Question: Find the matrix $\mathbf{M}$ such that $\mathbf{M} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$.

["Finding the Matrix $\mathbf{M}$: A Step-by-Step Guide to Solving Linear Matrix Equations", "When faced with a linear matrix equation like\n[\n\mathbf{M} \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix} = \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix},\n]\nmany learners wonder: how do we find the unknown matrix $\mathbf{M}$? This article explains the method clearly and provides a practical solution, making solving such equations intuitive and accessible.", "---", "### Understanding the Problem", "We are given:\n[\n\mathbf{M} A = B,\n]\nwhere\n[\nA = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix},\n]\nand we want to solve for $\mathbf{M}$.", "To isolate $\mathbf{M}$, we need to "undo" multiplication by $A$ on the left. Since matrix multiplication is associative and matrices have inverses when invertible, we multiply both sides on the left by $A^{-1}$:", "[\n\mathbf{M} A = B \quad \Rightarrow \quad A^{-1} (\mathbf{M} A) = A^{-1} B \quad \Rightarrow \quad (A^{-1} \mathbf{M}) A = A^{-1} B \quad \Rightarrow \quad \mathbf{M} = A^{-1} B.\n]", "Thus,\n$$\n\mathbf{M} = A^{-1} B.\n$$", "---", "### Step 1: Check if $A$ is Invertible", "We must ensure $A$ is invertible. A $2\ imes 2$ matrix\n[\n\mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix}\n]\nhas determinant\n[\n\det(\mathbf{A}) = ad - bc.\n]", "For our $A$:\n[\na = 1, ; b = 2, ; c = 3, ; d = 4,\n]\n[\n\det(A) = (1)(4) - (2)(3) = 4 - 6 = -2 <br/>\neq 0.\n]", "Since the determinant is nonzero, $A$ is invertible.", "---", "### Step 2: Compute $A^{-1}$", "The inverse of a $2\ imes 2$ matrix is given by:\n[\n\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}.\n]", "Substitute values:\n[\nA^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}.\n]", "---", "### Step 3: Multiply $A^{-1}$ by $B$", "Now compute\n[\n\mathbf{M} = A^{-1} B = \begin{pmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}.\n]", "Perform matrix multiplication:", "- First row, first column:\n $(-2)(5) + (1)(7) = -10 + 7 = -3$", "- First row, second column:\n $(-2)(6) + (1)(8) = -12 + 8 = -4$", "- Second row, first column:\n $\left(\frac{3}{2}\right)(5) + \left(-\frac{1}{2}\right)(7) = \frac{15}{2} - \frac{7}{2} = \frac{8}{2} = 4$", "- Second row, second column:\n $\left(\frac{3}{2}\right)(6) + \left(-\frac{1}{2}\right)(8) = 9 - 4 = 5$", "So,\n[\n\mathbf{M} = \begin{pmatrix} -3 & -4 \ 4 & 5 \end{pmatrix}.\n]", "---", "### Step 4: Verification", "To confirm correctness, multiply\n[\n\mathbf{M} A = \begin{pmatrix} -3 & -4 \ 4 & 5 \end{pmatrix} \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix} = \begin{pmatrix} (-3)(1) + (-4)(3) & (-3)(2) + (-4)(4) \ (4)(1) + (5)(3) & (4)(2) + (5)(4) \end{pmatrix} = \begin{pmatrix} -3 -12 & -6 -16 \ 4 + 15 & 8 + 20 \end{pmatrix} = \begin{pmatrix} -15 & -22 \ 19 & 28 \end{pmatrix}.\n]", "Oops! That doesn’t match $B = \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}$.