5.3. Determinants via Row Reduction

In this section we will first consider the effect of row operations on the value of a determinant. This leads the way to a more efficient way to compute \(n\times n\) determinants.

It also leads the way to two very important properties of determinants, namely

• The product rule: \(\det{(AB)} = \det{A}\cdot\det{B}\).

• The matrix \(A\) is invertible if and only if \(\det{A} \neq 0\) .

5.3.1. How Row Operations affect a Determinant

We have seen in Section 5.2 that the cofactor expansion of an \(n \times n\) determinant works best using a row (or a column) with many, preferably \(n-1\), zeros. When solving a linear system, or finding the inverse of a matrix, we have seen how to create zeros via row reduction. The important thing: row reducing an augmented matrix does not alter the solution(s) of the corresponding linear system. The next proposition describes the effect of row operations on a determinant.

Proposition 5.3.1 (How row operations affect a determinant)

For the determinant of an \(n\times n\) matrix \(A\) the following rules apply.

1. If a row of \(A\) is scaled with a factor \(c\), the determinant is scaled with a factor \(c\).

2. If a multiple of one row of \(A\) is added to another row, the determinant does not change.

3. When two rows of \(A\) are swapped, the determinant changes sign.

We postpone the proof till the end of this section and first look at examples and a few consequences.

Example 5.3.1

The following identities show what happens with a \(4\times 4\) determinant when

1. the second row is scaled with a factor \(c\):

   \[\begin{split} \left|\begin{array}{rrrr} a_{11} &a_{12} &a_{13} &a_{14} \\ ca_{21} &ca_{22} &ca_{23} &ca_{24} \\ a_{31} &a_{32} &a_{33} &a_{34} \\ a_{41} &a_{42} &a_{43} &a_{44} \end{array} \right| = c \left|\begin{array}{rrrr} a_{11} &a_{12} &a_{13} &a_{14} \\ a_{21} &a_{22} &a_{23} &a_{24} \\ a_{31} &a_{32} &a_{33} &a_{34} \\ a_{41} &a_{42} &a_{43} &a_{44} \end{array} \right|, \end{split}\]

2. the first row is added \((-k)\) times to the third row:

   \[\begin{split} \left|\begin{array}{llll} a_{11} &a_{12} &a_{13} &a_{14} \\ a_{21} &a_{22} &a_{23} &a_{24} \\ a_{31}-ka_{11} &a_{32}-ka_{12} &a_{33}-ka_{13} &a_{34}-ka_{14} \\ a_{41} &a_{42} &a_{43} &a_{44} \end{array} \right| = \left|\begin{array}{rrrr} a_{11} &a_{12} &a_{13} &a_{14} \\ a_{21} &a_{22} &a_{23} &a_{24} \\ a_{31} &a_{32} &a_{33} &a_{34} \\ a_{41} &a_{42} &a_{43} &a_{44} \end{array} \right|, \end{split}\]

3. the first and the fourth row are swapped:

   \[\begin{split} \left|\begin{array}{rrrr} a_{41} &a_{42} &a_{43} &a_{44} \\ a_{21} &a_{22} &a_{23} &a_{24} \\ a_{31} &a_{32} &a_{33} &a_{34} \\ a_{11} &a_{12} &a_{13} &a_{14} \end{array} \right| = - \left|\begin{array}{rrrr} a_{11} &a_{12} &a_{13} &a_{14} \\ a_{21} &a_{22} &a_{23} &a_{24} \\ a_{31} &a_{32} &a_{33} &a_{34} \\ a_{41} &a_{42} &a_{43} &a_{44} \end{array} \right|. \end{split}\]

Note that these properties can be expressed using elementary matrices (cf. Section 3.2) .

Example 5.3.2

Let \(A\) be an arbitrary \(4\times 4\) matrix, and \(E_1, E_2\) and \(E_3\) the elementary matrices corresponding to the row operations in Example 5.3.1. So

\[\begin{split} E_1 = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & c & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] , \quad E_2 = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -k & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] , \quad E_3 = \left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right] . \end{split}\]

Then we have

\[ \det{(E_1A)} = c \det{A}, \quad \det{(E_2A)} = \det{A}, \quad \det{(E_3A)} = -\det{A}. \]

Since

\[ \det{E_1} = c, \quad \det{E_2} = 1, \quad \det{E_3} = -1, \]

we see that in all three cases we have that

(5.3.1)\[\det{(E_iA)} = \det{E_i} \cdot \det{A}.\]

These are the basics for the general product rule we will see later, that states that

\[ \text{det}(AB) = \text{det}(A) \text{det} (B) \]

for arbitrary \(n\times n\) matrices \(A\) and \(B\).

