Q.If $A = \begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{bmatrix}$, then find the matrix $X$, such that $2A + 3X = 5B$.

Concept: Scalar Multiplication — multiplying a matrix by a constant multiplies every entry.

We have $2A + 3X = 5B$. Isolate $3X$:

Now compute $5B$ and $2A$ entry by entry:

Subtract:

Finally, divide by 3 (scalar multiplication):

The core idea here is scalar multiplication — matrices obey the same algebraic rules as numbers when it comes to addition and multiplication by a constant. So you can solve for an unknown matrix $X$ exactly as you would solve $2a + 3x = 5b$ for a number $x$. The only difference is that the operations are performed entry by entry.

Let’s walk through it.

1. Isolate the term containing $X$. Start with $2A + 3X = 5B$. Subtract $2A$ from both sides:

2. Divide both sides by 3. Since scalar multiplication is just multiplying every entry, dividing by 3 means multiplying by $\frac{1}{3}$:

3. Compute $5B$ and $2A$ separately. Multiply each entry of $B$ by 5:

Multiply each entry of $A$ by 2:

4. Subtract $2A$ from $5B$. Subtract corresponding entries:

5. Multiply by $\frac{1}{3}$ to get $X$. Divide every entry by 3:

Method: Solving a matrix equation for an unknown matrix using scalar algebra

Use this for equations like $2A + 3X = 5B$ where $X$ is an unknown matrix.

Steps

Step 1: Isolate the term containing $X$ as in scalar algebra.

Treat $A, B, X$ like numbers for $+$, $-$, and scalar multiples: $3X = 5B - 2A$.

Step 2: Compute the scalar multiples and combine entry-wise.

Form $5B$ and $2A$ by scaling every entry, then subtract corresponding entries.

Step 3: Divide by the scalar to get $X$.

Multiply the result by $\tfrac{1}{3}$ (every entry), leaving fractional entries in exact form.

Common Mistakes

Mistake 1: Dividing only some entries by the scalar.

Why it's wrong: $\tfrac13(5B-2A)$ divides every entry by $3$; a partial division gives a wrong matrix. Correct approach: apply the $\tfrac13$ to all entries.

Mistake 2: Sign slips in the subtraction.

Why it's wrong: e.g. $10 - (-4) = 14$, not $6$. Correct approach: subtract each entry of $2A$ from the matching entry of $5B$, keeping signs.

Mistake 3: Leaving improper or rounded fractions.

Why it's wrong: entries like $-\tfrac{10}{3}$ should stay exact, not be rounded. Correct approach: keep fractions in exact form.