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Mathematics Question on Matrices

Let AA be a 3×33 \times 3 real matrix such that**
A(1 0 1)=2(1 0 1),A(0 1 1)=4(−1 0 1),A(1 1 0)=2(1 1 0).A \begin{pmatrix} 1 \\\ 0 \\\ 1 \end{pmatrix} = 2 \begin{pmatrix} 1 \\\ 0 \\\ 1 \end{pmatrix}, \quad A \begin{pmatrix} 0 \\\ 1 \\\ 1 \end{pmatrix} = 4 \begin{pmatrix} -1 \\\ 0 \\\ 1 \end{pmatrix}, \quad A \begin{pmatrix} 1 \\\ 1 \\\ 0 \end{pmatrix} = 2 \begin{pmatrix} 1 \\\ 1 \\\ 0 \end{pmatrix}.
Then, the system (A−3I)(x y z)=(1 2 3)(A - 3I) \begin{pmatrix} x \\\ y \\\ z \end{pmatrix} = \begin{pmatrix} 1 \\\ 2 \\\ 3 \end{pmatrix} has

A

unique solution

B

exactly two solutions

C

no solution

D

infinitely many solutions

Answer

unique solution

Explanation

Solution

Define the matrix AA with elements. Let

A=(x1y1z1 x2y2z2 x3y3z3).A = \begin{pmatrix} x_1 & y_1 & z_1 \\\ x_2 & y_2 & z_2 \\\ x_3 & y_3 & z_3 \end{pmatrix}.

Use the given conditions to form equations. Using the dot product notation for matrix multiplication, we have the following conditions:

From

A(1 0 1)=(2 0 1):A \begin{pmatrix} 1 \\\ 0 \\\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\\ 0 \\\ 1 \end{pmatrix} :

(x1 y1 z1)⋅(1 0 1)=2,(x2 y2 z2)⋅(1 0 1)=0,(x3 y3 z3)⋅(1 0 1)=1.(x_1 \ y_1 \ z_1) \cdot \begin{pmatrix} 1 \\\ 0 \\\ 1 \end{pmatrix} = 2, \quad (x_2 \ y_2 \ z_2) \cdot \begin{pmatrix} 1 \\\ 0 \\\ 1 \end{pmatrix} = 0, \quad (x_3 \ y_3 \ z_3) \cdot \begin{pmatrix} 1 \\\ 0 \\\ 1 \end{pmatrix} = 1.

Expanding each dot product:

x1+z1=2,x2+z2=0,x3+z3=1.(1)x_1 + z_1 = 2, \quad x_2 + z_2 = 0, \quad x_3 + z_3 = 1. \quad (1)

From

A(1 1 1)=(4 0 1):A \begin{pmatrix} 1 \\\ 1 \\\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\\ 0 \\\ 1 \end{pmatrix} :

(x1 y1 z1)⋅(1 1 1)=4,(x2 y2 z2)⋅(1 1 1)=0,(x3 y3 z3)⋅(1 1 1)=1.(x_1 \ y_1 \ z_1) \cdot \begin{pmatrix} 1 \\\ 1 \\\ 1 \end{pmatrix} = 4, \quad (x_2 \ y_2 \ z_2) \cdot \begin{pmatrix} 1 \\\ 1 \\\ 1 \end{pmatrix} = 0, \quad (x_3 \ y_3 \ z_3) \cdot \begin{pmatrix} 1 \\\ 1 \\\ 1 \end{pmatrix} = 1.

Expanding each dot product:

x1+y1+z1=4,x2+y2+z2=0,x3+y3+z3=1.(2)x_1 + y_1 + z_1 = 4, \quad x_2 + y_2 + z_2 = 0, \quad x_3 + y_3 + z_3 = 1. \quad (2)

From

A(0 1 1)=(2 1 0):A \begin{pmatrix} 0 \\\ 1 \\\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\\ 1 \\\ 0 \end{pmatrix} :

(x1 y1 z1)⋅(0 1 1)=2,(x2 y2 z2)⋅(0 1 1)=1,(x3 y3 z3)⋅(0 1 1)=0.(x_1 \ y_1 \ z_1) \cdot \begin{pmatrix} 0 \\\ 1 \\\ 1 \end{pmatrix} = 2, \quad (x_2 \ y_2 \ z_2) \cdot \begin{pmatrix} 0 \\\ 1 \\\ 1 \end{pmatrix} = 1, \quad (x_3 \ y_3 \ z_3) \cdot \begin{pmatrix} 0 \\\ 1 \\\ 1 \end{pmatrix} = 0.

Expanding each dot product:

y1+z1=2,y2+z2=1,y3+z3=0.(3)y_1 + z_1 = 2, \quad y_2 + z_2 = 1, \quad y_3 + z_3 = 0. \quad (3)

Solve for elements of AA. Using equations (1), (2), and (3), we can solve for the individual elements of AA:

From equation (2): x1+y1+z1=4x_1 + y_1 + z_1 = 4 and y1+z1=2y_1 + z_1 = 2 from equation (3). Substitute z1=2−y1z_1 = 2 - y_1 into equation (1) to find x1,y1,z1x_1, y_1, z_1.

Similarly, solve for x2,y2,z2x_2, y_2, z_2 and x3,y3,z3x_3, y_3, z_3 to complete the matrix AA.

Set up the system:

(A−3I)(x y z)=(1 2 3).(A - 3I) \begin{pmatrix} x \\\ y \\\ z \end{pmatrix} = \begin{pmatrix} 1 \\\ 2 \\\ 3 \end{pmatrix}.

Now, calculate A−3IA - 3I and substitute to find the unique solution for the system.

Therefore, the answer is: unique solution.