The next two examples illustrate the practical use of the three rules involving row operations.

Example 5.3.3

\[\begin{split} \left|\begin{array}{rrr} 5 & 10 & 15 \\ 3 & 8 & 6 \\ 2 & -2 & 5 \end{array} \right|\stackrel{(1)}{=} 5 \left|\begin{array}{rrr} 1 & 2 & 3 \\ 3 & 8 & 6 \\ 2 & -2 & 5 \end{array} \right|\stackrel{(2)}{=} 5 \left|\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 2 & -3 \\ 0 & -6 & -1 \end{array} \right|\stackrel{(3)}{=} 5 \cdot 1 \cdot \left|\begin{array}{rr} 2 & -3 \\ -6 & -1 \end{array} \right|. \end{split}\]

The steps involved are:

1. (1) take out a factor \(5\) from the first row,
2. (2) subtract the first row \(3\) times from the second row and \(2\) times from the third row (or: add it \((-3)\) times and \((-2)\) times respectively)
3. (3) expand along the first column.

Evaluating the \(2\times 2\) determinant at the end leads to the answer \(-100\).

Can you describe the row operations and cofactor expansions in the following computation?

\[\begin{split} \begin{array}{rcl} \left|\begin{array}{rrrr} 3 & 2 & 4 & 3 \\ 1 & 2 & 3 & -1 \\ 4&3&8&2 \\ 2&5&7&-4 \end{array} \right|&=& \left|\begin{array}{rrrr} 0 & -4 & -5 & 6 \\ 1 & 2 & 3 & -1 \\ 0 & -5 & -4 & 6 \\0 &1 & 1 & -2 \end{array} \right|= (-1)\cdot \left|\begin{array}{rrr} -4 & -5 & 6 \\ -5 & -4 & 6 \\ 1 & 1 & -2 \end{array} \right| \\ &=& (-1)\cdot \left|\begin{array}{rrr} 0 & -1 & -2 \\ 0 & 1 & -4 \\ 1 & 1 & -2 \end{array} \right|= - \left|\begin{array}{rr} -1 & -2 \\ 1 & -4 \end{array} \right| = -6. \end{array} \end{split}\]

Remark 5.3.1

Because of Proposition 5.2.3 that states

\[ \det{A} = \text{det}\big(A^T\big) \]

every rule involving row operations may be transformed into a rule about column operations. It is here that computing a determinant differs strikingly from the reduction of an (augmented) matrix to an echelon matrix. Another, more subtle difference is that a row operation applied to a matrix leads to an equivalent matrix, which we denote by the symbol \(\sim\), whereas row or column operations on a determinant give equal values all the time. So then we write \(=\).

In the next example column operations are used.

Example 5.3.4

\[\begin{split} \left|\begin{array}{rrrr} 1 & 1 & 1 & -1 \\ 2 & 4 & 5 & 3 \\4 & 5 & 2 & -1 \\ 5 & 7 & 4 & -2 \end{array} \right|= \left|\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 2 & 2 & 3 & 5 \\4 & 1 & -2 & 3 \\ 5 & 2 & -1 & 3 \end{array} \right|= \left|\begin{array}{rrr} 2 & 3 & 5 \\ 1 & -2 & 3 \\ 2 & -1 & 3 \end{array} \right|= \ldots \end{split}\]

And do you see what is happening here?

\[\begin{split} \left|\begin{array}{rrrr} 1 & 2 & 3 & 4 \\ -2 & 2 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ -5 & 0 & 0 & 5 \end{array} \right|= \left|\begin{array}{rrrr} 10& 2 & 3 & 4 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 5 \end{array} \right|= 10 \left|\begin{array}{rrrr} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \end{array} \right|= 100. \end{split}\]

An interesting consequence of rule (3) of Proposition 5.3.1 is the following.

Corollary 5.3.1

If a matrix \(A\) has two equal rows (or columns), then \(\det{A} = 0\).

Proof. Suppose the \(i\)th and the \(j\)th row of \(A\) are equal, and let \(\det{A} = d\). Let \(B\) be the matrix \(A\) with the \(i\)th and \(j\)th row interchanged.

