How to prove that if a $3 \times 3$ matrix has two equal rows, it has no inverse?

Question

In my maths classes in school we have said that if a matrix has two equal rows then it has no inverse.

I can see this by calculating that the determinant of any $3 \times 3$ matrix with two equal rows is always $0$.

I would like to know how if you have some $3 \times 3$ matrix with two equal rows, you can prove that the columns are linearly dependent, and that the matrix therefore has no inverse. I am going off the definition of linear independence being that you cannot have a linear combination that equals the zero vector.

PS. I am a total beginner to linear algebra so I may have completely messed up some simple stuff in the question.

Answer 1

So if a function is not injective it is not invertible. Create the function $f(x)=Ax$ with $A$ and $n \times n$ matrix and $x$ a vector in $\mathbb{R}^n$. By the definition of linear dependence we know that there exists some vectors $x_0 \neq 0$ so that $Ax_0=0$ since one row is a linear combination of the other. However $A0=0$ as well so we have $f(x_0)=f(0)$ and since $x_0 \neq 0$ we know $f(x)$ cannot be injective, and so is not invertible.

Answer 2

The argument can be simply proven. Let us say that the $i$-th row and the $j$-th row of an arbitrary square matrix $X$ are equal for $i\ne j$. Let us say that this matrix, has an inverse $X^{-1}$. According to the definition of matrix multiplication, the entry of the $p$-th row and $q$-th column of $XX^{-1}$ is equal to the inner product of the $p$-th row of $X$ and the $q$-th column of $X^{-1}$. Now consider the $i$-th row of $X$ and the $i$-th column of $X^{-1}$. Since $XX^{-1}=I$, the inner product of the $i$-th row of $X$ and the $i$-th column of $X^{-1}$ should be unity. But this leads to contradiction, since the entry of the $j$-th column and the $i$-th row of $XX^{-1}$ would have to be unity, as opposed to zero. This simple argument, proves that square matrices with two distinct equal rows do not have inverse.

Answer 3

Let consider wlog

$$A=\begin{bmatrix}a&b&c\\a&b&c\\d&e&f\end{bmatrix} \implies Ax =\begin{bmatrix}a&b&c\\a&b&c\\d&e&f\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}ax_1+bx_2+cx_3\\ax_1+bx_2+cx_3\\dx_1+ex_2+fx_3\end{bmatrix}=\begin{bmatrix}s\\s\\t\end{bmatrix}$$

that is for any vector $x$ the product $Ax$ is a vector with two equal components, therefore column space of $A$ doesn't span $\mathbb R^3$ but at most a subspace with dimension $2$, therefore $A$ is not invertible.

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As an alternative, given the system

$$Ax =b \iff \begin{bmatrix}a&b&c\\a&b&c\\d&e&f\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}$$

by elimination we obtain the equivalent

$$\begin{bmatrix}a&b&c\\0&0&0\\d&e&f\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}b_1\\b_2-b_1\\b_3\end{bmatrix}$$

which has not solutions when $b_2-b_1\neq 0 \iff b_1\neq b_2$.

Answer 4

Let $\{e_1,e_2,e_3\}$ be the standard basis of $F^3$ (where $F$ is the field over which the matrix is defined). If two different rows $i$ and $j$ of a $3\times3$ matrix $A$ are identical to each other, then $(e_i-e_j)^TA=0$. If $A$ has an inverse, we will arrive at the contradiction that $$0=\left((e_i-e_j)^TA\right)A^{-1}=(e_i-e_j)^T(AA^{-1})=(e_i-e_j)^TI_3=(e_i-e_j)^T\neq0.$$ More generally, if an $n\times n$ matrix $A$ has linearly dependent rows, so that $\sum_{i=1}^nx_ia_i=0$ (here $a_i$ denotes the $i$-th row of $A$) for some coefficients $x_1,x_2,\ldots,x_n$ that are not all zero, then $x^TA=0$ where $x^T= (x_1,x_2,\ldots,x_n)\ne0 $. Hence $A^{-1}$ cannot exist, otherwise $$0=(x^TA)A^{-1}=x^T(AA^{-1})=x^TI_n=x^T\neq0.$$