Q.Find the value of $x$ if $\begin{bmatrix} 1 & x & 1 \end{bmatrix}\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = O$.

Concept: Matrix Multiplication Compatibility — the product is defined only when the inner dimensions match, and here we multiply a $1 \times 3$ row, a $3 \times 3$ matrix, and a $3 \times 1$ column to get a $1 \times 1$ scalar (zero).

Step 1: Multiply the row vector by the matrix first:

Step 2: Multiply this $1 \times 3$ result by the column vector: …

We start with the expression

where $O$ here means the zero scalar (the number 0). The product is a $1\times1$ matrix, i.e., a number.

Why multiply in this order?

Matrix multiplication is associative, so we can either multiply the row vector with the matrix first, or the matrix with the column vector first. Both give the same final scalar. We'll do the first multiplication: row vector times matrix, which yields another row vector. Then multiply that row vector by the column vector to get the scalar.

1. Multiply the row vector by the matrix Let

The product $\mathbf{r}M$ is a $1\times3$ row vector. Each entry is the dot product of $\mathbf{r}$ with the corresponding column of $M$.

• First column: $1\cdot1 + x\cdot2 + 1\cdot15 = 1 + 2x + 15 = 2x + 16$.
• Second column: $1\cdot3 + x\cdot5 + 1\cdot3 = 3 + 5x + 3 = 5x + 6$.
• Third column: $1\cdot2 + x\cdot1 + 1\cdot2 = 2 + x + 2 = x + 4$.

So

2. Multiply this row vector by the column vector The column vector is

The product $\mathbf{r}M \cdot \mathbf{c}$ is the dot product:

Compute each term:

• First term: $2x + 16$.
• Second term: $2(5x+6) = 10x + 12$.
• Third term: $x(x+4) = x^2 + 4x$.

Sum them: …

Method: Evaluating a row-matrix-column triple product

Use this whenever a $1\times n$ row, an $n\times n$ matrix, and an $n\times1$ column are multiplied to give a single number.

Steps

Step 1: Use associativity to choose an order.

$(RM)C=R(MC)$, so multiply left-to-right for convenience.

Step 2: Multiply row by matrix.

$R_{1\times n}M_{n\times n}$ is still a $1\times n$ row, not a scalar yet.

Step 3: Multiply that row by the column. …

Common Mistakes

Mistake 1: Treating the row-times-matrix result as a scalar.

Why it's wrong: $\begin{bmatrix}1&x&1\end{bmatrix}$ times the $3\times3$ matrix is still a $1\times3$ row; only after multiplying by the $3\times1$ column do you get a number. Correct approach: carry the intermediate row, then contract with the column.

Mistake 2: Misreading $O$ as the zero matrix. …