Sean Fitzpatrick |
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University of Lethbridge |
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Compute the products:
Assuming each product below is defined,
\(A(BC)=(AB)C\)
\(A(B+C)=AB+AC\)
\((A+B)C = AC+BC\)
\(A(kB)=(kA)B = k(AB)\) for any scalar \(k\)
\(I_mA = A\) and \(AI_n = A\)
For real numbers \(a\) and \(b\text{,}\) if \(a\neq 0\) and \(ax=b\text{,}\) we know \(x=\frac{b}{a}\text{.}\)
If \(A\) is a matrix, \(\vec{b}\) is a vector, and \(A\vec{x}=\vec{b}\text{,}\) can we similarly solve for \(\vec{x}\text{?}\)
Short answer: no. Longer answer: sometimes. Sort of.
A square (\(n\times n\)) matrix \(A\) is invertible if \(AB=I_n=BA\) for some \(B\text{.}\)
We call \(B\) the inverse of \(A\text{,}\) and write \(B=A^{-1}\text{.}\)
For \(A = \bbm 2\amp -1\\-1\amp 1\ebm\text{,}\) \(D = \bbm 1\amp 1\\1\amp 2\ebm\text{,}\) we saw that \(AD = I_2\) and \(DA=I_2\text{.}\) So \(D=A^{-1}\text{.}\)
Suppose \(A\bbm x\\y\ebm = \bbm 7\\-5\ebm\text{.}\) How can we use \(D\) to solve for \(\vec{x}=\bbm x\\y\ebm\text{?}\)
Given \(A\text{,}\) an \(m\times n\) matrix, and \(B\text{,}\) an \(n\times p\) matrix, write \(B = \bbm B_1\amp B_2 \amp \cdots B_p\ebm\text{,}\) where the \(B_i\) are columns of \(B\text{.}\) Then
Suppose \(A,B\) are both \(n\times n\text{,}\) and \(AB = I_n\text{.}\) Then
Let \(A\) be an \(n\times n\) matrix.
Form the augmented matrix \([\begin{array}{c|c}A\amp I_n\end{array}]\)
Use row operations until you reach \([\begin{array}{c|c}R\amp B\end{array}]\text{,}\) where \(R\) is the RREF of \(A\text{.}\)
If \(R\) has a row of zeros, then \(A\) is not invertible.
If \(R=I_n\text{,}\) then \(B=A^{-1}\text{.}\)
Compute the inverses (if possible) of:
Let \(A\) be an \(n\times n\) matrix. The following statements are all equivalent to the statement “\(A\) is invertible”:
The rank of \(A\) is \(n\text{.}\)
The RREF of \(A\) is equal to \(I_n\text{.}\)
The system \(A\vec{x}=\vec{b}\) is consistent for every \(n\times 1\) vector \(\vec{b}\text{.}\)
The only solution to the system \(A\vec{x}=\vec{0}\) is \(\vec{x}=\vec{0}\text{.}\)
There is a matrix \(B\) such that \(AB=I_n\text{.}\)
There is a matrix \(C\) such that \(CA=I_n\text{.}\)
Let \(A\) and \(B\) be invertible matrices.
The inverse of \(A\) is unique.
\(A^{-1}\) is also invertible, and \((A^{-1})^{-1}=A\text{.}\)
\(AB\) is invertible, and \((AB)^{-1}=B^{-1}A^{-1}\text{.}\)
For any nonzero scalar \(k\text{,}\) \(kA\) is invertible, and \((kA)^{-1}=\frac1k A^{-1}\text{.}\)
For any \(n\times 1\) vector \(\vec{b}\text{,}\) the unique solution to \(A\vec{x}=\vec{b}\) is \(\vec{x}=A^{-1}\vec{b}\text{.}\)
Show that if \(A,B,C\) are invertible, then \((ABC)^{-1}=C^{-1}B^{-1}A^{-1}\text{.}\)
Show that if \(A^3 = 4I_n\text{,}\) then \(A\) is invertible.
Show that if \(A^2-3A+2I_n=0\text{,}\) then \(A\) is invertible.
Simplify the expression \(B(AB)^{-1}(AB)^2B^{-1}\text{.}\)