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Elementary Linear Algebra: For University of Lethbridge Math 1410

Section 6.4 Properties of the Determinant

In the previous section we learned how to compute the determinant. In this section we learn some of the properties of the determinant, and this will allow us to compute determinants more easily. In the next section we will see one application of determinants.
We start with a theorem that gives us more freedom when computing determinants.
We alluded to this fact way back after Example 6.3.6. We had just learned what cofactor expansion was and we practised along the second row and down the third column. Later, we found the determinant of this matrix by computing the cofactor expansion along the first row. In all three cases, we got the number 0. This wasn’t a coincidence. The above theorem states that all three expansions were actually computing the determinant.
How does this help us? By giving us freedom to choose any row or column to use for the expansion, we can choose a row or column that looks “most appealing.” This usually means “it has lots of zeros.” We demonstrate this principle below.

Example 6.4.2. Computing a \(4\times 4\) determinant.

Find the determinant of
\begin{equation*} \tta = \bbm 1 \amp 2 \amp 0\amp 9\\2\amp -3\amp 0\amp 5\\7\amp 2\amp 3\amp 8\\-4\amp 1\amp 0\amp 2\ebm\text{.} \end{equation*}
Solution.
Our first reaction may well be “Oh no! Not another \(4\times 4\) determinant!” However, we can use cofactor expansion along any row or column that we choose. The third column looks great; it has lots of zeros in it. The cofactor expansion along this column is
\begin{align*} \det(A) \amp = a_{1,3}C_{1,3} + a_{2,3}C_{2,3} + a_{3,3}C_{3,3}+a_{4,3}C_{4,3}\\ \amp = 0\cdot C_{1,3} + 0\cdot C_{2,3} + 3\cdot C_{3,3} + 0\cdot C_{4,3}\text{.} \end{align*}
The wonderful thing here is that three of our cofactors are multiplied by 0. We won’t bother computing them since they will not contribute to the determinant. Thus
\begin{align*} \det(A) \amp = 3\cdot C_{3,3}\\ \amp = 3\cdot (-1)^{3+3}\cdot \bvm 1\amp 2\amp 9\\2\amp -3\amp 5\\-4\amp 1\amp 2\evm \\ \amp = 3\cdot(-147) \quad\quad\quad \scriptstyle{\begin{pmatrix}\text{we computed the determinant of the}\\ 3\times 3 \text{ matrix without showing our work;}\\ \text{it is } -147\end{pmatrix}}\\ \displaystyle \amp = -447\text{.} \end{align*}
Wow. That was a lot simpler than computing all that we did in Example 6.3.10. Of course, in that example, we didn’t really have any shortcuts that we could have employed. Our next example involves a \(5\times 5\) determinant. At first, this looks like trouble, until we realize that the matrix is triangular. As we’ll see, this makes our job much easier.

Example 6.4.3. Computing the determinant of a \(5\times 5\) (triangular) matrix.

