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Mathematics Question on Determinants

AA is a 3×33 \times 3 matrix where its first row is (1 0 0)(1 \,0 \,0) , second row is (2 1 0)(2 \,1 \,0) and third row is (3 2 1).P,Q(3 \,2 \,1). P, Q and RR are column matrices such that AP=(1 0 0)T,AQ=(2 3 0)TAP = (1 \,0 \,0)^T, AQ = (2 \,3 \,0)^T and AR=(0 0 1)TAR = (0\, 0\, 1)^T . If P,QP, Q and RR are three columns of matrix UU , then ∣U∣=| U| =

A

00

B

11

C

33

D

99

Answer

33

Explanation

Solution

We have, A=[100 210 321]A=\left[\begin{matrix}1&0&0\\\ 2&1&0\\\ 3&2&1\end{matrix}\right]
Let P=[x1 y1 z1]P=\left[\begin{matrix}x_{1}\\\ y_{1}\\\ z_{1}\end{matrix}\right], Q=[x2 y2 z2]Q=\left[\begin{matrix}x^{2}\\\ y^{2}\\\ z^{2}\end{matrix}\right]
and R=[x3 y3 z3]R=\left[\begin{matrix}x_{3}\\\ y_{3}\\\ z_{3}\end{matrix}\right]
Now, AP=[x1 2x1+y1 3x1+2y1+z1]=[1 0 0]AP=\left[\begin{matrix}x_{1}\\\ 2x_{1}+y_{1}\\\ 3x_{1}+2y_{1} +z_{1}\end{matrix}\right]=\left[\begin{matrix}1\\\ 0\\\ 0\end{matrix}\right]
⇒x1=1,y1=−2\Rightarrow x_{1}=1, y_{1}=-2 and z1=1z_{1}=1
Again, AQ=[x2 2x2+y2 3x22y2+z2]=[2 3 0]AQ=\left[\begin{matrix}x_{2}\\\ 2x_{2}+y_{2}\\\ 3x_{2} 2y_{2} +z_{2}\end{matrix}\right]=\left[\begin{matrix}2\\\ 3\\\ 0\end{matrix}\right]
⇒x2=2\Rightarrow x_{2}=2,
y2=−1y_{2}=-1
and z2=−4z_{2}=-4
∴AR=[x3 2x3+y3 3x3+2v3+z3]=[0 0 1]\therefore AR=\left[\begin{matrix}x_{3}\\\ 2x_{3} +y_{3}\\\ 3x_{3}+2v_{3} +z_{3}\end{matrix}\right]=\left[\begin{matrix}0\\\ 0\\\ 1\end{matrix}\right]
⇒x3=0,y3=0\Rightarrow x_{3}=0, y_{3}=0 and z3=1z_{3}=1
So, U=[120 −2−10 1−41]U=\left[\begin{matrix}1&2&0\\\ -2&-1&0\\\ 1&-4&1\end{matrix}\right]
∣U∣=1(−1+4)\left|U\right|=1\left(-1+4\right)
=3=3