On the one hand, \(B = A\), so

\[ \det{B} = \det{A} = d, \]

on the other hand, because of Proposition 5.3.1, Rule 2, we have

\[ \det{B} = -\det{A} = -d. \]

We may conclude \(d = -d\), which is only possible if

\[ d = \det{A} = 0. \]

Exercise 5.3.1

Give an alternative proof of Corollary 5.3.1 using Rule i. and Rule ii. of Proposition 5.3.1.

Solution to Exercise 5.3.1 (click to show)

Suppose \(A\) is a matrix with two equal rows, say row \(i\) and row \(j\) are equal.

If we subtract the \(i\)th row from the \(j\)th row, we get a matrix \(A_2\) with \(j\)th row equal to zero, and with det\((A_2) = \) det\((A)\). If we take the factor \(0\) out, we see that det\((A_2) = 0\).

For instance, with a \(4\times 4\) matrix with equal second and fourth row we would have

\[\begin{split} \begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ \color{blue}a_{21} & \color{blue}a_{22} & \color{blue}a_{23} & \color{blue}a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ \color{blue}a_{21} & \color{blue}a_{22} & \color{blue}a_{23} & \color{blue}a_{24} \end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ \color{blue}0 & \color{blue}0 &\color{blue} 0 & \color{blue}0 \end{vmatrix} = {\color{blue}0}\begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ \ast & \ast &\ast &\ast \end{vmatrix} = 0. \end{split}\]

5.3.2. Determinants versus Invertibility

With the knowledge built so far we can show the important property that was already hinted at in Section 5.2.

Theorem 5.3.1

For any square matrix \(A\):

\[ A \,\,\text{is invertible} \quad \iff \quad \det{A} \neq 0. \]

The proof is – we think – quite instructive.

Proof. In the previous section we have already seen that the statement is true for triangular matrices.

Now suppose \(A\) is any \(n \times n\) matrix. Via row reduction \(A\) can be brought to echelon form \(F\), and for a square matrix the echelon form is an upper triangular matrix (with possibly zeros on the diagonal).

From Proposition 5.2.2 we know that for a triangular matrix \(F\) we have
‘\(F\) is invertible’ is equivalent to ‘\(\det{F} \neq 0\)’.

The row operations transforming \(A\) to \(F\) can be performed by multiplications with elementary matrices \(E_1, \ldots E_k\).

We have seen (Equation (5.3.1) in Example 5.3.2) that

\[ \det{(E_iA)} = \det{E_i} \cdot \det{A}. \]

Furthermore, the determinant of an elementary matrix is nonzero. Namely, for a row scaling it is equal to \(c\), for a row swap it is equal to \((-1)\), and for adding a multiple of a row to another row it is equal to \(1\). Hence, if \(m\) of the row operations are row scalings and \(\ell\) of the row operations are row swaps, then

\[ \det{\left(E_kE_{k-1}\cdots E_1A\right)}= c_1\cdots c_m \cdot (-1)^{\ell} \det{A} = \alpha \det{A}, \]

with \(\alpha \neq 0\).

So if \(E_kE_{k-1}\cdots E_1A = F\), where \(F\) is an echelon matrix, we see that

\[ \det{A} \neq 0 \quad \iff \quad \det{F}\neq 0. \]

We conclude that

\[ A \,\text{is invertible} \,\,\iff\,\, F \,\text{is invertible} \,\,\iff\,\, \det{F} \neq 0\,\,\iff\,\, \det{A} \neq 0. \]

Theorem 5.3.2

For two \(n\times n\) matrices \(A\) and \(B\) it always holds that

\[ \det{(AB)} = \det{A}\cdot\det{B}. \]

The proof combines the property that for elementary matrices \(A\) it is already shown that \(\det{(EA)} = \det{E}\cdot\det{A}\) (Equation (5.3.1)). For more details you can push on \(\vee\) below.

Proof of Theorem 5.3.2.