Find the determinant of
\begin{equation*} \tta = \bbm 1 \amp 2 \amp 3\amp 4\amp 5\\0\amp 6\amp 7\amp 8\amp 9\\0\amp 0\amp 10\amp 11\amp 12\\0\amp 0\amp 0\amp 13\amp 14\\0\amp 0\amp 0\amp 0\amp 15\ebm\text{.} \end{equation*}
Solution.
Since we can expand along any row or column, things are not as bad as they might at first seem. In fact, this problem is very easy. What row or column should we choose to find the determinant along? There are two obvious choices: the first column or the last row. Both have 4 zeros in them. We choose the first column. We omit most of the cofactor expansion, since most of it is just 0:
\begin{equation*} \det(A) = 1\cdot (-1)^{1+1}\cdot\bvm 6 \amp 7 \amp 8\amp 9\\0\amp 10\amp 11\amp 12\\0\amp 0\amp 13\amp 14\\0\amp 0\amp 0\amp 15\evm\text{.} \end{equation*}
Similarly, this determinant is not bad to compute; we again choose to use cofactor expansion along the first column. Note: technically, this cofactor expansion is \(6\cdot(-1)^{1+1}A_{1,1}\text{;}\) we are going to drop the \((-1)^{1+1}\) terms from here on out in this example (it will show up a lot...).
\begin{equation*} \det(A) = 1\cdot 6\cdot\bvm 10 \amp 11 \amp 12\\0\amp 13\amp 14\\0\amp 0\amp 15\evm\text{.} \end{equation*}
You can probably can see a trend. We’ll finish out the steps without explaining each one.
\begin{align*} \det(A) \amp = 1\cdot6\cdot10\cdot\bvm 13\amp 14\\0\amp 15\evm\\ \amp = 1\cdot6\cdot10\cdot13\cdot15\\ \amp = 11700\text{.} \end{align*}
We see that the final determinant is the product of the diagonal entries. This works for any triangular matrix (and since diagonal matrices are triangular, it works for diagonal matrices as well). This is an important enough idea that we’ll put it into a box.

Key Idea 6.4.4. The Determinant of Triangular Matrices.

The determinant of a triangular matrix is the product of its diagonal elements.
It is now again time to start thinking like a mathematician. Remember, mathematicians see something new and often ask “How does this relate to things I already know?” So now we ask, “If we change a matrix in some way, how is its determinant changed?”
The standard way that we change matrices is through elementary row operations. If we perform an elementary row operation on a matrix, how will the determinant of the new matrix compare to the determinant of the original matrix?
Let’s experiment first and then we’ll officially state what happens.

Example 6.4.5. Row operations and determinants.

Let
\begin{equation*} \tta = \bbm 1 \amp 2\\3 \amp 4\ebm\text{.} \end{equation*}
Let \(\ttb\) be formed from \(\tta\) by doing one of the following elementary row operations:
  1. \(\displaystyle 2R_1+R_2\rightarrow R_2\)
  2. \(\displaystyle 5R_1 \rightarrow R_1\)
  3. \(\displaystyle R_1\leftrightarrow R_2\)
Find \(\det(A)\) as well as \(\det(B)\) for each of the row operations above.
Solution.
It is straightforward to compute \(\det(A) = -2\text{.}\)
Let \(\ttb\) be formed by performing the first row operation on \(\tta\text{;}\) thus
\begin{equation*} \ttb = \bbm 1 \amp 2\\5 \amp 8\ebm\text{.} \end{equation*}
It is clear that \(\det(B) = -2\text{,}\) the same as \(\det(A)\text{.}\)
Now let \(\ttb\) be formed by performing the second elementary row operation on \(\tta\text{;}\) that is,
\begin{equation*} \ttb = \bbm 5 \amp 10\\3 \amp 4\ebm\text{.} \end{equation*}
We can see that \(\det(B) = -10\text{,}\) which is \(5\cdot\det(A)\text{.}\)
Finally, let \(\ttb\) be formed by the third row operation given; swap the two rows of \(\tta\text{.}\) We see that
\begin{equation*} \ttb = \bbm 3 \amp 4\\1 \amp 2\ebm \end{equation*}
and that \(\det(B) = 2\text{,}\) which is \((-1)\cdot\det(A)\text{.}\)
We’ve seen in the above example that there seems to be a relationship between the determinants of matrices “before and after” being changed by elementary row operations. Certainly, one example isn’t enough to base a theory on, and we have not proved anything yet. Regardless, the following theorem is true.
Let’s put this theorem to use in a couple of examples.

Example 6.4.7. Using row operations to compute a determinant.

