Calculate \(\det(\ABld)\) for the following matrices.
\(\ABld = \displaystyle \left( \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right)\)
\(\ABld = \displaystyle \left( \begin{array}{cc} 5 & 7 \\ -3 & 2 \end{array} \right)\)
\(\ABld = \displaystyle \left( \begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & 1 \\ 4 & -5 & 3 \end{array} \right)\)
\(\ABld = \displaystyle \left( \begin{array}{ccc} 1 & 1 & \pi \\ 2 & -1 & 1 \\ 1 & -\pi & 6 \end{array} \right)\)
Calculate \(\det(B)\) for the following matrices.
\(\BBld = \displaystyle \left( \begin{array}{ccc} 2 & 3 & 4 \\ 0 & -5 & 7 \\ 0 & 0 & 19 \end{array} \right)\)
\(\BBld = \displaystyle \left( \begin{array}{ccc} 1 & 4 & 0 \\ 0 & 0 & 2 \\ 4 & -2 & 5 \end{array} \right)\)
\(\BBld = \displaystyle \left( \begin{array}{ccc} 3 & 3 & 3 \\ -1 & -1 & -1 \\ 10 & 4 & 1 \end{array} \right)\)
\(\BBld = \displaystyle \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right)\)
Decide whether the following systems have nonzero solutions.
\(\displaystyle \left\{ \begin{array}{l} 2x + 4y = 0 \\ 3x + 2y = 0 \end{array} \right.\)
\(\displaystyle \left\{ \begin{array}{r} 10x + 5y + 7z = 0 \\ 6x + y + 9z = 0 \\ 2x + 2y - z = 0 \end{array} \right.\)
\(\displaystyle \left\{ \begin{array}{r} x_1 + 2x_2 = 0 \\ x_2 + 2x_3 = 0 \\ x_3 + 2x_1 = 0 \end{array} \right.\)
Decide whether the following systems have unique solutions.
\(\displaystyle \left\{ \begin{array}{r} 4x + 10y = 6 \\ -2x - 5y = -3 \end{array} \right.\)
\(\displaystyle \left\{ \begin{array}{r} x+y = 0 \\ y+z = 0 \\ x+y+z = 1 \end{array} \right.\)
\(\displaystyle \left\{ \begin{array}{r} 2x_1 - 3x_2 + 5x_3 = 8 \\ 3x_1 -3x_2 +9x_3 = 12 \\ x_1 - 2x_2 + 2x_3 = 6 \end{array} \right.\)
As suggested in the text, show that the determinant of a lower triangular matrix is the product of its diagonal entries. Is the same true of upper triangular matrices?
Suppose that the system of equations
\[\left\{ \begin{array}{r} a_{11}z_1 + a_{12}z_2 + \cdots + a_{1n}z_n = 0 \\ a_{21}z_1 + a_{22}z_2 + \cdots + a_{2n}z_n = 0 \\ \vdots \hspace{0.17in} \\ a_{n1}z_1 + a_{n2}z_2 + \cdots + a_{nn}z_n = 0 \end{array} \right.\] has a nonzero solution \((z_1, \cdots, z_n)\). Let \((b_1, \cdots, b_n)\) be \(n\) real numbers (this will be our forcing terms), and suppose that the list of numbers \((y_1, \cdots, y_n)\) satisfies the equations \[\left\{ \begin{array}{r} a_{11}y_1 + a_{12}y_2 + \cdots + a_{1n}y_n = b_1 \\ a_{21}y_1 + a_{22}y_2 + \cdots + a_{2n}y_n = b_2 \\ \vdots \hspace{0.17in} \\ a_{n1}y_1 + a_{n2}y_2 + \cdots + a_{nn}y_n = b_n \end{array} \right. \tag{$\star$}\] The goal of this problem is to show that the solution \((y_1, \cdots, y_n)\) is necessarily not unique. To do this, we’ll actually produce infinitely many other solutions of (\(\star\)). Let \(\alpha\) be any real number, and let \(x_i = y_i + \alpha z_i\). Show that \((x_1, \cdots, x_n)\) also satisfies (\(\star\)). [This was a step in the proof of Theorems 2.16 and 2.17.]
