Let \[ A = \begin{bmatrix} 1 & -1 \\ 2 & 2 \\ 1 & 0 \end{bmatrix}, \qquad B = \begin{bmatrix} 3 & 1 \\ -4 & 4 \end{bmatrix}. \] Is there a matrix \(C\) such that \(CA = B\)?
Solution
1. \(C\) must be a \(2 \times 3\) matrix (so that \(CA\) is \(2 \times 2\)).
\(B\) is \(2 \times 2\) and \(A\) is \(3 \times 2\), so \(C\) is \(2 \times 3\).
2. \(CA = B\) means each row of \(B\) is a linear combination of the rows of \(A\) (with coefficients given by the corresponding row of \(C\)).
the rows of \(CA\) are linear combinations of the rows of \(A\).
3. The rows of \(A\) are \(r_1 = (1, -1)\), \(r_2 = (2, 2)\), \(r_3 = (1, 0)\); the rows of \(B\) are \(b_1 = (3, 1)\) and \(b_2 = (-4, 4)\).
read off the rows.
4. \(b_1 = (3, 1)\) is in the row space of \(A\).
\((3, 1) = 1 \cdot (1, -1) + 1 \cdot (2, 2) + 0 \cdot (1, 0) = (1+2, -1+2) = (3, 1)\).
5. \(b_2 = (-4, 4)\) is in the row space of \(A\).
solve \(c_1(1,-1) + c_2(2,2) + c_3(1,0) = (-4, 4)\), i.e. \(c_1 + 2c_2 + c_3 = -4\) and \(-c_1 + 2c_2 = 4\); taking \(c_1 = 0\), \(c_2 = 2\), \(c_3 = -8\) gives \(0 \cdot (1,-1) + 2 \cdot (2,2) - 8 \cdot (1,0) = (4,4) - (8,0) = (-4, 4)\).
6. Hence both rows of \(B\) are in the row space of \(A\), so there is a matrix \(C\) with \(CA = B\).
4 and 5.
7. Explicitly, \(C = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & -8 \end{bmatrix}\).
row 1 of \(C\) gives \(b_1 = 1 \cdot r_1 + 1 \cdot r_2 + 0 \cdot r_3\); row 2 gives \(b_2 = 0 \cdot r_1 + 2 \cdot r_2 - 8 \cdot r_3 = (4, 4) - (8, 0) = (-4, 4)\).
8. Q.E.D.
yes, such a \(C\) exists (<1>7).