0 \\ 1 \end{gathered}))|=2((|\begin{gathered} 1 \\ 0 \\ 1 \end{gathered}))|,A((|\begin{gathered} -1 \\ 0 \\ 1 \end{gathered}))|=4((|\begin{gathered} -1 \\ 0 \\ 1 \end{gathered}))|,A((|\begin{gathered} 0 \\ 1 \\ 0 \end{gathered}))|=2((|\begin{gathered} 0 \\ 1 \\ 0 \end{gathered}))|
A
((
∣
1
0
1
))
∣
=
2
((
∣
1
0
1
))
∣
,
A
((
∣
−
1
0
1
))
∣
=
4
((
∣
−
1
0
1
))
∣
,
A
((
∣
0
1
0
))
∣
=
2
((
∣
0
1
0
))
∣
Then, the system
(
A
−
3
I
)
(
(
∣
x
y
z
)
)
∣
=
(
(
∣
1
2
3
)
)
∣
(A-3I)((|\begin{gathered} x \\ y \\ z \end{gathered}))|=((|\begin{gathered} 1 \\ 2 \\ 3 \end{gathered}))|
(
A
−
3
I
)
((
∣
x
y
z
))
∣
=
((
∣
1
2
3
))
∣
has :
A
exactly two solutions
B
infinitely many solutions
C
unique solution
D
no solution
Check answer
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Solutions · 1
Post your solution
Official solution
A
=
[
[
∣
x
1
y
1
z
1
x
2
y
2
z
2
x
3
y
3
z
3
]
]
∣
A=[[|\begin{gathered} {x}_{1} \quad {y}_{1} \quad {z}_{1} \\ {x}_{2} \quad {y}_{2} \quad {z}_{2} \\ {x}_{3} \quad {y}_{3} \quad {z}_{3} \end{gathered}]]|
A
=
[[
∣
x
1
y
1
z
1
x
2
y
2
z
2
x
3
y
3
z
3
]]
∣
A
[
[
∣
1
0
1
]
]
∣
=
[
[
∣
2
0
2
]
]
∣
A[[|\begin{gathered} 1 \\ 0 \\ 1 \end{gathered}]]|=[[|\begin{gathered} 2 \\ 0 \\ 2 \end{gathered}]]|\,
A
[[
∣
1
0
1
]]
∣
=
[[
∣
2
0
2
]]
∣
∴
[
[
∣
x
1
+
z
1
x
2
+
z
2
x
3
+
z
3
]
]
∣
=
[
[
∣
2
0
2
]
]
∣
\therefore [[|\begin{gathered} {x}_{1}+{z}_{1} \\ {x}_{2}+{z}_{2} \\ {x}_{3}+{z}_{3} \end{gathered}]]|=[[|\begin{gathered} 2 \\ 0 \\ 2 \end{gathered}]]|
∴
[[
∣
x
1
+
z
1
x
2
+
z
2
x
3
+
z
3
]]
∣
=
[[
∣
2
0
2
]]
∣
∴
x
1
+
z
1
=
2
x
2
+
z
2
=
0
x
3
+
z
3
=
0
\begin{gathered} \therefore {x}_{1}+{z}_{1} =2\, \\ {x}_{2}+{z}_{2} =0\, \\ {x}_{3}+{z}_{3} =0\, \end{gathered}
∴
x
1
+
z
1
=
2
x
2
+
z
2
=
0
x
3
+
z
3
=
0
A
[
[
∣
−
1
0
1
]
]
∣
=
[
[
∣
−
4
0
4
]
]
∣
∴
[
[
∣
−
x
1
+
z
1
−
x
2
+
z
2
−
x
3
+
z
3
]
]
∣
=
[
[
∣
4
0
4
]
]
∣
\begin{gathered} A[[|\begin{array}{|c|}-1 \\ 0 \\ 1\end{array}]]|=[[|\begin{array}{|c|}-4 \\ 0 \\ 4\end{array}]]| \\ \therefore [[|\begin{array}{|c|}-{x}_{1}+{z}_{1} \\ -{x}_{2}+{z}_{2} \\ -{x}_{3}+{z}_{3}\end{array}]]|=[[|\begin{array}{|c|}4 \\ 0 \\ 4\end{array}]]| \end{gathered}
A
[[
∣
−
1
0
1
]]
∣
=
[[
∣
−
4
0
4
]]
∣
∴
[[
∣
−
x
1
+
z
1
−
x
2
+
z
2
−
x
3
+
z
3
]]
∣
=
[[
∣
4
0
4
]]
∣
⇒
−
x
1
+
z
1
=
−
4
−
x
2
+
x
2
=