Let \(\ABld\) be an \(n\times n\) matrix, say \[\ABld = \left( \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{array} \right) \,,\] and suppose that \(\ABld\) has a column of zeros. For concreteness let it be the first column: \(a_{11} = a_{21} = \cdots = a_{n1} = 0\). Show that the system \[\left\{ \begin{array}{r} a_{11} x_{1} + a_{12} x_{2} + \cdots + a_{1n} x_{n} = 0 \\ a_{21} x_{1} + a_{22} x_{2} + \cdots + a_{2n} x_{n} = 0 \\ \vdots \; \\ a_{n1} x_{1} + a_{n2} x_{2} + \cdots + a_{nn} x_{n} = 0 \end{array} \right.\] has a nonzero solution by explicitly giving one. [Note. This implies by Theorem A.2 that \(\det(A) = 0\).]
Let \(\ABld\) be an \(n{\times}n\) matrix as in the previous problem, say \[\ABld = \left( \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{array} \right) \,,\] and this time suppose that \(\ABld\) has a row of zeros. For concreteness let it be the first row: \(a_{11}=a_{12}=\cdots=a_{1n}=0\). Explain why \(\det(\ABld)=0\).
[Note. This time the solution to the system of linear equations is not straightforward to write down, but we know from this problem that one is out there somewhere.]
One might naïvely hope that the determinant function is linear, i.e., \(\det(\ABld+\BBld)=\det(\ABld)+\det(\BBld)\). For \(2{\times}2\) matrices, this is the hope that \[\det\!\left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right) + \det\!\left( \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array} \right) \stackrel{?}{=} \det\!\left( \begin{array}{cc} a_{11} + b_{11} & a_{12} + b_{12} \\ a_{21} + b_{21} & a_{22} + b_{22} \end{array} \right) \,.\] Unfortunately, this is flagrantly false. Find two \(2\times 2\) matrices \(\ABld\) and \(\BBld\) for which \(\det(\ABld+\BBld)\neq\det(\ABld)+\det(\BBld)\).
One might also naïvely hope that you can pull scalars through a determinant in the sense that \(\det(k\ABld) = k\det(A)\). For \(2{\times}2\) matrices, this is the hope that \[\det\!\left( \begin{array}{cc} k a_{11} & k a_{12} \\ k a_{21} & k a_{22} \end{array} \right) \stackrel{?}{=} k \det\!\left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right) \,.\] This too is false, but this time something similar is true. What is the relationship between \(\det(k\ABld)\) and \(\det(\ABld)\) if \(\ABld\) is a \(2{\times}2\) matrix? (What if \(\ABld\) is an \(n{\times}n\) matrix?)
Calculate the determinant for the following matrices:
\(\XBld = \displaystyle \left( \begin{array}{cc} x & y \\ 0 & t \end{array} \right) \,,\)
\(\YBld = \displaystyle \left( \begin{array}{cc} \alpha & 1 \\ 2 & 5 \end{array} \right) \,,\)
where \(\alpha\), \(x\), \(y\), and \(t\) are scalar parameters. Under what conditions is the determinant nonzero for each matrix?
Calculate the determinant for the following matrices:
\(\MBld = \displaystyle \left( \begin{array}{ccc} 0 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & \alpha \end{array} \right) \,, \)
\(\NBld = \displaystyle \left( \begin{array}{ccc} 2 & 2 & \frac{5}{7} \\ a & b & c \\ 4 & 4 & \frac{10}{7} \end{array} \right) \,, \)
\(\PBld = \displaystyle \left (\begin{array}{ccc} 1 & 2 & 3 \\ 0 & \beta & 0 \\ 1 & 0 & 5 \end{array} \right) \,,\)
where \(a, b, c, e, f, \alpha\), and \(\beta\) are scalar parameters.
Under what conditions is the determinant in (c) nonzero?
Decide under which conditions the following systems have unique solutions (here, \(a, b,\) and \(\alpha\) are scalars, while \(x\) and \(y\) are the unknowns).
\[\begin{aligned} a \cdot x + b \cdot y = 0 , \\ x + y = 1.\end{aligned}\]
\[\begin{aligned} x + y = 0 ,\\ y + z = 0 ,\\ \alpha \cdot x + y + z = 0 . \end{aligned}\]