Q.Let $A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$, $B = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix}$, $C = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix}$. Find a matrix $D$ such that $CD - AB = O$.

Idea: $CD - AB = O$ means $CD = AB$; since $C$ is invertible, $D = C^{-1}(AB)$.

Step 1 — compute $AB$.

Step 2 — invert $C$. $\det C = (2)(8)-(5)(3) = 1$, so

Step 3 — solve. …

The equation $CD - AB = O$ rearranges to $CD = AB$. Because $C$ sits on the left of the unknown $D$, we undo it by left-multiplying both sides by $C^{-1}$: $C^{-1}(CD) = C^{-1}(AB)$, i.e. $D = C^{-1}(AB)$. So the plan is: compute $AB$, invert $C$, multiply.

Step 1 — compute $AB$

• $(1,1):\ 2\cdot5+(-1)\cdot7 = 3$
• $(1,2):\ 2\cdot2+(-1)\cdot4 = 0$
• $(2,1):\ 3\cdot5+4\cdot7 = 43$
• $(2,2):\ 3\cdot2+4\cdot4 = 22$

So $AB = \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix}$.

Step 2 — invert $C$

so $C$ is invertible. For a $2\times2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ the inverse is $\tfrac{1}{\det}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$; here $\det=1$, so

Step 3 — solve for $D$

• $(1,1):\ 8\cdot3+(-5)\cdot43 = 24-215 = -191$
• $(1,2):\ 8\cdot0+(-5)\cdot22 = -110$
• $(2,1):\ -3\cdot3+2\cdot43 = -9+86 = 77$ …

Method: Solving a matrix equation for an unknown matrix using an inverse

Use this when an equation like $CD - AB = O$ must be solved for the matrix $D$ and the coefficient matrix is square and invertible.

Steps

Step 1: Rearrange to isolate the unknown's term.

$CD - AB = O \Rightarrow CD = AB$.

Step 2: Left-multiply by the inverse of the coefficient on the correct side. …

Common Mistakes

Mistake 1: Multiplying by $C^{-1}$ on the wrong side.

Why it's wrong: because $C$ is on the left of $D$, you must left-multiply: $D = C^{-1}(AB)$, not $(AB)C^{-1}$. Matrix multiplication is not commutative. Correct approach: apply the inverse on the same side as the coefficient.

Mistake 2: Errors in the $2\times2$ inverse formula. …