Let
\begin{equation*} \tta = \bbm 1 \amp 2 \amp 1\\0\amp 1\amp 1\\1\amp 1\amp 1\ebm\text{.} \end{equation*}
Compute \(\det(A)\text{,}\) then find the determinants of the following matrices by inspection using Theorem 6.4.6.
\begin{equation*} \ttb = \bbm 1 \amp 1 \amp 1\\1\amp 2\amp 1\\0\amp 1\amp 1\ebm\quad \ttc = \bbm 1\amp 2\amp 1\\0\amp 1\amp 1\\7\amp 7\amp 7\ebm\quad \ttd = \bbm 1\amp -1\amp -2\\0\amp 1\amp 1\\1\amp 1\amp 1\ebm\text{.} \end{equation*}
Solution.
Computing \(\det(A)\) by cofactor expansion down the first column or along the second row seems like the best choice, utilizing the one zero in the matrix. We can quickly confirm that \(\det(A) = 1\text{.}\)
To compute \(\det(B)\text{,}\) notice that the rows of \(\tta\) were rearranged to form \(\ttb\text{.}\) There are different ways to describe what happened; saying \(R_1\leftrightarrow R_2\) was followed by \(R_1\leftrightarrow R_3\) produces \(\ttb\) from \(\tta\text{.}\) Since there were two row swaps, \(\det(B) = (-1)(-1)\det(A) = \det(A) = 1\text{.}\)
Notice that \(\ttc\) is formed from \(\tta\) by multiplying the third row by 7. Thus \(\det(C) = 7\cdot\det(A) = 7\text{.}\)
It takes a little thought, but we can form \(\ttd\) from \(\tta\) by the operation \(-3R_2+R_1\rightarrow R_1\text{.}\) This type of elementary row operation does not change determinants, so \(\det(D) = \det(A)\text{.}\)

Example 6.4.8. Effect of elementary row operations on the determinant.

The matrix \(\ttb\) was formed from \(\tta\) using the following elementary row operations, though not necessarily in this order. Find \(\det(A)\text{.}\)
\begin{equation*} B = \bbm 1 \amp 2 \amp 3\\0\amp 4\amp 5\\0\amp 0\amp 6\ebm \end{equation*}
\begin{gather*} 2R_1\rightarrow R_1\\ \frac 13 R_3\rightarrow R_3\\ R_1 \leftrightarrow R_2\\ 6R_1+R_2 \rightarrow R_2 \end{gather*}
Solution.
It doesn’t take too much work to compute \(\det\ttb=24\text{.}\) In looking at our list of elementary row operations, we see that only the first three have an effect on the determinant. Therefore
\begin{equation*} 24 = \det(B) = 2\cdot \frac13 \cdot (-1)\cdot \det(A) \end{equation*}
and hence
\begin{equation*} \det(A) = -36\text{.} \end{equation*}
In the previous example, we may have been tempted to “rebuild” \(\tta\) using the elementary row operations and then computing the determinant. This can be done, but in general it is a bad idea; it takes too much work and it is too easy to make a mistake.
Let’s continue to think like mathematicians; mathematicians tend to remember “problems” they’ve encountered in the past, and when they learn something new, in the backs of their minds they try to apply their new knowledge to solve their old problem. (This is why mathematicians rarely smile: they are remembering their problems)
What “problem” did we recently uncover? We stated in the last chapter that even computers could not compute the determinant of large matrices with cofactor expansion. How then can we compute the determinant of large matrices?
We just learned two interesting and useful facts about matrix determinants. First, the determinant of a triangular matrix is easy to compute: just multiply the diagonal elements. Secondly, we know that given any square matrix, we can use elementary row operations to put the matrix in triangular form. We can then find the determinant of the new matrix (which is easy), and adjust that number by recalling what elementary operations we performed.

Example 6.4.9. Reducing a determinant to triangular form.

