Click the arrow to see answer.
\[\begin{bmatrix}1&3\\5&1\end{bmatrix},\quad \begin{bmatrix}0&3\\-2&1\end{bmatrix},\quad \begin{bmatrix}0&1\\-2&5\end{bmatrix}\]
Solved Problems for Chapter 7
Let \(A = \left [ \begin{array}{rrr} 1 & 2 & 4 \\ 0 & 1 & 3 \\ -2 & 5 & 1 \end{array} \right ]\). When doing cofactor expansion along the top row, we encounter three minor
matrices. What are they?
Find the following determinant by (a) expanding along the first row, (b) second
column.
\[\begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 2 & 1 & 1 \end{vmatrix}\]
Click the arrow to see answer.
\[\begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 2 & 1 & 1 \end{vmatrix}=1\begin{vmatrix}1 & 3 \\1 & 1\end{vmatrix}-2\begin{vmatrix}2 & 3\\2 & 1\end{vmatrix}+1\begin{vmatrix}2 & 1\\2 & 1\end{vmatrix}=6\]
\[\begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 2 & 1 & 1 \end{vmatrix}=-2\begin{vmatrix}2 & 3 \\2 & 1\end{vmatrix}+1\begin{vmatrix}1 & 1\\2 & 1\end{vmatrix}-1\begin{vmatrix}1 & 1\\2 & 3\end{vmatrix}=6\]
Compute the determinant by co-factor expansion. Pick the easiest row or column
to use.
\[\begin{vmatrix} 1 & 0 & 0 & 1 \\ 2 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ 2 & 1 & 3 & 1 \end{vmatrix}\]
Click the arrow to see answer.
Expand along the third row.
\[\begin{vmatrix} 1 & 0 & 0 & 1 \\ 2 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ 2 & 1 & 3 & 1 \end{vmatrix}=-2\begin{vmatrix} 1 & 0 & 0\\2 & 1 & 1\\2 & 1 & 3 \end{vmatrix}=-2\left (1\begin{vmatrix} 1 & 1\\1 & 3 \end{vmatrix}\right )=-2(2)=-4\]
An operation is done to get from the first matrix to the second. Identify what was
done and tell how it will affect the value of the determinant.
\begin{equation*} \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] \rightarrow \cdots \rightarrow \left [ \begin{array}{cc} a & c \\ b & d \end{array} \right ] \end{equation*}
Click the arrow to see answer.
It does not change the determinant. This was just
taking the transpose.
An operation is done to get from the first matrix to the second. Identify what was
done and tell how it will affect the value of the determinant.
\begin{equation*} \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] \rightarrow \cdots \rightarrow \left [ \begin{array}{cc} c & d \\ a & b \end{array} \right ] \end{equation*}
Click the arrow to see answer.
In this case two rows were switched and so the
resulting determinant is \(-1\) times the first.
An operation is done to get from the first matrix to the second. Identify what was
done and tell how it will affect the value of the determinant.
\begin{equation*} \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] \rightarrow \cdots \rightarrow \left [ \begin{array}{cc} a & b \\ a+c & b+d \end{array} \right ] \end{equation*}
Click the arrow to see answer.
The determinant is unchanged. It was just the first
row added to the second.
An operation is done to get from the first matrix to the second. Identify what was
done and tell how it will affect the value of the determinant.
\begin{equation*} \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] \rightarrow \cdots \rightarrow \left [ \begin{array}{cc} a & b \\ 2c & 2d \end{array} \right ] \end{equation*}
Click the arrow to see answer.
The second row was multiplied by 2 so the
determinant of the result is 2 times the original determinant.
An operation is done to get from the first matrix to the second. Identify what was
done and tell how it will affect the value of the determinant.
\begin{equation*} \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] \rightarrow \cdots \rightarrow \left [ \begin{array}{cc} b & a \\ d & c \end{array} \right ] \end{equation*}
Click the arrow to see answer.
In this case the two columns were switched
so the determinant of the second is \(-1\) times the determinant of the first.
Let \(A\) be an \(n\times n\) matrix and suppose there are \(n-1\) rows such that all rows are linear
combinations of these \(n-1\) rows. Show \(\det \left ( A\right ) =0\).
Click the arrow to see answer.
If the determinant is nonzero, then it will remain
nonzero with row operations applied to the matrix. In this case, you can obtain a
row of zeros by doing row operations. Thus the determinant must be zero.
Construct \(2\times 2\) matrices \(A\) and \(B\) to illustrate the property that \(\det A \det B = \det (AB)\).