We already know that the identity holds if \(A\) is an elementary matrix. It will also hold if \(A\) is not invertible, as in that case \(AB\) is also not invertible, and we know that the determinant of any non-invertible matrix is zero. So then

\[ \det{(AB)} = 0 = \det{A}\cdot\det{B}. \]

Hence suppose that the matrix \(A\) is invertible. In that case (cf. Theorem 3.4.1) \(A\) can be written as a product of elementary matrices.

\[ A = E_1E_2\ldots E_k. \]

So then, step by step we find that

\[\begin{split} \begin{array}{rl} \det{A} & = \det{(E_1E_2\cdots E_k)} = \det{(E_1(E_2\cdots E_k))} = \\ & = \det{E_1} \det{(E_2\cdots E_k)} = \ldots = \det{E_1} \det{E_2} \cdots \det{E_k}, \end{array} \end{split}\]

and also

\[\begin{split} \begin{array}{rl} \det{(AB)} & = \det{(E_1E_2\cdots E_kB)} = \det{E_1} \det{(E_2\cdots E_kB)} \\ & = \ldots = \\ & = \det{E_1} \det{E_2} \cdots \det{E_k} \det{B} \\ & = \det{A}\det{B}. \end{array} \end{split}\]

Corollary 5.3.2

If the matrix \(A\) is invertible, then \(\text{det}\big(A^{-1}\big)= \dfrac{1}{\det{A}}\).

Proof. We can combine the three properties

i. \(AA^{-1} = I\),   ii. \(\det{(AA^{-1})} = \det{A}\det{\left(A^{-1}\right)}\)   and   iii. \(\det{I} = 1\).

As follows:

\[ \text{det}(A)\text{det}\left(A^{-1}\right) = \text{det}(AA^{-1}) = \text{det}(I) = 1, \]

so indeed

\[ \text{det}\left(A^{-1}\right) = \dfrac{1}{\text{det}(A)}. \]

Exercise 5.3.2

For each of the following statements decide whether they are true or false. In case true, give an argument, in case false, give a counterexample.

1. For each \(n \times n\) matrix \(A\) it holds that

   
   
   \[ \text{det}\big(A^k\big)= \big(\det{A}\big)^k. \]

2. For each two \(n \times n\) matrices \(A\) and \(B\) it holds that

   
   
   \[ \det{(A+B)} = \det{A}+\det{B}. \]

3. For each \(n \times n\) matrix \(A\) it holds that

   
   
   \[ \det{(-A)} = -\det{A}. \]

Solution to Exercise 5.3.2 (click to show)

We treat the statements one by

1. For each \(n \times n\) matrix \(A\) it holds that \(\text{det}\big(A^k\big)= \big(\det{A}\big)^k\). This is true, and follows from repeatedly using the property \(\det(AB) = \det(A)\det(B)\). Namely,

   
   
   \[ \det(A^k) = \det(A\cdot A \cdot A \cdots A) = \det(A)\cdot\det(A)\cdot\det(A)\cdots\det(A). \]

2. For each two \(n \times n\) matrices \(A\) and \(B\) it holds that

   
   
   \[ \det{(A+B)} = \det{A}+\det{B}. \]

   This statement is false. A trivial counterexample is given by \(A = B = I_n\), for \(n \geq 2\). Namely, for these matrices we see that

   
   
   \[ \det{A} + \det{B} = 1 + 1 = 2 \neq \det{(A+B)} = \det{(2I)} = 2^n. \]

3. For each \(n \times n\) matrix \(A\) it holds that

   
   
   \[ \det{(-A)} = -\det{A}. \]

   This is not true in general. The correct statement would be: for an \(n \times n\) matrix \(A\)
   

   \[ \det{(-A)} = (-1)^n\det{(A)}. \]

   One way to see this is true is to write \(-A = (-I)A\) and use the product rule (Theorem 5.3.2).

We will conclude this section, for the interested reader, with a proof of the properties of Proposition 5.3.1. In fact we will prove the column version, and we add one related rule that will be of use both immediately in the proof and also later on.

Proposition 5.3.2

Suppose \(A\) is an \(n\times n\) matrix for which the \(k\)th column is the sum of two vectors in \(\R^n\). So

\[ \vect{a}_k = \vect{b} +\vect{c}. \]

Then

(5.3.2)\[\begin{split}\begin{array}{l} \det{[\vect{a}_1 \,\, \ldots \,\, \vect{b}+\vect{c} \,\, \ldots \,\, \vect{a}_n]} = \\ \qquad \qquad \qquad \det{[\vect{a}_1 \,\, \ldots \,\, \vect{b} \,\, \ldots \,\, \vect{a}_n]} + \det{[\vect{a}_1 \,\, \ldots \,\, \vect{c} \,\, \ldots \,\, \vect{a}_n]} \end{array}\end{split}\]

So, click on \(\vee\) on the right for the proof of Proposition 5.3.1 and Proposition 5.3.2.

