# Linear Algebra/Exploration/Solutions

## Solutions

This exercise is recommended for all readers.
Problem 1

Evaluate the determinant of each.

1. ${\displaystyle {\begin{pmatrix}3&1\\-1&1\end{pmatrix}}}$
2. ${\displaystyle {\begin{pmatrix}2&0&1\\3&1&1\\-1&0&1\end{pmatrix}}}$
3. ${\displaystyle {\begin{pmatrix}4&0&1\\0&0&1\\1&3&-1\end{pmatrix}}}$
1. ${\displaystyle 4}$
2. ${\displaystyle 3}$
3. ${\displaystyle -12}$
Problem 2

Evaluate the determinant of each.

1. ${\displaystyle {\begin{pmatrix}2&0\\-1&3\end{pmatrix}}}$
2. ${\displaystyle {\begin{pmatrix}2&1&1\\0&5&-2\\1&-3&4\end{pmatrix}}}$
3. ${\displaystyle {\begin{pmatrix}2&3&4\\5&6&7\\8&9&1\end{pmatrix}}}$
1. ${\displaystyle 6}$
2. ${\displaystyle 21}$
3. ${\displaystyle 27}$
This exercise is recommended for all readers.
Problem 3

Verify that the determinant of an upper-triangular ${\displaystyle 3\!\times \!3}$ matrix is the product down the diagonal.

${\displaystyle \det({\begin{pmatrix}a&b&c\\0&e&f\\0&0&i\end{pmatrix}})=aei}$

Do lower-triangular matrices work the same way?

For the first, apply the formula in this section, note that any term with a ${\displaystyle d}$, ${\displaystyle g}$, or ${\displaystyle h}$ is zero, and simplify. Lower-triangular matrices work the same way.

This exercise is recommended for all readers.
Problem 4

Use the determinant to decide if each is singular or nonsingular.

1. ${\displaystyle {\begin{pmatrix}2&1\\3&1\end{pmatrix}}}$
2. ${\displaystyle {\begin{pmatrix}0&1\\1&-1\end{pmatrix}}}$
3. ${\displaystyle {\begin{pmatrix}4&2\\2&1\end{pmatrix}}}$
1. Nonsingular, the determinant is ${\displaystyle -1}$.
2. Nonsingular, the determinant is ${\displaystyle -1}$.
3. Singular, the determinant is ${\displaystyle 0}$.
Problem 5

Singular or nonsingular? Use the determinant to decide.

1. ${\displaystyle {\begin{pmatrix}2&1&1\\3&2&2\\0&1&4\end{pmatrix}}}$
2. ${\displaystyle {\begin{pmatrix}1&0&1\\2&1&1\\4&1&3\end{pmatrix}}}$
3. ${\displaystyle {\begin{pmatrix}2&1&0\\3&-2&0\\1&0&0\end{pmatrix}}}$
1. Nonsingular, the determinant is ${\displaystyle 3}$.
2. Singular, the determinant is ${\displaystyle 0}$.
3. Singular, the determinant is ${\displaystyle 0}$.
This exercise is recommended for all readers.
Problem 6

Each pair of matrices differ by one row operation. Use this operation to compare ${\displaystyle \det(A)}$ with ${\displaystyle \det(B)}$.

1. ${\displaystyle A={\begin{pmatrix}1&2\\2&3\end{pmatrix}}}$ ${\displaystyle B={\begin{pmatrix}1&2\\0&-1\end{pmatrix}}}$
2. ${\displaystyle A={\begin{pmatrix}3&1&0\\0&0&1\\0&1&2\end{pmatrix}}}$ ${\displaystyle B={\begin{pmatrix}3&1&0\\0&1&2\\0&0&1\end{pmatrix}}}$
3. ${\displaystyle A={\begin{pmatrix}1&-1&3\\2&2&-6\\1&0&4\end{pmatrix}}}$ ${\displaystyle B={\begin{pmatrix}1&-1&3\\1&1&-3\\1&0&4\end{pmatrix}}}$
1. ${\displaystyle \det(B)=\det(A)}$ via ${\displaystyle -2\rho _{1}+\rho _{2}}$
2. ${\displaystyle \det(B)=-\det(A)}$ via ${\displaystyle \rho _{2}\leftrightarrow \rho _{3}}$
3. ${\displaystyle \det(B)=(1/2)\cdot \det(A)}$ via ${\displaystyle (1/2)\rho _{2}}$
Problem 7

Show this.