Find the determinant of \(\tta\) by first putting \(\tta\) into a triangular form, where
\begin{equation*} \tta = \bbm 2 \amp 4 \amp -2\\-1\amp -2\amp 5\\3\amp 2\amp 1\ebm\text{.} \end{equation*}
Solution.
In putting \(\tta\) into a triangular form, we need not worry about getting leading 1s, but it does tend to make our life easier as we work out a problem by hand. So let’s scale the first row by \(1/2\text{:}\)
\begin{equation*} \frac 12R_1 \rightarrow R_1 \qquad \bbm 1\amp 2\amp -1\\-1\amp -2\amp 5\\3\amp 2\amp 1\ebm\text{.} \end{equation*}
Now let’s get 0s below this leading 1:
\begin{equation*} \begin{matrix}R_1 + R_2 \rightarrow R_2\\-3R_1+R_3\rightarrow R_3\end{matrix} \qquad \bbm 1\amp 2\amp -1\\0\amp 0\amp 4\\0\amp -4\amp 4\ebm\text{.} \end{equation*}
We can finish in one step; by interchanging rows 2 and 3 we’ll have our matrix in triangular form.
\begin{equation*} R_2 \leftrightarrow R_3 \qquad \bbm 1\amp 2\amp -1\\0\amp -4\amp 4\\0\amp 0\amp 4\ebm\text{.} \end{equation*}
Let’s name this last matrix \(\ttb\text{.}\) The determinant of \(\ttb\) is easy to compute as it is triangular; \(\det(\ttb) = -16\text{.}\) We can use this to find \(\det(\tta)\text{.}\)
Recall the steps we used to transform \(\tta\) into \(\ttb\text{.}\) They are:
\begin{equation*} \begin{matrix}\frac 12R_1 \rightarrow R_1\\R_1 + R_2 \rightarrow R_2\\-3R_1+R_3\rightarrow R_3\\R_2 \leftrightarrow R_3\end{matrix} \end{equation*}
The first operation multiplied a row of \(\tta\) by \(\frac 12\text{.}\) This means that the resulting matrix had a determinant that was \(\frac12\) the determinant of \(\tta\text{.}\)
The next two operations did not affect the determinant at all. The last operation, the row swap, changed the sign. Combining these effects, we know that
\begin{equation*} -16 = \det(\ttb)= (-1)\frac12\det(\tta)\text{.} \end{equation*}
Solving for \(\det(\tta)\) we have that \(\det(\tta)=32\text{.}\)
In practice, we don’t need to keep track of operations where we add multiples of one row to another; they simply do not affect the determinant. Also, in practice, these steps are carried out by a computer, and computers don’t care about leading 1s. Therefore, row scaling operations are rarely used. The only things to keep track of are row swaps, and even then all we care about are the number of row swaps. An odd number of row swaps means that the original determinant has the opposite sign of the triangular form matrix; an even number of row swaps means they have the same determinant.
If you find yourself needing to compute a determinant by hand (say, on an exam), it’s a good idea to keep the following principles in mind:
  1. Stick to row operations of the type \(R_i+kR_j\to R_i\) as much as possible: they don’t change the determinant.
  2. Getting all the way to triangular form isn’t really necessary. Use row operations of the above type to create as many zeros as possible in one of the columns, and then expand along that column.
To see how these principles work in practice, let’s repeat Example 6.4.9. This time we’ll focus on creating zeros, but we won’t worry about getting to triangular form. Since adding a multiple of one row to another doesn’t change the determinant, we can compute \(\det(A)\) with a string of equalities, as follows:
\begin{align*} \begin{vmatrix} 2\amp 4\amp -2\\-1\amp 2\amp 5\\3\amp 2\amp 1 \end{vmatrix} \amp = \begin{vmatrix} 0\amp 0\amp 8\\-1\amp 2\amp 5\\3\amp 2\amp 1 \end{vmatrix} \quad \text{ (Add } 2R_2 \text{ to } R_1)\\ \amp = 0+0+8(-1)^{1+3}\begin{vmatrix} -1\amp -2\\3\amp 2 \end{vmatrix} \quad \text{ (Expand along Row 1)}\\ \amp = 8((-1)(2)-(-2)(3))\\ \amp = 8(4)=32\text{.} \end{align*}
Of course, in this case we got lucky and ended up with two zeros in the first row after one row operation. However, had this not been the case, we would have simply done one more row operation (\(R_3+3R_2\to R_3\)) to create a second zero in the first column, and then done a cofactor expansion along that column.
For larger determinants, we can follow the same routine: create zeros in one column, expand along that column to get a smaller determinant, and repeat.
Let’s think some more like a mathematician. How does the determinant work with other matrix operations that we know? Specifically, how does the determinant interact with matrix addition, scalar multiplication, matrix multiplication, the transpose and the trace? We’ll again do an example to get an idea of what is going on, then give a theorem to state what is true.

Example 6.4.10. Determinants and matrix operations.

Let
\begin{equation*} \tta = \bbm 1 \amp 2\\3 \amp 4\ebm \ \text{and} \ \ttb = \bbm 2\amp 1\\ 3\amp 5\ebm\text{.} \end{equation*}
Find the determinants of the matrices \(\tta\text{,}\) \(\ttb\text{,}\) \(\tta-\ttb\text{,}\) \(3\tta\text{,}\) \(\tta\ttb\text{,}\) \(\ttat\text{,}\) and \(\ttai\text{.}\) Can you find any connections between these values?
Solution.
We can quickly compute that \(\det(A) = -2\) and \(\det(B) = 7\text{.}\)
\begin{align*} \det(\tta-\ttb) \amp = \det\left(\bbm 1\amp 2\\3\amp 4\ebm-\bbm 2\amp 1\\ 3\amp 5\ebm\right)\\ \amp = \bvm -1\amp 1\\0\amp -1\evm\\ \amp = 1\text{.} \end{align*}
It’s tough to find a connection between \(\det(\tta-\ttb)\text{,}\) \(\det(A)\) and \(\det(B)\text{.}\)
\begin{align*} \det(3\tta) \amp = \bvm 3\amp 6\\9\amp 12\evm\\ \amp = -18\text{.} \end{align*}
We can figure this one out; multiplying one row of \(\tta\) by 3 increases the determinant by a factor of 3; doing it again (and hence multiplying both rows by 3) increases the determinant again by a factor of 3. Therefore \(\det(3\tta) = 3\cdot3\cdot\det(A)\text{,}\) or \(3^2\cdot\det(A)\text{.}\)
\begin{align*} \det(\tta\ttb) \amp = \det\left(\bbm 1\amp 2\\3\amp 4\ebm\bbm 2\amp 1\\ 3\amp 5\ebm\right)\\ \amp = \bvm 8\amp 11\\18 \amp 23\evm\\ \amp = -14\text{.} \end{align*}
This one seems clear; \(\det(\tta\ttb) = \det(A)\det(B)\text{.}\)
\begin{align*} \det(A^T) \amp = \bvm 1\amp 3\\2\amp 4\evm\\ \amp = -2\text{.} \end{align*}
Obviously \(\det(A^T) = \det(A)\text{;}\) is this always going to be the case? If we think about it, we can see that the cofactor expansion along the first row of \(\tta\) will give us the same result as the cofactor expansion along the first column of \(\ttat\text{.}\)
\begin{align*} \det(A^{-1}) \amp = \bvm -2\amp 1\\ 3/2 \amp -1/2\evm\\ \amp = 1-3/2\\ \amp =-1/2\text{.} \end{align*}
It seems as though
\begin{equation*} \det(A^{-1}) = \frac{1}{\det(A)}\text{.} \end{equation*}
We now state a few theorems that confirm our conjectures from the previous example.
To see that Theorem 6.4.11 is true, note that if \(A\) is not invertible, then the reduced row echelon form \(R\) of \(A\) must have a row of zeros. Performing a cofactor expansion along this row, we immediately see that \(\det(R)=0\text{.}\) Since \(R\) is obtained from \(A\) by a series of elementary row operations, we know from Theorem 6.4.6 that \(\det(A)\) is a multiple of \(\det(R)\text{,}\) and thus \(\det(A)=0\text{.}\)
It follows from Theorem 6.4.11 (using the logical principle known as the contrapositive) that if \(\det(A)\neq 0\text{,}\) we’re guaranteed that \(A\) is invertible.
At this point, we naturally should ask whether or not the converse to Theorem 6.4.11 is true as well: suppose we know \(\det(A)=0\text{.}\) Does that imply that \(A\) is not invertible? (Or equivalently, if we know \(A\) is invertible, does this imply that \(\det(A)\neq 0\text{?}\)) The answer is yes, but to see this, we first need a more general result.
Proving that \(\det(AB)=\det(A)\det(B)\) is most easily done using elementary matrices. (See Section 4.6.) Recall that multiplying a matrix on the left by an elementary matrix is the same as doing the corresponding row operation: if \(A\) is any \(3\times 3\) matrix, then \(EA\) can be obtained from \(A\) using the same row operation used to create \(E\text{.}\)
Theorem 6.4.6 then tells us that
\begin{equation*} \det(EB) = \det(E)\det(B) \end{equation*}
for any matrix \(B\) and elementary matrix \(E\text{.}\) The rest boils down to two cases: either \(\det(A)=0\text{,}\) in which case \(A\) is not invertible, so neither is \(AB\text{,}\) and thus \(\det(AB)=0 = 0\det(B)\text{,}\) or \(\det(A)\neq 0\text{.}\) In the latter case, \(A\) is invertible, and can be written as a product of elementary matrices. We can then prove that \(\det(AB)=\det(A)\det(B)\) by applying Theorem 6.4.6 repeatedly.
From Theorem 6.4.12, we see that \(\det(AB)=\det(A)\det(B)\) for any matrices \(A\) and \(B\text{.}\) What does this tell us in the case of an invertible matrix? Recall that if \(A\) is invertible, then we can determine the inverse matrix \(A^{-1}\) such that
\begin{equation*} AA^{-1} = I_n\text{.} \end{equation*}
Now, the identity matrix is triangular, and all of its diagonal entries are equal to 1, so we immediately see that \(\det(I_n) = 1\text{.}\) Thus, taking the determinant of both sides of the above equation, we have
\begin{equation*} \det(AA^{-1}) = \det(A)\det(A^{-1})=1\text{.} \end{equation*}
We have a product of two numbers equal to one, which tells us that neither of these numbers can be zero. (Otherwise, the product would be zero as well.) Thus, if \(A\) is invertible, it must be the case that \(\det(A)\neq 0\text{,}\) so a matrix \(A\) is invertible if and only if \(\det(A)\neq 0\text{.}\)
As an added bonus, we can rearrange the above equation to give us one more property of the determinant:
Combining Theorems Theorem 6.4.11 and Theorem 6.4.13 allows us to add on to our Invertible Matrix Theorem.
This new addition to the Invertible Matrix Theorem is very useful; we’ll refer back to it in Chapter 7 when we discuss eigenvalues.
In the next section we’ll see how the determinant can be used to solve systems of linear equations.