Proof of Proposition 5.3.1 and Proposition 5.3.2.

For typographical reasons we will prove the three rules stated as column operations. For an \(n \times n\) matrix

\[ A = [\vect{a}_1 \,\,\vect{a}_2 \,\,\ldots\,\, \vect{a}_n] \]

the rules can then be formulated as

1. (1) \(\det{[\vect{a}_1 \, \vect{a}_2 \, \ldots \, c \vect{a}_k \, \ldots \, \vect{a}_n]} = c \det{A}\);
2. (2) \(\det{[\vect{a}_1 \, \ldots \, \vect{a}_k \, \ldots \, \vect{a}_j \, \ldots \, \vect{a}_n]} = - \det{A}\);
3. (3) \(\det{[\vect{a}_1 \, \ldots \, \vect{a}_j \, \ldots \, \vect{a}_k + c\vect{a}_j \, \ldots \, \vect{a}_n]} = \det{A}\).

So, let us consider them one by one.

1. (1) If a column is scaled with a factor \(c\), then the determinant is also scaled with a factor \(c\).

Suppose \(\tilde{A}\) is the result of scaling the \(k\)th column of \(A\) with a factor \(c\). Then expanding det\((\tilde{A})\) along its \(k\)th column, keeping in mind that \(\tilde{a}_{ik} = c {a}_{ik}\) and \(\tilde{A}_{ik} = {A}_{ik}\), yields

\[ \text{det}\big(\tilde{A}\big)= \sum_{i=1}^n (-1)^{i+k} \tilde{a}_{ik}\det{\tilde{A}_{ik}} =\sum_{i=1}^n (-1)^{i+k} c {a}_{ik}\det{{A}_{ik}}. \]

If we take out the constant factor \(c\) we see that

\[ \text{det}\big(\tilde{A}\big) = c\sum_{i=1}^n (-1)^{i+k} {a}_{ik}\det{{A}_{ik}} = c \det{A}. \]

Proposition 5.3.2 is proved in much the same way as rule (1) by expansion along the \(k\)th column.

2. (2) The proof of the swapping rule is more involved.

One way to set about is

• first settle the rule when the first two columns are interchanged;

• next consider the swapping of two arbitrary consecutive columns;

• finally note that any column swap is the result of an odd number of swaps of consecutive columns.

We will consider the swapping of the first two columns in complete detail. Let \(A^{\ast}\) be the result of swapping the first two columns of the matrix \(A\). The crucial thing to note here is that, for each value of \(i\),

\[ a_{i2}^{\ast} = a_{i1} \quad \text{and} \quad A_{i2}^{\ast} = A_{i1}. \]

To make this explicit for a \(4\times 4\) matrix:

\[\begin{split} \begin{array}{ccc} A = \left[\begin{array}{rrrr} a_{11} &a_{12} &a_{13} &a_{14} \\ a_{21} &a_{22} &a_{23} &a_{24} \\ a_{31} &a_{32} &a_{33} &a_{34} \\ a_{41} &a_{42} &a_{43} &a_{44} \end{array} \right] & \Longrightarrow & A^{\ast} = \left[\begin{array}{rrrr} a_{12} &a_{11} &a_{13} &a_{14} \\ a_{22} &a_{21} &a_{23} &a_{24} \\ a_{32} &a_{31} &a_{33} &a_{34} \\ a_{42} &a_{41} &a_{43} &a_{44} \end{array} \right] \\ \Big\Downarrow && \Big\Downarrow \\ A_{3,1} = \left[\begin{array}{rrr} a_{12} &a_{13} &a_{14} \\ a_{22} &a_{23} &a_{24} \\ a_{42} &a_{43} &a_{44} \end{array} \right] &= & A_{3,2}^{\ast} = \left[\begin{array}{rrr} a_{12} &a_{13} &a_{14} \\ a_{22} &a_{23} &a_{24} \\ a_{42} &a_{43} &a_{44} \end{array} \right] . \end{array} \end{split}\]

Expanding det\((A^{\ast})\) along its second column yields

\[ \text{det}\big(A^{\ast} \big)= \sum_{i=1}^n (-1)^{i+2} a_{i2}^{\ast}\text{det}\big(A_{i2}^{\ast} \big)= \sum_{j=1}^n (-1)^{i+2} {a}_{i1}\text{det}\big({A}_{i1}\big). \]

