Sean Fitzpatrick |
---|
University of Lethbridge |
![]() |
Compute the determinant of the given matrices, possibly after doing a row operation:
\(A = \bbm 1\amp 4\amp 0\\2\amp -3\amp 5\\0\amp -1\amp -2\ebm\)
\(B = \bbm 3\amp 0 \amp -2\amp 1\\ -2\amp 1 \amp 1 \amp 3\\ 0 \amp -1\amp 2 \amp -3\\ 4\amp 0\amp 1\amp 0\ebm\)
If \(B\) is obtained from \(A\) using the row operation \(R_i\leftrightarrow R_j\text{,}\) then \(\det B = -\det A\text{.}\)
If \(B\) is obtained from \(A\) using the row operation \(kR_i \to R_i\text{,}\) then \(\det B = k\det A\text{.}\)
If \(B\) is obtained from \(A\) using the row operation \(R_i+kR_j\to R_i\text{,}\) then \(\det B=\det A\text{.}\)
Note: these effects are most easily observed in elementary matrices!
Suppose \(B\) is obtained from \(A\) using the folloing row operations:
\(\frac{1}{4}R_1\to R_1\)
\(R_2-4R_1\to R_2\)
\(R_2\leftrightarrow R_3\)
\(R_3+3R_2\to R_3\)
If \(A\) is a \(4\times 4\) matrix and \(\det A = -3\text{,}\) what is the value of \(\det(2A)\text{?}\)
Let \(A\) and \(B\) be \(n\times n\) matrices. Then:
\(\det(AB) = \det(A)\det(B)\)
\(\det(A^T) = \det(A)\)
\(\det(kA)=k^n\det(A)\)
A matrix \(A\) is invertible if and only if \(\det(A)\neq 0\text{.}\) Furthermore, if \(A\) is invertible, then
Given that \(\det A = 3\) and \(\det B = -2\text{,}\) what is the value of:
\(\det(A^2B^3)\)
\(\det(B^{-1}AB)\)
\(\det(2AB^{-1})\)
What can you say about \(\det A\) if:
\(A^2=A\)
\(A^4=I\)
\(PA=P\text{,}\) where \(P\) is invertible.
Recall: given an \(n\times n\) matrix \(A\text{,}\) the \((i,j)\) cofactor is the number \(C_{ij}=(-1)^{i+j}\det M_{ij}\text{,}\) where \(M_{ij}\) is the \((i,j)\) minor.
The matrix of cofactors of \(A\) is the matrix \(\cof(A)\) whose \((i,j)\) entry is \(C_{ij}\text{.}\)
Example: find \(\cof(A)\) if \(A = \bbm 2\amp -1\amp 3\\0\amp 4\amp -2\\1\amp -1\amp 0\ebm\text{.}\)
The adjugate of an \(n\times n\) matrix \(A\) is given by \(\adj(A)=\cof(A)^T\text{.}\)
For any \(n\times n\) matrix \(A\text{,}\)
Use the formula \(\di A^{-1}=\frac{1}{\abs{A}}\adj(A)\) to compute the inverse of:
\(A = \bbm 2\amp 1\amp -3\\3\amp 0\amp 2\\0\amp 1\amp 4\ebm\)
\(A = \bbm 1\amp 0\amp x\\0\amp -x\amp 2\\x\amp 0\amp 3\ebm\text{.}\)
Suppose we have a system of \(n\) equations in \(n\) unknowns, written as \(A\vec{x}=\vec{b}\text{.}\)
If \(\det A =0\text{,}\) then \(A\) is not invertible, and this system has either no solution, or infinitely many solutions.
If \(\det A\neq 0\text{,}\) then
Result: if \(A_i\) denotes the matrix obtained by replacing column \(i\) of \(A\) by \(\vec{b}\text{,}\) then
Use Cramer's rule to solve the system: