
# 3.E: Exercises


## Exercise $$\PageIndex{1}$$

Find the determinants of the following matrices.

1. $$\left[\begin{array}{cc}1&3\\0&2\end{array}\right]$$
2. $$\left[\begin{array}{cc}0&3\\0&2\end{array}\right]$$
3. $$\left[\begin{array}{cc}4&3\\6&2\end{array}\right]$$

## Exercise $$\PageIndex{2}$$

Let $$A=\left[\begin{array}{ccc}1&2&4\\0&1&3\\-2&5&1\end{array}\right]$$. Find the following.

1. $$minor(A)_{11}$$
2. $$minor(A)_{21}$$
3. $$minor(A)_{32}$$
4. $$cof(A)_{11}$$
5. $$cof(A)_{21}$$
6. $$cof(A)_{32}$$

## Exercise $$\PageIndex{3}$$

Find the determinants of the following matrices.

1. $$\left[\begin{array}{ccc}1&2&3\\3&2&2\\0&9&8\end{array}\right]$$
2. $$\left[\begin{array}{ccc}4&3&2\\1&7&8\\3&-9&3\end{array}\right]$$
3. $$\left[\begin{array}{cccc}1&2&3&2\\1&3&2&3\\4&1&5&0\\1&2&1&2\end{array}\right]$$
1. The answer is $$31$$.
2. The answer is $$375$$.
3. The answer is $$-2$$.

## Exercise $$\PageIndex{4}$$

Find the following determinant by expanding along the first row and second column. $\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|\nonumber$

$\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|=6\nonumber$

## Exercise $$\PageIndex{5}$$

Find the following determinant by expanding along the first column and third row. $\left|\begin{array}{ccc}1&2&1\\1&0&1\\2&1&1\end{array}\right|\nonumber$

$\left|\begin{array}{ccc}1&2&1\\1&0&1\\2&1&1\end{array}\right|=2\nonumber$

## Exercise $$\PageIndex{6}$$

Find the following determinant by expanding along the second row and first column. $\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|\nonumber$

$\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|=6\nonumber$

## Exercise $$\PageIndex{7}$$

Compute the determinant by cofactor expansion. Pick the easiest row or column to use. $\left|\begin{array}{cccc}1&0&0&1\\2&1&1&0\\0&0&0&2\\2&1&3&1\end{array}\right|\nonumber$

$\left|\begin{array}{cccc}1&0&0&1\\2&1&1&0\\0&0&0&2\\2&1&3&1\end{array}\right|=-4\nonumber$

## Exercise $$\PageIndex{8}$$

Find the determinant of the following matrices.

1. $$A=\left[\begin{array}{cc}1&-34\\0&2\end{array}\right]$$
2. $$A=\left[\begin{array}{ccc}4&3&14\\0&-2&0\\0&0&5\end{array}\right]$$
3. $$A=\left[\begin{array}{cccc}2&3&15&0\\0&4&1&7\\0&0&-3&5\\0&0&0&1\end{array}\right]$$

## Exercise $$\PageIndex{9}$$

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. $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\to\cdots\to\left[\begin{array}{cc}a&c\\b&d\end{array}\right]\nonumber$

It does not change the determinant. This was just taking the transpose.

## Exercise $$\PageIndex{10}$$

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. $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\to\cdots\to\left[\begin{array}{cc}c&d\\a&b\end{array}\right]\nonumber$

In this case two rows were switched and so the resulting determinant is $$−1$$ times the first

## Exercise $$\PageIndex{11}$$

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. $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\to\cdots\to\left[\begin{array}{cc}a&b\\a+c&b+d\end{array}\right]\nonumber$

The determinant is unchanged. It was just the first row added to the second.

## Exercise $$\PageIndex{12}$$

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. $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\to\cdots\to\left[\begin{array}{cc}a&b\\2c&2d\end{array}\right]\nonumber$

The second row was multiplied by $$2$$ so the determinant of the result is $$2$$ times the original determinant.

## Exercise $$\PageIndex{13}$$

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. $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\to\cdots\to\left[\begin{array}{cc}b&a\\d&c\end{array}\right]\nonumber$

In this case the two columns were switched so the determinant of the second is $$−1$$ times the determinant of the first.

## Exercise $$\PageIndex{14}$$

Let $$A$$ be an $$r\times r$$ matrix and suppose there are $$r −1$$ rows (columns) such that all rows (columns) are linear combinations of these $$r −1$$ rows (columns). Show $$\det(A) = 0$$.

If the determinant is nonzero, then it will remain nonzero with row operations applied to the matrix. However, by assumption, you can obtain a row of zeros by doing row operations. Thus the determinant must have been zero after all.

