# Matricies Proof

#### Kbotz

a) SHOW that a matrix with a row of zeros cannot have an inverse

b) SHOW that a matrix with a column of zeros cannot have an inverse

how do i show this? i have no clue
help would be much appreciated

#### dwsmith

MHF Hall of Honor
a) SHOW that a matrix with a row of zeros cannot have an inverse

b) SHOW that a matrix with a column of zeros cannot have an inverse

how do i show this? i have no clue
help would be much appreciated
A matrix is invertible iff. the $$\displaystyle det\neq0$$. What happens if a column of row is alls zeros?

Definition:
The determinant of an nxn matrix A, denoted det(A), is a scalar associated with the matrix A that is defined inductively as
$$\displaystyle det(A)= \begin{cases} a_{11}, & \mbox{if }n=1 \\ a_{11}A_{11}+a_{12}A_{12}+\dots+a_{1n}A_{1n}, & \mbox{if }n>1 \end{cases}$$ where $$\displaystyle A_{1j}=(-1)^{1+j}det(M_{1j}),\ j=1,...,n$$ are the cofactors associated with the entries in the first row of A.

Leon, S. (2010). Linear algebra with applications. Upper Saddle River, NJ: Pearson.

a.
By expanding across the row of all zeros, each term of the cofactor expansion will have a factor of 0. Hence, the sum will equal $$\displaystyle 0=det(A)$$

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#### Kbotz

so if a matrix consists of a column of zeros, it's $$\displaystyle det=0$$. do i prove it by solving for the determinant?

#### dwsmith

MHF Hall of Honor
so if a matrix consists of a column of zeros, it's $$\displaystyle det=0$$. do i prove it by solving for the determinant?
That is what I would do, because when you solve the det of a matrix, you expand down the easiest row (usually the row with the most zeros).

#### Kbotz

if i was to do it with any 3x3 matrix, i use the expansion of minors method.
i can seem to do it for a column of zeros. how is it done for a row?

#### dwsmith

MHF Hall of Honor
if i was to do it with any 3x3 matrix, i use the expansion of minors method.
i can seem to do it for a column of zeros. how is it done for a row?
$$\displaystyle det(A)= \begin{cases} a_{11}, & \mbox{if }n=1 \\ a_{11}A_{11}+a_{12}A_{12}+\dots+a_{1n}A_{1n}, & \mbox{if }n>1 \end{cases}$$

$$\displaystyle det(A)= \begin{cases} a_{11}, & \mbox{if }n=1 \\ a_{11}A_{11}+a_{21}A_{21}+\dots+a_{n1}A_{n1}, & \mbox{if }n>1 \end{cases}$$

If we are expanding along the rows (first def) or the columns (second def), what are the values of $$\displaystyle a_{ij}$$?

$$\displaystyle det(A)=det(A^T)$$
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