If
[
−
4

3

−
3

1
]
𝑋
+
[
2

−
3

1

−
5
]
=
[
−
1

3

−
2

4
]
𝑋
,
which of the following matrices is
𝑋
?

Question

If
[
−
4
	
3

−
3
	
1
]
𝑋
+
[
2
	
−
3

1
	
−
5
]
=
[
−
1
	
3

−
2
	
4
]
𝑋
,
which of the following matrices is 
𝑋
?

BlackTom AI · Accepted Answer

We start by clearly restating the problem and the given matrices to keep the algebra organized.

Question: If
[
−4  3
−3  1
] X + [
2  −3
1  −5
] X = [
−1  3
−2  4
] X, which of the following matrices is X?

Answer options:
1) [
−1/2  1/2
−2/3  1
]
2) [
2/3  −1
1/9  −4/3
]
3) [
1/3  −1
5/9  −1
]
4) [−2  7/3
−1/3  2/3]

Now, approach the equation by moving all terms to one side and factoring out X. Let A = [−4  3; −3  1], B = [2  −3; 1  −5], and C = [−1  3; −2  4]. Then the equation becomes (A + B)X = CX, which is equivalent to (A + B − C)X = 0. Compute A + B and subtract C:
- A + B = [−4+2   3+ (−3); −3+1   1+(−5)] = [−2   0; −2  −4].
- A + B − C = [−2−(−1)   0−3; −2−(−2)   −4−4] = [−1  −3; 0  −8].
Thus the matrix equation is [−1  −3; 0  −8] X = 0.

Let X = [a  b; c  d]. Compute the product MX with M = [−1  −3; 0  −8]:
- First column: [−1·a + (−3)·c; 0·a + (−8)·c] = [−a − 3c; −8c]. This must be [0; 0], giving the equations −a − 3c = 0 and −8c = 0. From −8c = 0, we get c = 0, which then implies −a = 0, so a = 0.
- Second column: [−1·b + (−3)·d; 0·b + (−8)·d] = [−b − 3d; −8d]. This must be [0; 0], giving −b − 3d = 0 and −8d = 0. From −8d = 0, we get d = 0, which then implies −b = 0, so b = 0.
Therefore, the only solution is a = b = c = d = 0, i.e., X is the zero matrix.

Now, evaluate the given options in light of this result. For each candidate X, MX should be the zero matrix if it were a valid solution. However:
- Option 1: X = [ [−1/2, 1/2], [−2/3, 1] ]. Substituting into MX would produce nonzero entries in general, contradicting MX = 0 unless by coincidence the fractions align perfectly, which they do not.
- Option 2: X = [ [2/3, −1], [1/9, −4/3] ]. Similar quick check shows MX ≠ 0 due to the first column giving a nonzero vector unless very special cancellations occur, which they do not here.
- Option 3: X = [ [1/3, −1], [5/9, −1] ]. Again, the product MX does not yield the zero matrix.
- Option 4: X = [ [−2, 7/3], [−1/3, 2/3] ]. Yet again, MX is not the zero matrix.

The algebra above demonstrates that the system (A + B − C)X = 0 forces X to be the zero matrix. Since none of the provided options equals the zero matrix, none satisfy the equation MX = 0. In typical matrix-equation problems, this means the listed candidates do not include the valid solution, which is the zero matrix in this case. Any of the nonzero candidates would fail to satisfy the equation because M is invertible (its determinant is (−1)(−8) − (0)(−3) = 8 ≠ 0), so MX = 0 forces X = 0.

To summarize the reasoning path: construct M = A + B − C, solve MX = 0 by solving the linear system for the columns of X, find that all entries must be zero, and then check that none of the given 2x2 matrices equals the zero matrix, hence none are correct solutions.

If [ − 4 3 − 3 1 ] 𝑋 + [ 2 − 3 1 − 5 ] = [ − 1 3 − 2 4 ] 𝑋 , which of the following matrices is 𝑋 ?

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