${\displaystyle \det({\begin{pmatrix}1&1&1\\a&b&c\\a^{2}&b^{2}&c^{2}\end{pmatrix}})=(b-a)(c-a)(c-b)}$

Using the formula for the determinant of a ${\displaystyle 3\!\times \!3}$ matrix we expand the left side

${\displaystyle 1\cdot b\cdot c^{2}+1\cdot c\cdot a^{2}+1\cdot a\cdot b^{2}-b^{2}\cdot c\cdot 1-c^{2}\cdot a\cdot 1-a^{2}\cdot b\cdot 1}$

and by distributing we expand the right side.

${\displaystyle (bc-ba-ac+a^{2})\cdot (c-b)=c^{2}b-b^{2}c-bac+b^{2}a-ac^{2}+acb+a^{2}c-a^{2}b}$

Now we can just check that the two are equal. (Remark. This is the ${\displaystyle 3\!\times \!3}$ case of Vandermonde's determinant which arises in applications).

This exercise is recommended for all readers.
Problem 8

Which real numbers ${\displaystyle x}$ make this matrix singular?

${\displaystyle {\begin{pmatrix}12-x&4\\8&8-x\end{pmatrix}}}$

This equation

${\displaystyle 0=\det({\begin{pmatrix}12-x&4\\8&8-x\end{pmatrix}})=64-20x+x^{2}=(x-16)(x-4)}$

has roots ${\displaystyle x=16}$ and ${\displaystyle x=4}$.

Problem 9

Do the Gaussian reduction to check the formula for ${\displaystyle 3\!\times \!3}$ matrices stated in the preamble to this section.

${\displaystyle {\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}}}$ is nonsingular iff ${\displaystyle aei+bfg+cdh-hfa-idb-gec\neq 0}$

We first reduce the matrix to echelon form. To begin, assume that ${\displaystyle a\neq 0}$ and that ${\displaystyle ae-bd\neq 0}$.

${\displaystyle {\begin{array}{rcl}{\xrightarrow[{}]{(1/a)\rho _{1}}}\;{\begin{pmatrix}1&b/a&c/a\\d&e&f\\g&h&i\end{pmatrix}}&{\xrightarrow[{-g\rho _{1}+\rho _{3}}]{-d\rho _{1}+\rho _{2}}}&{\begin{pmatrix}1&b/a&c/a\\0&(ae-bd)/a&(af-cd)/a\\0&(ah-bg)/a&(ai-cg)/a\end{pmatrix}}\\&{\xrightarrow[{}]{(a/(ae-bd))\rho _{2}}}&{\begin{pmatrix}1&b/a&c/a\\0&1&(af-cd)/(ae-bd)\\0&(ah-bg)/a&(ai-cg)/a\end{pmatrix}}\end{array}}}$

This step finishes the calculation.

${\displaystyle {\xrightarrow[{}]{((ah-bg)/a)\rho _{2}+\rho _{3}}}{\begin{pmatrix}1&b/a&c/a\\0&1&(af-cd)/(ae-bd)\\0&0&(aei+bgf+cdh-hfa-idb-gec)/(ae-bd)\end{pmatrix}}}$

Now assuming that ${\displaystyle a\neq 0}$ and ${\displaystyle ae-bd\neq 0}$, the original matrix is nonsingular if and only if the ${\displaystyle 3,3}$ entry above is nonzero. That is, under the assumptions, the original matrix is nonsingular if and only if ${\displaystyle aei+bgf+cdh-hfa-idb-gec\neq 0}$, as required.

