Consider a system of linear equations PX=Q  where P ∈ R3x3 and Q∈R3x3. Suppose P has an LU decomposition, P=LU, where

Which of the following statement(s) is/are TRUE?

The correct statements are:
1. The system PX=Q can be solved by first solving LY=Q and then UX=Y.
2. If P is invertible, then both L and U are invertible.
3. If P is singular, then at least one of the diagonal elements of U is zero.

From the provided image, we have a system of linear equations where a matrix $P$ has an LU decomposition $P=LU$.
The lower triangular matrix is given by:
$L=\left[\begin{array}{ccc}1& 0& 0\\ l_{21}& 1& 0\\ l_{31}& l_{32}& 1\end{array}\right]$
and the upper triangular matrix is given by:
$U=\left[\begin{array}{ccc}{u}_{11}& u_{12}& u_{13}\\ 0& u_{22}& u_{23}\\ 0& 0& u_{33}\end{array}\right]$

Step 1: Analyzing the solution method of the system (Statement 1)
We are given the system:
$PX=Q$
Substituting $P=LU$ into the equation gives:
$\left(LU\right)X=Q$
By matrix associativity, we can write:
$L\left(UX\right)=Q$
Let us define an intermediate variable matrix $Y=UX$. The system splits into two sequential steps:
1. Solve the lower-triangular system $LY=Q$ for $Y$ using forward substitution.
2. Solve the upper-triangular system $UX=Y$ for $X$ using back substitution.
Hence, this statement is TRUE.

Step 2: Determinant relations and invertibility (Statement 2)
The determinant of the product of two square matrices is the product of their determinants:
$\det \left(P\right)=\det \left(LU\right)=\det \left(L\right)·\det \left(U\right)$
If $P$ is invertible, its determinant is non-zero:
$\det \left(P\right)\ne 0$
Consequently:
$\det \left(L\right)\ne 0\text{and}\det \left(U\right)\ne 0$
Because their determinants are non-zero, both $L$ and $U$ must be invertible.
Hence, this statement is TRUE.

Step 3: Singular matrix properties (Statement 3)
If $P$ is singular, its determinant is zero:
$\det \left(P\right)=0\Rightarrow \det \left(L\right)·\det \left(U\right)=0$
For a triangular matrix, the determinant is the product of the diagonal elements.
The diagonal elements of $L$ are all equal to 1, as shown in the image. Therefore:
$\det \left(L\right)=1·1·1=1$
Since $\det \left(L\right)=1\ne 0$, it must be that:
$\det \left(U\right)=0$
The determinant of $U$ is the product of its diagonal elements:
$\det \left(U\right)=u_{11}·u_{22}·u_{33}=0$
This product is zero if and only if at least one of the diagonal elements ($u_{11}$, $u_{22}$, or $u_{33}$) is zero.
Hence, this statement is TRUE.