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Mathematical Expression Editor
Challenge Problems for Chapter 7
Consider the matrix
\begin{equation*} A = \left [ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos t & -\sin t \\ 0 & \sin t & \cos t \end{array} \right ] \end{equation*}
Does there exist a value of \(t\) for which this matrix fails to have
an inverse? Explain.
Click the arrow to see answer.
No. It has a nonzero determinant for all \(t\).
If \(A,B,\) and \(C\) are each \(n\times n\) matrices and \(ABC\) is invertible, show why each of \(A,B,\) and \(C\) are
invertible.
Click the arrow to see answer.
This follows because \(\det \left ( ABC\right ) =\det \left ( A\right ) \det \left ( B\right ) \det \left ( C\right ) \) and if this product is nonzero, then each determinant in the
product is nonzero and so each of these matrices is invertible.
Suppose \(A,B\) are \(n\times n\) matrices and that \(AB=I.\) Show that then \(BA=I.\)
First explain why \(\det \left ( A\right ) ,\det \left ( B\right ) \) are
both nonzero. Then \(\left ( AB\right ) A=A\) and then show \(BA\left ( BA-I\right ) =0.\) Now use what is given to conclude \(A\left ( BA-I\right ) =0.\)
Suppose \(A\) is an upper triangular matrix. Show that \(A^{-1}\) exists if and only if all elements
of the main diagonal are non zero. Is it true that \(A^{-1}\) will also be upper triangular?
Explain. Could the same be concluded for lower triangular matrices?
The given
condition is what it takes for the determinant to be non zero. Recall that the
determinant of an upper triangular matrix is just the product of the entries on the
main diagonal.
Let \(A\), \(B\), and \(C\) denote \(n\times n\) matrices. Assume that \(\det A=-1\), \(\det B=2\), and \(\det C=3\). Evaluate
(a)
\(\det (A^3BC^TB^{-1})\)
(b)
\(\det (B^2C^{-1}AB^{-1}C^T)\)
(c)
\(\det (A^{-1}B^{-1}AB)\)
If \(A\) and \(B\) are \(n\times n\) matrices such that \(AB =-BA\), and if \(n\) is odd, show that either \(A\) or \(B\) has no inverse.
Show that no \(3\times 3\) matrix \(A\) exists such that \(A^2+I = O\). Find a \(2\times 2\) matrix \(A\) with this property.
Show that \(\det (A+B^T ) = \det (A^T +B)\) for any \(n\times n\) matrices \(A\) and \(B\).