We finish by running down what happens if the assumptions that were taken for convienence in the prior paragraph do not hold. First, if ${\displaystyle a\neq 0}$ but ${\displaystyle ae-bd=0}$ then we can swap

${\displaystyle {\begin{pmatrix}1&b/a&c/a\\0&0&(af-cd)/a\\0&(ah-bg)/a&(ai-cg)/a\end{pmatrix}}{\xrightarrow[{}]{\rho _{2}\leftrightarrow \rho _{3}}}{\begin{pmatrix}1&b/a&c/a\\0&(ah-bg)/a&(ai-cg)/a\\0&0&(af-cd)/a\end{pmatrix}}}$

and conclude that the matrix is nonsingular if and only if either ${\displaystyle ah-bg=0}$ or ${\displaystyle af-cd=0}$. The condition "${\displaystyle ah-bg=0}$ or ${\displaystyle af-cd=0}$" is equivalent to the condition "${\displaystyle (ah-bg)(af-cd)=0}$". Multiplying out and using the case assumption that ${\displaystyle ae-bd=0}$ to substitute ${\displaystyle ae}$ for ${\displaystyle bd}$ gives this.

${\displaystyle 0=ahaf-ahcd-bgaf+bgcd=ahaf-ahcd-bgaf+aegc=a(haf-hcd-bgf+egc)}$

Since ${\displaystyle a\neq 0}$, we have that the matrix is nonsingular if and only if ${\displaystyle haf-hcd-bgf+egc=0}$. Therefore, in this ${\displaystyle a\neq 0}$ and ${\displaystyle ae-bd=0}$ case, the matrix is nonsingular when ${\displaystyle haf-hcd-bgf+egc-i(ae-bd)=0}$.

The remaining cases are routine. Do the ${\displaystyle a=0}$ but ${\displaystyle d\neq 0}$ case and the ${\displaystyle a=0}$ and ${\displaystyle d=0}$ but ${\displaystyle g\neq 0}$ case by first swapping rows and then going on as above. The ${\displaystyle a=0}$, ${\displaystyle d=0}$, and ${\displaystyle g=0}$ case is easy— that matrix is singular since the columns form a linearly dependent set, and the determinant comes out to be zero.

Problem 10

Show that the equation of a line in ${\displaystyle \mathbb {R} ^{2}}$ thru ${\displaystyle (x_{1},y_{1})}$ and ${\displaystyle (x_{2},y_{2})}$ is expressed by this determinant.

${\displaystyle \det({\begin{pmatrix}x&y&1\\x_{1}&y_{1}&1\\x_{2}&y_{2}&1\end{pmatrix}})=0\qquad x_{1}\neq x_{2}}$

Figuring the determinant and doing some algebra gives this.

${\displaystyle {\begin{array}{rl}0&=y_{1}x+x_{2}y+x_{1}y_{2}-y_{2}x-x_{1}y-x_{2}y_{1}\\(x_{2}-x_{1})\cdot y&=(y_{2}-y_{1})\cdot x+x_{2}y_{1}-x_{1}y_{2}\\y&={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}\cdot x+{\frac {x_{2}y_{1}-x_{1}y_{2}}{x_{2}-x_{1}}}\end{array}}}$

Note that this is the equation of a line (in particular, in contains the familiar expression for the slope), and note that ${\displaystyle (x_{1},y_{1})}$ and ${\displaystyle (x_{2},y_{2})}$ satisfy it.

This exercise is recommended for all readers.
Problem 11

Many people know this mnemonic for the determinant of a ${\displaystyle 3\!\times \!3}$ matrix: first repeat the first two columns and then sum the products on the forward diagonals and subtract the products on the backward diagonals. That is, first write

${\displaystyle \left({\begin{array}{ccc|cc}h_{1,1}&h_{1,2}&h_{1,3}&h_{1,1}&h_{1,2}\\h_{2,1}&h_{2,2}&h_{2,3}&h_{2,1}&h_{2,2}\\h_{3,1}&h_{3,2}&h_{3,3}&h_{3,1}&h_{3,2}\end{array}}\right)}$

and then calculate this.