## Exercise $$\PageIndex{15}$$

Show $$\det(aA) = a^n \det(A)$$ for an $$n\times n$$ matrix $$A$$ and scalar $$a$$.

$$\det(aA) = \det(aIA) = \det(aI)\det(A) = a^n \det(A)$$. The matrix which has a down the main diagonal has determinant equal to $$a^n$$.

## Exercise $$\PageIndex{16}$$

Construct $$2\times 2$$ matrices $$A$$ and $$B$$ to show that the $$\det A\det B = \det(AB)$$.

$\begin{array}{c}\det\left(\left[\begin{array}{cc}1&2\\3&4\end{array}\right]\left[\begin{array}{cc}-1&2\\-5&6\end{array}\right]\right)=-8 \\ \det\left[\begin{array}{cc}1&2\\3&4\end{array}\right]\det\left[\begin{array}{cc}-1&2\\-5&6\end{array}\right]=-2\times 4=-8\end{array}\nonumber$

## Exercise $$\PageIndex{17}$$

Is it true that $$\det(A+B) = \det(A)+\det(B)$$? If this is so, explain why. If it is not so, give a counter example.

This is not true at all. Consider $$A=\left[\begin{array}{cc}1&0\\0&1\end{array}\right],\: B=\left[\begin{array}{cc}-1&0\\0&-1\end{array}\right]$$.

## Exercise $$\PageIndex{18}$$

An $$n\times n$$ matrix is called nilpotent if for some positive integer, $$k$$ it follows $$A^k = 0$$. If $$A$$ is a nilpotent matrix and $$k$$ is the smallest possible integer such that $$A^k = 0$$, what are the possible values of $$\det(A)$$?

It must be $$0$$ because $$0 = \det(0) = \det (A^k) = (\det(A))^k$$.

## Exercise $$\PageIndex{19}$$

A matrix is said to be orthogonal if $$A^TA = I$$. Thus the inverse of an orthogonal matrix is just its transpose. What are the possible values of $$\det(A)$$ if $$A$$ is an orthogonal matrix?

You would need $$\det (AA^T) = \det(A)\det (A^T) = \det(A)^2 = 1$$ and so $$\det(A) = 1$$, or $$-1$$.

## Exercise $$\PageIndex{20}$$

Let $$A$$ and $$B$$ be two $$n\times n$$ matrices. $$A ∼ B$$ ($$A$$ is similar to $$B$$) means there exists an invertible matrix $$P$$ such that $$A = P^{−1}BP$$. Show that if $$A ∼ B$$, then $$\det(A) = \det(B)$$.

$$\det(A) = \det(S^{−1}BS) = \det(S^{−1})\det(B)\det(S) = \det(B)\det(S^{−1}S) = \det(B)$$.

## Exercise $$\PageIndex{21}$$

Tell whether each statement is true or false. If true, provide a proof. If false, provide a counter example.

1. If A is a $$3\times 3$$ matrix with a zero determinant, then one column must be a multiple of some other column.
2. If any two columns of a square matrix are equal, then the determinant of the matrix equals zero.
3. For two $$n\times n$$ matrices $$A$$ and $$B$$, $$\det(A+B) = \det(A) +\det(B)$$.
4. For an $$n\times n$$ matrix $$A$$, $$\det(3A) = 3 \det(A)$$
5. If $$A^{−1}$$ exists then $$\det(A^{−1}) = \det(A)^{−1}$$.
6. If $$B$$ is obtained by multiplying a single row of $$A$$ by $$4$$ then $$\det(B) = 4 \det(A)$$.
7. For $$A$$ an $$n\times n$$ matrix, $$\det(−A) = (−1)^n \det(A)$$.
8. If $$A$$ is a real $$n\times n$$ matrix, then $$\det (A^TA) ≥ 0$$.
9. If $$A^k = 0$$ for some positive integer $$k$$, then $$\det(A) = 0$$.
10. If $$AX = 0$$ for some $$X\neq 0$$, then $$\det(A) = 0$$.
1. False. Consider $$\left[\begin{array}{ccc}1&1&2\\-1&5&4\\0&3&3\end{array}\right]$$
2. True.
3. False.
4. False.
5. True.
6. False.
7. True.
8. True.
9. True.
10. True.

