Q.Find $x$, if $\begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = O$.

Idea: The triple product is a single number (a quadratic in $x$); set it to $0$.

Step 1 — row times matrix.

since column 1 gives $x-2$, column 2 gives $-10$, column 3 gives $2x-5-3 = 2x-8$.

Step 2 — times the column. …

A $1\times3$ row, a $3\times3$ matrix and a $3\times1$ column multiply to a $1\times1$ number. Here $O$ means that scalar $0$, so we get one equation in $x$.

Step 1 — row times matrix

With $R=\begin{bmatrix} x & -5 & -1 \end{bmatrix}$ and $M=\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$, dot $R$ with each column of $M$:

• Column 1: $x(1)+(-5)(0)+(-1)(2) = x-2$
• Column 2: $x(0)+(-5)(2)+(-1)(0) = -10$
• Column 3: $x(2)+(-5)(1)+(-1)(3) = 2x-5-3 = 2x-8$

So $RM = \begin{bmatrix} x-2 & -10 & 2x-8 \end{bmatrix}$.

Step 2 — times the column

$\begin{bmatrix} x-2 & -10 & 2x-8 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = (x-2)(x) + (-10)(4) + (2x-8)(1).$ …

Method: Row $\times$ matrix $\times$ column giving an equation in the unknown

Use this when a (row)(matrix)(column) product with an unknown $x$ inside is set equal to $O$; the product collapses to a single scalar equation.

Steps

Step 1: Note the result is a $1\times 1$ scalar.

$(1\times n)(n\times n)(n\times 1) = 1\times 1$, so $O$ is the number $0$ and you obtain one equation.

Step 2: Collapse row $\times$ matrix first. …

Common Mistakes

Mistake 1: Dropping a root of the quadratic.

Why it's wrong: $x^2 = 48$ gives $x = \pm 4\sqrt{3}$, not just $4\sqrt3$. Correct approach: a squared unknown yields two roots — report both.

Mistake 2: Mishandling signs so the linear terms don't cancel. …