Click the arrow to see answer.
\[ \det \left ( \left [ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right ] \left [ \begin{array}{rr} -1 & 2 \\ -5 & 6 \end{array} \right ] \right ) = -8 \]
\[ \det \left [ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right ] \det \left [ \begin{array}{rr} -1 & 2 \\ -5 & 6 \end{array} \right ] = -2 \times 4 = -8 \]
An \(n\times n\) matrix is called nilpotent if for some positive integer, \(k\) it follows \(A^{k}=O.\) If \(A\) is a
nilpotent matrix and \(k\) is the smallest possible integer such that \(A^{k}=O,\) what are the possible
values of \(\det \left ( A\right )\)?
Click the arrow to see answer.
\(\det {A}=0\) because \(0=\det \left ( 0\right ) =\det \left ( A^{k}\right ) =\left ( \det \left ( A\right ) \right ) ^{k}.\)
A matrix is said to be orthogonal if \(A^{T}A=I.\) Thus the inverse of an orthogonal matrix
is just its transpose. What are the possible values of \(\det \left ( A\right ) \) if \(A\) is an orthogonal
matrix?
Click the arrow to see answer.
You would need \(\det \left ( AA^{T}\right ) =\det \left ( A\right ) \det \left ( A^{T}\right ) =\det \left ( A\right ) ^{2}=1\) and so \(\det \left ( A\right ) =1,\) or \(-1\).
Let \(A\) and \(B\) be two \(n\times n\) matrices. We say that \(A\) is similar to \(B\) and write \(A\sim B\) provided that
there exists an invertible matrix \(P\) such that \(A=P^{-1}BP.\) Show that if \(A\sim B,\) then \(\det \left ( A\right ) =\det \left ( B\right )\).
Click the arrow to see answer.
\(\det \left ( A\right ) =\det \left ( S^{-1}BS\right ) =\det \left ( S^{-1}\right ) \det \left ( B\right ) \det \left ( S\right ) =\det \left ( B\right ) \det \left ( S^{-1}S\right ) =\det \left ( B\right ) \).
Determine whether each statement is true or false. If true, provide a proof. If false,
provide a counter example.
- (a)
- If any two columns of a square matrix are equal, then the determinant of the matrix equals zero.
- (b)
- If \(A^{-1}\) exists then \(\det \left ( A^{-1}\right ) =\det \left ( A\right ) ^{-1}.\)
- (c)
- If \(A\) is a real \(n\times n\) matrix, then \(\det \left ( A^{T}A\right ) \geq 0.\)
- (d)
- If \(AX=0\) for some \(X \neq 0,\) then \(\det \left ( A\right ) =0.\)
Click the arrow to see answer.
All of the statements are true.
- (a)
- \(\det (A)=\det (A^T)\). If two columns of \(A\) are equal, then two rows of \(A^T\) are equal. Applying row operations to \(A^T\) will produce a matrix with a row of zeros.
- (b)
- Suppose \(A\) is invertible. Then
\[1=\det (I)=\det (AA^{-1})=\det (A)\det (A^{-1})\]Divide both right and left side by \(\det (A)\) to obtain the result.
- (c)
- This follows from the fact that \(\det (A)=\det (A^T)\).
- (d)
- If \(AX=0\) for some \(X \neq \vec {0}\) then there is a non-trivial linear combination of the columns of \(A\) that is equal to \(\vec {0}\). This means that the columns of \(A\) are linearly dependent. This implies that the rows are also linearly dependent (why?). Applying elementary row operations will lead us to a row of zeros.
Let
\begin{equation*} A = \left [ \begin{array}{rrr} 1 & 0 & 3 \\ 1 & 0 & 1 \\ 3 & 1 & 0 \end{array} \right ] \end{equation*}
Determine whether the matrix \(A\) has an inverse by finding whether the
determinant is non-zero.
Click the arrow to see answer.
\[ \det \left [ \begin{array}{rrr} 1 & 0 & 3 \\ 1 & 0 & 1 \\ 3 & 1 & 0 \end{array} \right ] = 2 \]
and so it has an inverse. The inverse turns out to
equal \[ \left [ \begin{array}{rrr} -\frac {1}{2} & \frac {3}{2} & 0 \\ \frac {3}{2} & -\frac {9}{2} & 1 \\ \frac {1}{2} & -\frac {1}{2} & 0 \end{array} \right ] \]
Bibliography
Some of the problems come from the end of Chapter 3 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)
Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 272–315.