0
−
x
3
+
z
3
=
4
\begin{gathered} \Rightarrow -{x}_{1}+{z}_{1}=-4\, \\ -{x}_{2}+{x}_{2}=0\, \\ -{x}_{3}+{z}_{3}=4 \end{gathered}
⇒
−
x
1
+
z
1
=
−
4
−
x
2
+
x
2
=
0
−
x
3
+
z
3
=
4
A
[
[
∣
0
1
0
]
]
∣
=
[
[
∣
0
2
0
]
]
∣
∴
[
[
∣
y
1
y
2
y
3
]
]
∣
=
[
[
∣
0
2
0
]
]
∣
\begin{gathered} A[[|\begin{array}{|c|}0 \\ 1 \\ 0\end{array}]]|=[[|\begin{array}{|c|}0 \\ 2 \\ 0\end{array}]]| \\ \therefore [[|\begin{array}{|c|}{y}_{1} \\ {y}_{2} \\ {y}_{3}\end{array}]]|=[[|\begin{array}{|c|}0 \\ 2 \\ 0\end{array}]]| \end{gathered}
A
[[
∣
0
1
0
]]
∣
=
[[
∣
0
2
0
]]
∣
∴
[[
∣
y
1
y
2
y
3
]]
∣
=
[[
∣
0
2
0
]]
∣
∴
y
1
=
0
,
y
2
=
2
,
y
3
=
0
∴
(
2
)
,
(
3
)
,
(
4
)
,
(
5
)
,
(
6
)
(
7
)
x
1
=
3
x
,
x
2
=
0
,
x
3
=
−
1
y
1
=
0
,
y
2
=
2
,
y
3
=
0
z
1
=
−
1
,
z
2
=
0
,
z
3
=
3
\begin{gathered} \therefore {y}_{1}=0,{y}_{2}=2,{y}_{3}=0 \\ \therefore (2),(3),(4),(5),(6)(7) \\ {x}_{1}=3x,{x}_{2}=0,{x}_{3}=-1 \\ {y}_{1}=0,{y}_{2}=2,{y}_{3}=0 \\ {z}_{1}=-1,{z}_{2}=0,{z}_{3}=3 \end{gathered}
∴
y
1
=
0
,
y
2
=
2
,
y
3
=
0
∴
(
2
)
,
(
3
)
,
(
4
)
,
(
5
)
,
(
6
)
(
7
)
x
1
=
3
x
,
x
2
=
0
,
x
3
=
−
1
y
1
=
0
,
y
2
=
2
,
y
3
=
0
z
1
=
−
1
,
z
2
=
0
,
z
3
=
3
∴
A
=
[
[
∣
3
0
−
1
0
2
0
−
1
0
3
]
]
∣
∴
(
A
−
31
)
[
[
∣
x
y
z
]
]
∣
=
[
[
∣
−
1
2
3
]
]
∣
∴
[
[
∣
0
0
−
1
0
−
1
0
−
1
0
0
]
]
∣
[
[
∣
x
y
z
]
]
∣
=
[
[
∣
−
1
2
3
]
]
∣
[
[
∣
−
z
−
y
−
x
]
]
∣
=
[
[
∣
1
2
3
]
]
∣
[
z
=
−
1
]
,
[
y
=
−
2
]
,
[
x
=
−
3
]
\begin{gathered} \therefore A=[[|\begin{array}{|c|}3 \quad 0 \quad -1 \\ 0 \quad 2 \quad 0 \\ -1 \quad 0 \quad 3\end{array}]]| \\ \therefore (A-31)[[|\begin{array}{|c|}x \\ y \\ z\end{array}]]|=[[|\begin{array}{|c|}-1 \\ 2 \\ 3\end{array}]]| \\ \therefore [[|\begin{array}{|c|}0 \quad 0 \quad -1 \\ 0 \quad -1 \quad 0 \\ -1 \quad 0 \quad 0\end{array}]]|[[|\begin{array}{|c|}x \\ y \\ z\end{array}]]|=[[|\begin{array}{|c|}-1 \\ 2 \\ 3\end{array}]]| \\ [[|\begin{array}{|c|}-z \\ -y \\ -x\end{array}]]|=[[|\begin{array}{|c|}1 \\ 2 \\ 3\end{array}]]| \\ [z=-1],[y=-2],[x=-3] \end{gathered}
∴
A
=
[[
∣
3
0
−
1
0
2
0
−
1
0
3
]]
∣
∴
(
A
−
31
)
[[
∣
x
y
z
]]
∣
=
[[
∣
−
1
2
3
]]
∣
∴
[[
∣
0
0
−
1
0
−
1
0
−
1
0
0
]]
∣
[[
∣
x
y
z
]]
∣
=
[[
∣
−
1
2
3
]]
∣
[[
∣
−
z
−
y
−
x
]]
∣
=
[[
∣
1
2
3
]]
∣
[
z
=
−
1
]
,
[
y
=
−
2
]
,
[
x
=
−
3
]
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