Question

Consider two systems of 3 linear equations in 3 unknowns $AX = B$ and $CX = D$. If $AX = B$ has the unique solution $X = D$ and $CX = D$ has the unique solution $X = B$, then the solution of $(A - C^{-1})X = 0$ is

• A. $B$
• B. $D$
• C. $B + D$
• D. $B - D$

Hint:

For homogeneous systems, substitute known solutions from related systems and use matrix algebra to verify. Check if the matrix is invertible for uniqueness.

Answer 1

The Correct Option is B

We are given two systems of linear equations:

\(AX = B\) and \(CX = D\).

The unique solution of \(AX = B\) is \(X = D\), and the unique solution of \(CX = D\) is \(X = B\).

We need to find the solution of the equation \((A - C^{-1})X = 0\). Let's analyze these conditions:

• Since \(AX = B\) has the unique solution \(X = D\), substituting gives:

\(AD = B\) (Equation 1)

• Since \(CX = D\) has the unique solution \(X = B\), substituting gives:

\(CB = D\) (Equation 2)

Now consider the equation \((A - C^{-1})X = 0\):

• Solving for \(X\), we have: \(AX = C^{-1}X\)

This implies that \(AX = C^{-1}CX = X\), since \(C^{-1}C = I\) (where \(I\) is the identity matrix). Therefore, this reduces to finding \(X\) such that:

\((A - I)X = 0\)

Considering that \(AD = B\) and \(CB = D\), we can deduce the following by substituting \(X = D\):

• The equation \((A - C^{-1})X = 0\) becomes \((A - I)D = 0\), which holds because substituting \(X = D\) satisfies both initial equation conditions for a unique solution.

Thus, the solution to \((A - C^{-1})X = 0\) is: \(D\)