Exercises Exercises

Exercise Group.

Find the determinant of the given matrix using cofactor expansion along any row or column you choose.
1.
\(\bbm 1\amp 2\amp 3\\ -5\amp 0\amp 3\\ 4\amp 0\amp 6\ebm\)
2.
\(\bbm -4\amp 4\amp -4\\ 0\amp 0\amp -3\\ -2\amp -2\amp -1\ebm \)
3.
\(\bbm -4\amp 1\amp 1\\ 0\amp 0\amp 0\\ -1\amp -2\amp -5\ebm\)
4.
\(\bbm 0\amp -3\amp 1\\ 0\amp 0\amp 5\\ -4\amp 1\amp 0\ebm\)
5.
\(\bbm -2\amp -3\amp 5\\ 5\amp 2\amp 0\\ -1\amp 0\amp 0\ebm\)
6.
\(\bbm -2\amp -2\amp 0\\ 2\amp -5\amp -3\\ -5\amp 1\amp 0\ebm\)
7.
\(\bbm -3\amp 0\amp -5\\ -2\amp -3\amp 3\\ -1\amp 0\amp 1\ebm\)
8.
\(\bbm 0\amp 4\amp -4\\ 3\amp 1\amp -3\\ -3\amp -4\amp 0\ebm\)
9.
\(\bbm 5\amp -5\amp 0\amp 1\\ 2\amp 4\amp -1\amp -1\\ 5\amp 0\amp 0\amp 4\\ -1\amp -2\amp 0\amp 5\ebm\)
10.
\(\bbm -1\amp 3\amp 3\amp 4\\ 0\amp 0\amp 0\amp 0\\ 4\amp -5\amp -2\amp 0\\ 0\amp 0\amp 2\amp 0\ebm\)
11.
\(\bbm -5\amp -5\amp 0\amp -2\\ 0\amp 0\amp 5\amp 0\\ 1\amp 3\amp 3\amp 1\\ -4\amp -2\amp -1\amp -5\ebm\)
12.
\(\bbm -1\amp 0\amp -2\amp 5\\ 3\amp -5\amp 1\amp -2\\ -5\amp -2\amp -1\amp -3\\ -1\amp 0\amp 0\amp 0\ebm\)
13.
\(\bbm 4\amp 0\amp 5\amp 1\amp 0\\ 1\amp 0\amp 3\amp 1\amp 5\\ 2\amp 2\amp 0\amp 2\amp 2\\ 1\amp 0\amp 0\amp 0\amp 0\\ 4\amp 4\amp 2\amp 5\amp 3\ebm\)
14.
\(\bbm 2\amp 1\amp 1\amp 1\amp 1\\ 4\amp 1\amp 2\amp 0\amp 2\\ 0\amp 0\amp 1\amp 0\amp 0\\ 1\amp 3\amp 2\amp 0\amp 3\\ 5\amp 0\amp 5\amp 0\amp 4\ebm \)

Exercise Group.

A matrix \(\ttm\) and \(\det(\ttm)\) are given. Matrices \(\tta\text{,}\) \(\ttb\) and \(\ttc\) are formed by performing operations on \(\ttm\text{.}\) Determine the determinants of \(\tta\text{,}\) \(\ttb\) and \(\ttc\) using Theorem 6.4.6 and Theorem 6.4.12, and indicate the operations used to form \(\tta\text{,}\) \(\ttb\) and \(\ttc\text{.}\)
