Q.If $\begin{bmatrix} 2 & 1 & 3 \end{bmatrix}\begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = A$, find $A$.

Concept: Matrix Equation Solving — multiply the row vector by the matrix, then multiply the result by the column vector.

Step 1: Multiply the row vector by the $3 \times 3$ matrix.

Let $R = \begin{bmatrix} 2 & 1 & 3 \end{bmatrix}$ and $M = \begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}$.

$R M = \begin{bmatrix} 2(-1)+1(-1)+3(0) & 2(0)+1(1)+3(1) & 2(-1)+1(0)+3(1) \end{bmatrix}$

$= \begin{bmatrix} -2 -1 + 0 & 0 + 1 + 3 & -2 + 0 + 3 \end{bmatrix} = \begin{bmatrix} -3 & 4 & 1 \end{bmatrix}$. …

The product is a $1\times3$ row times a $3\times3$ matrix times a $3\times1$ column, so the result is a single number.

Step 1 — Row vector times the matrix. With $R=\begin{bmatrix} 2 & 1 & 3 \end{bmatrix}$ and $M=\begin{bmatrix} -1 & 0 & -1 \\-1 & 1 & 0 \\0 & 1 & 1 \end{bmatrix}$, each entry of $RM$ is $R$ dotted with a column of $M$: …

Method: Evaluating a row-matrix-column triple product to a scalar

Use this when a $1\times3$ row, a $3\times3$ matrix, and a $3\times1$ column are multiplied.

Steps

Step 1: Use associativity.

Multiply left-to-right (or right-to-left) — both give the same scalar.

Step 2: Row times matrix stays a row. …

Common Mistakes

Mistake 1: Stopping at the $1\times3$ intermediate result.

Why it's wrong: the row times the $3\times3$ matrix gives $\begin{bmatrix}-3&4&1\end{bmatrix}$, still a row — you must multiply by the column to reach the scalar $-4$. Correct approach: carry out both multiplications.

Mistake 2: Multiplying in an inconsistent order. …