## Exercise $$\PageIndex{22}$$

Find the determinant using row operations to first simplify. $\left|\begin{array}{ccc}1&2&1\\2&3&2\\-4&1&2\end{array}\right|\nonumber$

$\left|\begin{array}{ccc}1&2&1\\2&3&2\\-4&1&2\end{array}\right|=-6\nonumber$

## Exercise $$\PageIndex{23}$$

Find the determinant using row operations to first simplify. $\left|\begin{array}{ccc}2&1&3\\2&4&2\\1&4&-5\end{array}\right|\nonumber$

$\left|\begin{array}{ccc}2&1&3\\2&4&2\\1&4&-5\end{array}\right|=-32\nonumber$

## Exercise $$\PageIndex{24}$$

Find the determinant using row operations to first simplify. $\left|\begin{array}{cccc}1&2&1&2\\3&1&-2&3\\-1&0&3&1\\2&3&2&-2\end{array}\right|\nonumber$

One can row reduce this using only row operation 3 to $\left[\begin{array}{cccc}1&2&1&2\\0&-5&-5&-3 \\ 0&0&2&\frac{9}{5} \\ 0&0&0&-\frac{63}{10}\end{array}\right]\nonumber$ and therefore, the determinant is $$-63$$. $\left|\begin{array}{cccc}1&2&1&2\\3&1&-2&3\\-1&0&3&1\\2&3&2&-2\end{array}\right|=63\nonumber$

## Exercise $$\PageIndex{25}$$

Find the determinant using row operations to first simplify. $\left|\begin{array}{cccc}1&4&1&2\\3&2&-2&3\\-1&0&3&3\\2&1&2&-2\end{array}\right|\nonumber$

One can row reduce this using only row operation 3 to $\left[\begin{array}{cccc}1&4&1&2\\0&-10&-5&-3 \\ 0&0&2&\frac{19}{5} \\ 0&0&0&-\frac{211}{20}\end{array}\right]\nonumber$ Thus the determinant is given by $\left|\begin{array}{cccc}1&4&1&2\\3&2&-2&3\\-1&0&3&3\\2&1&2&-2\end{array}\right|=211\nonumber$

## Exercise $$\PageIndex{26}$$

Let $A=\left[\begin{array}{ccc}1&2&3\\0&2&1\\3&1&0\end{array}\right]\nonumber$ Determine whether the matrix $$A$$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse which involves the cofactor matrix.

$$\det\left[\begin{array}{ccc}1&2&3\\0&2&1\\3&1&0\end{array}\right]=-13$$ and so it has an inverse. This inverse is \begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0&2\\3&1\end{array}\right| \\ -\left|\begin{array}{cc}2&3\\1&0\end{array}\right| &\left|\begin{array}{cc}1&3\\3&0\end{array}\right| &-\left|\begin{array}{cc}1&2\\3&1\end{array}\right| \\ \left|\begin{array}{cc}2&3\\2&1\end{array}\right|&-\left|\begin{array}{cc}1&3\\0&1\end{array}\right|&\left|\begin{array}{cc}1&2\\0&2\end{array}\right|\end{array}\right]^T &=\frac{1}{-13}\left[\begin{array}{ccc}-1&3&-6 \\ 3&-9&5 \\ -4&-1&2\end{array}\right]^T \\ &=\left[\begin{array}{ccc}\frac{1}{13}&-\frac{3}{13}&\frac{4}{13} \\ -\frac{3}{13}&\frac{9}{13}&\frac{1}{13} \\ \frac{6}{13}&-\frac{5}{13}&-\frac{2}{13}\end{array}\right]\end{aligned}

## Exercise $$\PageIndex{27}$$

Let $A=\left[\begin{array}{ccc}1&2&0\\0&2&1\\3&1&1\end{array}\right]\nonumber$ Determine whether the matrix $$A$$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

$$\det\left[\begin{array}{ccc}1&2&0\\0&2&1\\3&1&1\end{array}\right]=7$$ so it has an inverse. This inverse is $$\frac{1}{7}\left[\begin{array}{ccc}1&3&-6\\-2&1&5\\2&-1&2\end{array}\right]^T=\left[\begin{array}{ccc}\frac{1}{7}&-\frac{2}{7}&\frac{2}{7} \\ \frac{3}{7}&\frac{1}{7}&-\frac{1}{7} \\ -\frac{6}{7}&\frac{5}{7}&\frac{2}{7}\end{array}\right]$$

## Exercise $$\PageIndex{28}$$

Let $A=\left[\begin{array}{ccc}1&3&3\\2&4&1\\0&1&1\end{array}\right]\nonumber$ Determine whether the matrix $$A$$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

$\det\left[\begin{array}{ccc}1&3&3\\2&4&1\\0&1&1\end{array}\right]=3\nonumber$ so it has an inverse which is $\left[\begin{array}{ccc}1&0&-3 \\ -\frac{2}{3}&\frac{1}{3}&\frac{5}{3} \\ \frac{2}{3}&-\frac{1}{3}&-\frac{2}{3}\end{array}\right]\nonumber$