15.
\(\ttm=\bbm 9\amp 7\amp 8\\ 1\amp 3\amp 7\\ 6\amp 3\amp 3\ebm \text{,}\) \(\det(\ttm)=45\text{.}\)
  1. \(\displaystyle \tta = \bbm 18 \amp 14 \amp 16\\ 1\amp 3\amp 7\\ 6\amp 3\amp 3\ebm\)
  2. \(\displaystyle \ttb = \bbm 9\amp 7\amp 8\\ 1\amp 3\amp 7\\ 96\amp 73\amp 83\ebm\)
  3. \(\displaystyle \ttc = \bbm 9\amp 1\amp 6\\ 7\amp 3\amp 3\\ 8\amp 7\amp 3\ebm\)
16.
\(\ttm=\bbm 0\amp 3\amp 5\\ 3\amp 1\amp 0\\ -2\amp -4\amp -1\ebm\text{,}\) \(\det(\ttm)=-41\text{.}\)
  1. \(\displaystyle \tta = \bbm 0 \amp 3 \amp 5\\ -2\amp -4\amp -1\\ 3\amp 1\amp 0\ebm\)
  2. \(\displaystyle \ttb = \bbm 0\amp 3\amp 5\\ 3\amp 1\amp 0\\ 8\amp 16\amp 4\ebm\)
  3. \(\displaystyle \ttc = \bbm 3\amp 4\amp 5\\ 3\amp 1\amp 0\\ -2\amp -4\amp -1\ebm\)
17.
\(\ttm=\bbm 5\amp 1\amp 5\\ 4\amp 0\amp 2\\ 0\amp 0\amp 4\ebm\text{,}\) \(\det(\ttm)=-16\text{.}\)
  1. \(\displaystyle \tta = \bbm 0 \amp 0 \amp 4\\ 5\amp 1\amp 5\\ 4\amp 0\amp 2\ebm\)
  2. \(\displaystyle \ttb = \bbm -5\amp -1\amp -5\\ -4\amp 0\amp -2\\ 0\amp 0\amp 4\ebm\)
  3. \(\displaystyle \ttc = \bbm 15\amp 3\amp 15\\ 12\amp 0\amp 6\\ 0\amp 0\amp 12\ebm\)
18.
\(\ttm=\bbm 5\amp 4\amp 0\\ 7\amp 9\amp 3\\ 1\amp 3\amp 9\ebm\text{,}\) \(\det(\ttm)=120\text{.}\)
  1. \(\displaystyle \tta = \bbm 1 \amp 3 \amp 9\\ 7\amp 9\amp 3\\ 5\amp 4\amp 0\ebm\)
  2. \(\displaystyle \ttb = \bbm 5\amp 4\amp 0\\ 14\amp 18\amp 6\\ 3\amp 9\amp 27\ebm\)
  3. \(\displaystyle \ttc = \bbm -5\amp -4\amp 0\\ -7\amp -9\amp -3\\ -1\amp -3\amp -9\ebm\)