Noting that \((-1)^{i+2} = (-1)\cdot (-1)^{i+1}\) and taking out the common factor \((-1)\) from the sum yields

\[ \text{det}\big(A^{\ast} \big)= - \sum_{i=1}^n (-1)^{i+1} {a}_{i1}\text{det}\big({A}_{i1}\big)= - \text{det}\big({A}\big). \]

The same argument works for the interchanging of two arbitrary consecutive columns.
As stated the swapping of two arbitrary columns can be accomplished via an odd number of ‘consecutive swaps’, so then the determinant changes sign an odd number of times. And for an odd number \(n\) we have that \((-1)^n = -1\).
Convince yourself of this!

Lastly we have to prove

3. (3) If the multiple of one column of \(A\) is added to another column, the determinant does not change.

First Rule (2) implies, as in Corollary 5.3.1, that a determinant with two equal columns has the value 0.

We then proceed as follows for Rule (3):

\[\begin{split} \begin{array}{l} \det{[\vect{a}_1 \, \ldots \, \vect{a}_j \, \ldots \, \vect{a}_k + c\vect{a}_j \, \ldots \, \vect{a}_n]} = \\ \quad = \det{[\vect{a}_1 \, \ldots \, \vect{a}_j \, \ldots \, \vect{a}_k \, \ldots \, \vect{a}_n]} + \det{[\vect{a}_1 \, \ldots \, \vect{a}_j \, \ldots \, c \vect{a}_j \,\ldots \, \vect{a}_n]} \\ \quad = \det{[\vect{a}_1 \, \ldots \, \vect{a}_j \, \ldots \, \vect{a}_k \, \ldots \, \vect{a}_n]} + c \det{[\vect{a}_1 \, \ldots \, \vect{a}_j \, \ldots \, \vect{a}_j \, \ldots \, \vect{a}_n]}\\ \quad = \det{[\vect{a}_1 \, \ldots \, \vect{a}_j \, \ldots \, \vect{a}_k \, \ldots \, \vect{a}_n]} + 0. \end{array} \end{split}\]

This settles all matters.

5.3.3. Grasple Exercises

Grasple Exercise 5.3.1

Relatively simple 4x4 determinant

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Grasple Exercise 5.3.2

Slightly less simple 4x4 determinant

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Grasple Exercise 5.3.3

To compute a 5x5 determinant with entries in {-2,-1,0,1,2}

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Grasple Exercise 5.3.4

Finding a structured 5x5 determinant in a smart way

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Grasple Exercise 5.3.5

Checking linear (in)dependence of \(\vect{a}_1,\vect{a}_2,\vect{a}_3\) in \(\mathbb{R}^3\) via determinants.

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Grasple Exercise 5.3.6

Checking linear (in)dependence of \(\vect{a}_1,\vect{a}_2,\vect{a}_3,\vect{a}_4\) in \(\mathbb{R}^4\) via determinants.

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Grasple Exercise 5.3.7

Checking invertibility of a matrix \(A\) via det(\(A\)).

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Grasple Exercise 5.3.8

Find \(h\) (in matrix \(A\)) such that \(A\) is invertible.

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Grasple Exercise 5.3.9

Finding a parameter \(\alpha\) for which a 4x4 determinant has a given value

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Grasple Exercise 5.3.10

To compute det\(\left(PBP^{-1}\right)\), for given \(P\) and \(B\).

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Grasple Exercise 5.3.11

To compute det\(\left(A^3\right)\) for a given matrix \(A\)

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Grasple Exercise 5.3.12

To find det\(\left(kA^TB^{-1}\right)\), for 3x3 matrices \(A\) and \(B\)

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Grasple Exercise 5.3.13

What can det(\(A\)) be, if \(A^2 = kA\)?

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Grasple Exercise 5.3.14

What about det(\(A+B\)) = det(\(A\)) + det(\(B\))?

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Grasple Exercise 5.3.15

(True/False) det\((A) = 0 \iff A\) has a zero row?

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Grasple Exercise 5.3.16

What happens to a determinant under certain row or column operation(s)?

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Grasple Exercise 5.3.17

What happens to det(\(A\)) if the last column becomes the first?

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Grasple Exercise 5.3.18

What happens to det(\(A\)) if the order of the rows is reversed?

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