\nWhy?", "Wait — our earlier calculation of $\mathbf{M}$ must be rechecked.", "Let’s double-check the multiplication step.", "---", "Double verification of $ \mathbf{M} = A^{-1}B $:", "[\n\mathbf{M} = \begin{pmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}\n]", "- $(-2)(5) + (1)(7) = -10 + 7 = -3$ ✅\n- $(-2)(6) + (1)(8) = -12 + 8 = -4$ ✅\n- $\left(\frac{3}{2}\right)(5) + \left(-\frac{1}{2}\right)(7) = \frac{15 - 7}{2} = \frac{8}{2} = 4$ ✅\n- $\left(\frac{3}{2}\right)(6) + \left(-\frac{1}{2}\right)(8) = \frac{18 - 8}{2} = \frac{10}{2} = 5$ ✅", "So matrix is correct:\n[\n\mathbf{M} = \begin{pmatrix} -3 & -4 \ 4 & 5 \end{pmatrix}\n]", "Now recompute product $ \mathbf{M}A $:", "- $ (-3)(1) + (-4)(3) = -3 -12 = -15 $ → should be 5? ❌\n- $ (-3)(2) + (-4)(4) = -6 -16 = -22 $ → should be 6? ❌", "Something is off — but wait! This suggests a fundamental issue.", "Wait — recall: we derived\n$$\n\mathbf{M} = A^{-1}B\n\Rightarrow \mathbf{M}A = A^{-1}BA.\n$$", "But we must ensure our inversion and multiplication is straightforward.", "Let’s recompute $A^{-1}$ carefully:", "[\n\det = -2, \quad \mathbf{A}^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} \quad \ ext{✅}\n]", "Now $B = \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}$", "Compute first row of $ \mathbf{M}A $:", "- $(-2)(5) + (1)(7) = -10 + 7 = -3$\n- $(-2)(6) + (1)(8) = -12 + 8 = -4$", "Second row:", "- $\frac{3}{2}(5) + (-\frac{1}{2})(7) = \frac{15}{2} - \frac{7}{2} = \frac{8}{2} = 4$", "- $\frac{3}{2}(6) + (-\frac{1}{2})(8) = \frac{18}{2} - \frac{8}{2} = 9 - 4 = 5$", "So $ \mathbf{M} = \begin{pmatrix} -3 & -4 \ 4 & 5 \end{pmatrix} $", "Now compute $ \mathbf{M}A $:", "- $(-3)(1) + (-4)(3) = -3 -12 = -15 <br/>\ne 5$", "But $B_{11} = 5$. Contradiction.", "Ah! Here’s the key insight: matrix multiplication is not commutative, and even if $ \mathbf{M} = A^{-1}B $, then $ \mathbf{M}A $ must equal $B$. But our calculation does not confirm this.", "But wait — by construction:\n$$\n\mathbf{M} = A^{-1}B \Rightarrow \mathbf{M}A = A^{-1}BA\n\Rightarrow \ ext{This equals } B \ ext{ only if } AB = BA, \ ext{ which is not required.}\n$$", "So what went wrong?", "The correct derivation is solid:\n[\n\mathbf{M} \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix} = \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}\n\Rightarrow\n\begin{pmatrix} a & b \ c & d \end{pmatrix}\n\begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}\n= \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}\n]", "This gives the system of equations:", "1. $a(1) + b(3) = 5 \Rightarrow a + 3b = 5$\n2. $a(2) + b(4) = 6 \Rightarrow 2a + 4b = 6$\n3. $c(1) + d(3) = 7 \Rightarrow c + 3d = 7$\n4. $c(2) + d(4) = 8 \Rightarrow 2c + 4d = 8$", "Solve equations 1 and 2:", "From (1): $a = 5 - 3b$\nSubstitute into (2):\n$2(5 - 3b) + 4b = 6 \Rightarrow 10 - 6b + 4b = 6 \Rightarrow 10 - 2b = 6 \Rightarrow -2b = -4 \Rightarrow b = 2$\nThen $a = 5 - 6 = -1$", "Now equations 3 and 4:", "(3): $c + 3d = 7$\n(4): $2c + 4d = 8$ → divide by 2: $c + 2d = 4$", "Subtract:\n$(c + 3d) - (c + 2d) = 7 - 4 \Rightarrow d = 3$\nThen $c = 4 - 2d = 4 - 6 = -2$", "Thus,\n$$\n\mathbf{M} = \begin{pmatrix} -1 & 2 \ -2 & 3 \end{pmatrix}\n$$", "Now verify:", "[\n\begin{pmatrix} -1 &"]