Exercise Group.

Matrices \(\tta\) and \(\ttb\) are given. Verify part 3 of Theorem 6.4.12 by computing \(\det(\tta)\text{,}\) \(\det(\ttb)\) and \(\det(\tta\ttb)\text{.}\)
19.
\(\tta = \bbm 2 \amp 0\ 1 \amp 2\ebm\text{,}\) \(\ttb = \bbm 0\amp -4\\ 1\amp 3\ebm\text{.}\)
20.
\(\tta = \bbm 3 \amp -1\ 4 \amp 1\ebm\text{,}\) \(\ttb = \bbm -4\amp -1\\ -5\amp 3\ebm\text{.}\)
21.
\(\tta = \bbm -4 \amp 4\ 5 \amp -2\ebm\text{,}\) \(\ttb = \bbm -3\amp -4\\ 5\amp -3\ebm\text{.}\)
22.
\(\tta = \bbm -3 \amp -1\ 2 \amp -3\ebm\text{,}\) \(\ttb = \bbm 0\amp 0\\ 4\amp -4\ebm\text{.}\)

Exercise Group.

Find the determinant of the given matrix.
23.
\(\bbm 3\amp 2\amp 3\\ -6\amp 1\amp -10\\ -8\amp -9\amp -9\ebm\)
24.
\(\bbm 8\amp -9\amp -2\\ -9\amp 9\amp -7\\ 5\amp -1\amp 9\ebm\)
25.
\(\bbm -4\amp 3\amp -4\\ -4\amp -5\amp 3\\ 3\amp -4\amp 5\ebm\)
26.
\(\bbm 1\amp -2\amp 1\\ 5\amp 5\amp 4\\ 4\amp 0\amp 0\ebm\)
27.
\(\bbm 1\amp -4\amp 1\\ 0\amp 3\amp 0\\ 1\amp 2\amp 2\ebm\)
28.
\(\bbm 3\amp -1\amp 0\\ -3\amp 0\amp -4\\ 0\amp -1\amp -4\ebm\)
29.
\(\bbm -5\amp 0\amp -4\\ 2\amp 4\amp -1\\ -5\amp 0\amp -4\ebm\)
30.
\(\bbm 1\amp 0\amp 0\\ 0\amp 1\amp 0\\ -1\amp 1\amp 1\ebm\)