## Exercise $$\PageIndex{29}$$

Let $A=\left[\begin{array}{ccc}1&2&3\\0&2&1\\2&6&7\end{array}\right]\nonumber$ Determine whether the matrix $$A$$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

## Exercise $$\PageIndex{30}$$

Let $A=\left[\begin{array}{ccc}1&0&3\\1&0&1\\3&1&0\end{array}\right]\nonumber$ Determine whether the matrix $$A$$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

$\det\left[\begin{array}{ccc}1&0&3\\1&0&1\\3&1&0\end{array}\right]=2\nonumber$ and so it has an inverse. The inverse turns out to equal $\left[\begin{array}{ccc}-\frac{1}{2}&\frac{3}{2}&0 \\ \frac{3}{2}&-\frac{9}{2}&1 \\ \frac{1}{2}&-\frac{1}{2}&0\end{array}\right]\nonumber$

## Exercise $$\PageIndex{31}$$

For the following matrices, determine if they are invertible. If so, use the formula for the inverse in terms of the cofactor matrix to find each inverse. If the inverse does not exist, explain why.

1. $$\left[\begin{array}{cc}1&1\\1&2\end{array}\right]$$
2. $$\left[\begin{array}{ccc}1&2&3\\0&2&1\\4&1&1\end{array}\right]$$
3. $$\left[\begin{array}{ccc}1&2&1\\2&3&0\\0&1&2\end{array}\right]$$
1. $$\left|\begin{array}{cc}1&1\\1&2\end{array}\right|=1$$
2. $$\left|\begin{array}{ccc}1&2&3\\0&2&1\\4&1&1\end{array}\right|=-15$$
3. $$\left|\begin{array}{ccc}1&2&1\\2&3&0\\0&1&2\end{array}\right|=0$$

## Exercise $$\PageIndex{32}$$

Consider the matrix $A=\left[\begin{array}{ccc}1&0&0\\0&\cos t&-\sin t \\ 0&\sin t&\cos t\end{array}\right]\nonumber$ Does there exist a value of $$t$$ for which this matrix fails to have an inverse? Explain.

No. It has a nonzero determinant for all $$t$$

## Exercise $$\PageIndex{33}$$

Consider the matrix $A=\left[\begin{array}{ccc}1&t&t^2 \\ 0&1&2t \\ t&0&2\end{array}\right]\nonumber$ Does there exist a value of $$t$$ for which this matrix fails to have an inverse? Explain.

$\det\left[\begin{array}{ccc}1&t&t^2 \\ 0&1&2t \\ t&0&2\end{array}\right]=t^3+2\nonumber$ and so it has no inverse when $$t=-\sqrt[3]{2}$$

## Exercise $$\PageIndex{34}$$

Consider the matrix $A=\left[\begin{array}{ccc}e^t &\cosh t&\sinh t \\ e^t&\sinh t&\cosh t \\ e^t&\cosh t&\sinh t\end{array}\right]\nonumber$ Does there exist a value of t for which this matrix fails to have an inverse? Explain.

$\det\left[\begin{array}{ccc}e^t&\cosh t&\sinh t \\ e^t&\sinh t&\cosh t \\ e^t&\cosh t&\sinh t\end{array}\right]=0\nonumber$ and so this matrix fails to have a nonzero determinant at any value of $$t$$.

## Exercise $$\PageIndex{35}$$

Consider the matrix $A=\left[\begin{array}{ccc}e^t &e^{-t}\cos t&e^{-t}\sin t \\ e^t&-e^{-t}\cos t-e^{-t}\sin t &-e^{-t}\sin t+e^{-t}\cos t \\ e^t&2e^{-t}\sin t&-2e^{-t}\cos t\end{array}\right]\nonumber$ Does there exist a value of t for which this matrix fails to have an inverse? Explain.

$\det\left[\begin{array}{ccc}e^t&e^{-t}\cos t&e^{-t}\sin t \\ e^t&-e^{-t}\cos t-e^{-t}\sin t&-e^{-t}\sin t+e^{-t}\cos t \\ e^t&2e^{-t}\sin t&-2e^{-t}\cos t\end{array}\right]=5e^{-t}\neq 0\nonumber$ and so this matrix is always invertible.

## Exercise $$\PageIndex{36}$$

Show that if $$\det(A)\neq 0$$ for $$A$$ an $$n\times n$$ matrix, it follows that if $$AX = 0$$, then $$X = 0$$.

