Two Matrices are Nonsingular if and only if the Product is Nonsingular
Problem 562
An $n\times n$ matrix $A$ is called nonsingular if the only vector $\mathbf{x}\in \R^n$ satisfying the equation $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$.
Using the definition of a nonsingular matrix, prove the following statements.
(a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular.
(b) Let $A$ and $B$ be $n\times n$ matrices and suppose that the product $AB$ is nonsingular. Then:
The matrix $B$ is nonsingular.
The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.)
(a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular.
Let $\mathbf{v}\in \R^n$ and suppose that $(AB)\mathbf{x}=\mathbf{0}$.
Our goal is to prove that $\mathbf{x}=\mathbf{0}$.
Let $\mathbf{y}:=B\mathbf{x} \in \R^n$. Then we have
\begin{align*}
A\mathbf{y}=A(B\mathbf{x})=(AB)\mathbf{x}=\mathbf{0}.
\end{align*}
Since $A$ is nonsingular, this implies that the vector $\mathbf{y}=\mathbf{0}$.
Hence we have $\mathbf{y}=B\mathbf{x}=\mathbf{0}$.
Since $B$ is nonsingular, this further implies that $\mathbf{x}=\mathbf{0}$.
It follows that if $(AB)\mathbf{x}=\mathbf{0}$, then we must have $\mathbf{x}=\mathbf{0}$.
By definition, this means that the matrix $AB$ is nonsingular.
(b)-1. If $AB$ is nonsingular, then $B$ is nonsingular.
Suppose that $B\mathbf{x}=\mathbf{0}$. We prove that $\mathbf{x}=\mathbf{0}$.
Since $B\mathbf{x}=\mathbf{0}$, it yields that
\begin{align*}
(AB)\mathbf{x}=A(B\mathbf{x})=A\mathbf{0}=\mathbf{0}.
\end{align*}
As the matrix $AB$ is nonsingular, it follows from $(AB)\mathbf{x}=\mathbf{0}$ that $\mathbf{x}=\mathbf{0}$.
This proves that the matrix $B$ is nonsingular.
(b)-2. If $AB$ is nonsingular, then $A$ is nonsingular.
By part (1), we know that $B$ is nonsingular, hence it is invertible.
The inverse matrix $B^{-1}$ and the matrix $AB$ are both nonsingular.
Hence it follows from part (a) that the product of $AB$ and $B^{-1}$ is also nonsingular.
Thus,
\[A=(AB)B^{-1}\]
is a nonsingular matrix.
If $M, P$ are Nonsingular, then Exists a Matrix $N$ such that $MN=P$
Suppose that $M, P$ are two $n \times n$ non-singular matrix. Prove that there is a matrix $N$ such that $MN = P$.
Proof.
As non-singularity and invertibility are equivalent, we know that $M$ has the inverse matrix $M^{-1}$.
Let us think backwards. Suppose that […]
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(a) Find a $3\times 3$ nonsingular matrix $A$ satisfying $3A=A^2+AB$, where \[B=\begin{bmatrix}
2 & 0 & -1 \\
0 &2 &-1 \\
-1 & 0 & 1
\end{bmatrix}.\]
(b) Find the inverse matrix of $A$.
Solution
(a) Find a $3\times 3$ nonsingular matrix $A$.
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Determine the values of $x$ so that the matrix
\[A=\begin{bmatrix}
1 & 1 & x \\
1 &x &x \\
x & x & x
\end{bmatrix}\]
is invertible.
For those values of $x$, find the inverse matrix $A^{-1}$.
Solution.
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Prove the following statements.
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(b) $A$ is similar to itself.
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For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix.
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1 & 3 & -2 \\
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Quiz 4: Inverse Matrix/ Nonsingular Matrix Satisfying a Relation
(a) Find the inverse matrix of
\[A=\begin{bmatrix}
1 & 0 & 1 \\
1 &0 &0 \\
2 & 1 & 1
\end{bmatrix}\]
if it exists. If you think there is no inverse matrix of $A$, then give a reason.
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\[A^3=A^2B-3A^2,\]
where […]
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