Understanding Solutions of Linear Equations

We are given a system of 3 linear equations with 3 unknowns represented in matrix form as \(AX = B\). Here, \(A\) is a \(3 \times 3\) matrix, \(X\) is a \(3 \times 1\) column vector of unknowns, and \(B\) is a \(3 \times 1\) column vector.

We are told that \(X_1\) and \(X_2\) are two distinct solutions to this system. This means:

• \(AX_1 = B\)
• \(AX_2 = B\)
• \(X_1 \neq X_2\) (since they are distinct solutions)

We are also given that the linear combination \(aX_1 + bX_2\) is also a solution to the system \(AX = B\), where \(a\) and \(b\) are real numbers. This means:

\(A(aX_1 + bX_2) = B\)

Applying Matrix Properties to Find the Condition

We can use the properties of matrix multiplication, specifically linearity, to expand the left side of the equation \(A(aX_1 + bX_2) = B\):

\(A(aX_1 + bX_2) = A(aX_1) + A(bX_2)\)

We can factor out the scalars \(a\) and \(b\):

\(A(aX_1) + A(bX_2) = a(AX_1) + b(AX_2)\)

Now, substitute the known relationships \(AX_1 = B\) and \(AX_2 = B\) into this expression:

\(a(AX_1) + b(AX_2) = aB + bB\)

So, the equation \(A(aX_1 + bX_2) = B\) becomes:

\(aB + bB = B\)

We can factor out the vector \(B\) from the left side:

\((a + b)B = B\)

Rearranging the equation, we get:

\((a + b)B - B = 0\)

\((a + b - 1)B = 0\)

Analyzing the Result for System Solutions

The equation \((a + b - 1)B = 0\) holds if either \((a + b - 1) = 0\) or \(B = 0\).

Let's consider if \(B\) can be the zero vector. If \(B = 0\), the system is homogeneous, \(AX = 0\). In this case, if \(X_1\) and \(X_2\) are solutions, any linear combination \(aX_1 + bX_2\) is also a solution for any real numbers \(a\) and \(b\) (the set of solutions forms a vector subspace). However, the question asks for a specific condition on \(a\) and \(b\) for the combination to be a solution, implying that this is not always true for *any* \(a\) and \(b\).

Furthermore, if \(B = 0\), the equation \((a + b - 1)B = 0\) would be satisfied for any values of \(a\) and \(b\), which contradicts the nature of the options provided (which give specific relations between \(a\) and \(b\)). The fact that a combination is given as a solution under specific conditions suggests we are likely dealing with a non-homogeneous system where \(B \neq 0\).

If \(B \neq 0\), then for the equation \((a + b - 1)B = 0\) to be true, the scalar multiplier must be zero.

\(a + b - 1 = 0\)

This simplifies to:

\(a + b = 1\)

This condition \(a+b=1\) ensures that the linear combination \(aX_1 + bX_2\) acts as a solution to the non-homogeneous system \(AX=B\), using the property that \(AX_1=B\) and \(AX_2=B\).

Comparing with Options

We found the condition \(a + b = 1\).

Let's check the given options:

1. a = b
2. a + b = 1
3. a + b = 0
4. a - b = 1

Our derived condition matches the second option.

Revision Table: System of Linear Equations

Additional Information: Linear Combinations of Solutions

For a linear system \(AX = B\):

• If the system is homogeneous (\(B = 0\)), the set of all solutions is a vector space (the null space of \(A\)). Any linear combination of solutions is also a solution. That is, if \(X_1\) and \(X_2\) are solutions to \(AX = 0\), then \(A(aX_1 + bX_2) = a(AX_1) + b(AX_2) = a(0) + b(0) = 0\), so \(aX_1 + bX_2\) is also a solution for any \(a, b\).
• If the system is non-homogeneous (\(B \neq 0\)), the set of all solutions is an affine subspace. If \(X_p\) is a particular solution, the general solution is of the form \(X_p + X_h\), where \(X_h\) is any solution to the homogeneous system \(AX = 0\).

In our problem, \(X_1\) and \(X_2\) are distinct solutions to \(AX = B\) (where \(B \neq 0\)). We can write \(X_1 = X_p + X_{h1}\) and \(X_2 = X_p + X_{h2}\), where \(X_p\) is a particular solution and \(X_{h1}, X_{h2}\) are distinct solutions to the homogeneous system \(AX=0\). The linear combination is \(aX_1 + bX_2 = a(X_p + X_{h1}) + b(X_p + X_{h2}) = (a+b)X_p + aX_{h1} + bX_{h2}\).

For this combination to be a solution to \(AX = B\), it must be of the form \(X_p + X_h'\), where \(X_h'\) is a solution to the homogeneous system. So, \((a+b)X_p + (aX_{h1} + bX_{h2}) = X_p + X_h'\).

Since \(aX_{h1} + bX_{h2}\) is a solution to the homogeneous system \(AX=0\) (as \(X_{h1}, X_{h2}\) are), this simplifies to:

\((a+b)X_p + \text{homogeneous solution} = X_p + \text{another homogeneous solution}\)

This structure only holds for a non-homogeneous system if \(a+b\) acts as the scalar for the particular solution. The simplest way for this equality of structure to hold, especially if \(X_p\) is non-zero, is if \(a+b = 1\).

Alternatively, the derivation \((a+b-1)B = 0\) directly shows that for a non-homogeneous system (\(B \neq 0\)), the condition must be \(a+b-1 = 0\), which is \(a+b=1\). This confirms our earlier result.