If $$\det(A) \neq 0$$, then $$A^{−1}$$ exists and so you could multiply on both sides on the left by $$A^{−1}$$ and obtain that $$X = 0$$.

## Exercise $$\PageIndex{37}$$

Suppose $$A,B$$ are $$n\times n$$ matrices and that $$AB = I$$. Show that then $$BA = I$$. Hint: First explain why $$\det(A)$$, $$\det(B)$$ are both nonzero. Then $$(AB)A = A$$ and then show $$BA(BA−I) = 0$$. From this use what is given to conclude $$A(BA−I) = 0$$. Then use Exercise $$\PageIndex{36}$$.

You have $$1 = \det(A)\det(B)$$. Hence both $$A$$ and $$B$$ have inverses. Letting $$X$$ be given, $A(BA−I)X = (AB)AX −AX = AX −AX = 0\nonumber$ and so it follows from the above problem that $$(BA−I)X = 0$$. Since $$X$$ is arbitrary, it follows that $$BA = I$$.

## Exercise $$\PageIndex{38}$$

Use the formula for the inverse in terms of the cofactor matrix to find the inverse of the matrix $A=\left[\begin{array}{ccc}e^t&0&0 \\ 0&e^t\cos t&e^t\sin t \\ 0&e^t\cos t-e^t\sin t&e^t\cos t+e^t\sin t\end{array}\right]\nonumber$

$\det\left[\begin{array}{ccc}e^t&0&0 \\ 0&e^t\cos t&e^t\sin t \\ 0&e^t\cos t-e^t\sin t&e^t\cos t+e^t\sin t\end{array}\right]=e^{3t}.\nonumber$ Hence the inverse is $\begin{array}{l}e^{-3t}\left[\begin{array}{ccc}e^{2t}&0&0 \\ 0&e^{2t}\cos t+e^{2t}\sin t&-(e^{2t}\cos t-e^{2t}\sin )t \\ 0&-e^{2t}\sin t&e^{2t}\cos (t)\end{array}\right]^T \\ =\left[\begin{array}{ccc}e^{-t}&0&0 \\ 0&e^{-t}(\cos t+\sin t)&-(\sin t)e^{-t} \\ 0&-e^{-t}(\cos t-\sin t)&(\cos t)e^{-t}\end{array}\right] \end{array}\nonumber$

## Exercise $$\PageIndex{39}$$

Find the inverse, if it exists, of the matrix $A=\left[\begin{array}{ccc}e^t&\cos t&\sin t \\ e^t&-\sin t&\cos t \\ e^t&-\cos t&-\sin t\end{array}\right]\nonumber$

$\begin{array}{l} \left[\begin{array}{ccc}e^t&\cos t&\sin t \\ e^t&-\sin t&\cos t \\ e^t&-\cos t&-\sin t\end{array}\right]^{-1} \\ =\left[\begin{array}{ccc}\frac{1}{2}e^{-t}&0&\frac{1}{2}e^{-t} \\ \frac{1}{2}\cos t+\frac{1}{2}\sin t&-\sin t&\frac{1}{2}\sin t-\frac{1}{2}\cos t \\ \frac{1}{2}\sin t-\frac{1}{2}\cos t&\cos t&-\frac{1}{2}\cos t-\frac{1}{2}\sin t\end{array}\right]\end{array}\nonumber$

## Exercise $$\PageIndex{40}$$

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

## Exercise $$\PageIndex{41}$$

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.

This follows because $$\det(ABC) = \det(A)\det(B)\det(C)$$ and if this product is nonzero, then each determinant in the product is nonzero and so each of these matrices is invertible.

## Exercise $$\PageIndex{42}$$

Decide if this statement is true or false: Cramer’s rule is useful for finding solutions to systems of linear equations in which there is an infinite set of solutions.

False.

## Exercise $$\PageIndex{43}$$

Use Cramer’s rule to find the solution to \begin{aligned}x+2y&=1 \\ 2x-y&=2\end{aligned}

Solution is: $$[x = 1, y = 0]$$

## Exercise $$\PageIndex{44}$$

Use Cramer’s rule to find the solution to $\begin{array}{c}x+2y+z=1 \\ 2x-y-z=2 \\ x+z=1\end{array}\nonumber$

Solution is: $$[x = 1, y = 0,z = 0]$$. For example, $y=\frac{\left|\begin{array}{ccc}1&1&1\\2&2&-1\\1&1&1\end{array}\right|}{\left|\begin{array}{ccc}1&2&1\\2&-1&-1\\1&0&1\end{array}\right|}=0\nonumber$