Q.If $A = \begin{bmatrix}1 & 5\\ 7 & 12\end{bmatrix}$ and $B = \begin{bmatrix}9 & 1\\ 7 & 8\end{bmatrix}$, find a matrix $C$ such that $3A + 5B + 2C$ is a null matrix.

Concept: Matrix Equation Solving — treat the matrix equation like a scalar equation, solving for $C$ by isolating it.

We are given:

where $O$ is the $2 \times 2$ zero matrix.

Step 1: Isolate $2C$:

Step 2: Compute $-3A$ and $-5B$:

Step 3: Add them:

$-3A - 5B = \begin{bmatrix} -48 & -20 \\ -56 & -76 \end{bmatrix}$ …

The core idea here is that matrix equations obey the same algebraic rules as ordinary numbers — addition, subtraction, and scalar multiplication all work termwise. The only difference is that matrix multiplication is not commutative, but here we only need addition and scalar multiplication, so it's straightforward.

We are told that $3A + 5B + 2C$ equals the null matrix (all entries zero). That means:

where $O = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$.

1. Isolate the term with $C$. Subtract $3A$ and $5B$ from both sides:

2. Divide both sides by 2. Since 2 is a scalar, dividing means multiplying by $\frac12$:

This is the key formula. Now we just compute $3A + 5B$ entry by entry.

3. Compute $3A$:

4. Compute $5B$:

5. Add them: …

Method: Solving a linear matrix equation for an unknown matrix

When an equation like $3A+5B+2C=O$ must be solved for a matrix $C$, treat it much like a scalar linear equation: isolate the unknown matrix, then evaluate the right-hand side by scalar multiplication and matrix addition (all entrywise).

Steps

Step 1: Isolate the unknown matrix algebraically.

Move the known terms across; because matrix addition is commutative and associative, ordinary rearrangement is valid:

Step 2: Compute each scalar multiple.

Multiply every entry of $A$ by its scalar, and every entry of $B$ by its scalar.

Step 3: Add the resulting matrices entrywise. …

Common Mistakes

Mistake 1: Forgetting to divide the whole matrix by the leading coefficient.

Why it's wrong: from $2C=-3A-5B$ you must halve every entry; reporting $-3A-5B$ as $C$ is off by a factor of $2$. Correct approach: divide each entry by $2$.

Mistake 2: Sign errors when moving $3A$ and $5B$ across.

Why it's wrong: they become $-3A$ and $-5B$; keeping them positive gives the wrong matrix. Correct approach: negate each term you move to the other side. …