# Compute Determinant of a Matrix Using Linearly Independent Vectors

## Problem 193

Let $A$ be a $3 \times 3$ matrix.
Let $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are linearly independent $3$-dimensional vectors. Suppose that we have
$A\mathbf{x}=\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, A\mathbf{y}=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, A\mathbf{z}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}.$

Then find the value of the determinant of the matrix $A$.

Contents

We give two solutions.

## Solution 1.

Let $B$ be the $3\times 3$ matrix whose columns are the vectors $\mathbf{x},\mathbf{y}, \mathbf{z}$, that is,
$B=[\mathbf{x} \mathbf{y} \mathbf{z}].$

Then we have
$AB=\begin{bmatrix} 1 & 0 & 1 \\ 0 &1 &1 \\ 1 & 0 & 1 \end{bmatrix}.$

Then we have
$\det(A)\det(B)=\det(AB)=\begin{vmatrix} 1 & 0 & 1 \\ 0 &1 &1 \\ 1 & 0 & 1 \end{vmatrix}=0.$ (If two rows are equal, then the determinant is zero. Or you may compute the determinant by the second column cofactor expansion.)

Note that the column vectors of $B$ are linearly independent, and hence $B$ is nonsingular matrix. Thus the $\det(B)\neq 0$.
Therefore the determinant of $A$ must be zero.

## Solution 2.

Since
$\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}+\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},$

we have
$A\mathbf{x}+A\mathbf{y}=A\mathbf{z}.$ It follows that we have
$A(\mathbf{x}+\mathbf{y}-\mathbf{z})=\mathbf{0}.$

Since the vectors $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are linearly independent, the linear combination $\mathbf{x}+\mathbf{y}-\mathbf{z} \neq \mathbf{0}$.
Hence the matrix $A$ is singular, and the determinant of $A$ is zero.

(Recall that a matrix $A$ is singular if and only if there exist nonzero vector $\mathbf{v}$ such that $A\mathbf{u}=\mathbf{0}$.)

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