${\displaystyle {\begin{array}{l}h_{1,1}h_{2,2}h_{3,3}+h_{1,2}h_{2,3}h_{3,1}+h_{1,3}h_{2,1}h_{3,2}\\\quad -h_{3,1}h_{2,2}h_{1,3}-h_{3,2}h_{2,3}h_{1,1}-h_{3,3}h_{2,1}h_{1,2}\end{array}}}$
1. Check that this agrees with the formula given in the preamble to this section.
2. Does it extend to other-sized determinants?
1. The comparison with the formula given in the preamble to this section is easy.
2. While it holds for ${\displaystyle 2\!\times \!2}$ matrices
${\displaystyle {\begin{array}{rl}\left({\begin{array}{cc|c}h_{1,1}&h_{1,2}&h_{1,1}\\h_{2,1}&h_{2,2}&h_{2,1}\end{array}}\right)&={\begin{array}{l}h_{1,1}h_{2,2}+h_{1,2}h_{2,1}\\\quad -h_{2,1}h_{1,2}-h_{2,2}h_{1,1}\end{array}}\\&=h_{1,1}h_{2,2}-h_{1,2}h_{2,1}\end{array}}}$
it does not hold for ${\displaystyle 4\!\times \!4}$ matrices. An example is that this matrix is singular because the second and third rows are equal
${\displaystyle {\begin{pmatrix}1&0&0&1\\0&1&1&0\\0&1&1&0\\-1&0&0&1\end{pmatrix}}}$
but following the scheme of the mnemonic does not give zero.
${\displaystyle \left({\begin{array}{cccc|ccc}1&0&0&1&1&0&0\\0&1&1&0&0&1&1\\0&1&1&0&0&1&1\\-1&0&0&1&-1&0&0\end{array}}\right)={\begin{array}{l}1+0+0+0\\\quad -(-1)-0-0-0\end{array}}}$
Problem 12

The cross product of the vectors

${\displaystyle {\vec {x}}={\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}\qquad {\vec {y}}={\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}}$

is the vector computed as this determinant.

${\displaystyle {\vec {x}}\times {\vec {y}}=\det({\begin{pmatrix}{\vec {e}}_{1}&{\vec {e}}_{2}&{\vec {e}}_{3}\\x_{1}&x_{2}&x_{3}\\y_{1}&y_{2}&y_{3}\end{pmatrix}})}$

Note that the first row is composed of vectors, the vectors from the standard basis for ${\displaystyle \mathbb {R} ^{3}}$. Show that the cross product of two vectors is perpendicular to each vector.

The determinant is ${\displaystyle (x_{2}y_{3}-x_{3}y_{2}){\vec {e}}_{1}+(x_{3}y_{1}-x_{1}y_{3}){\vec {e}}_{2}+(x_{1}y_{2}-x_{2}y_{1}){\vec {e}}_{3}}$. To check perpendicularity, we check that the dot product with the first vector is zero

${\displaystyle {\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}\cdot {\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\x_{3}y_{1}-x_{1}y_{3}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}}=x_{1}x_{2}y_{3}-x_{1}x_{3}y_{2}+x_{2}x_{3}y_{1}-x_{1}x_{2}y_{3}+x_{1}x_{3}y_{2}-x_{2}x_{3}y_{1}=0}$

and the dot product with the second vector is also zero.

${\displaystyle {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}\cdot {\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\x_{3}y_{1}-x_{1}y_{3}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}}=x_{2}y_{1}y_{3}-x_{3}y_{1}y_{2}+x_{3}y_{1}y_{2}-x_{1}y_{2}y_{3}+x_{1}y_{2}y_{3}-x_{2}y_{1}y_{3}=0}$
Problem 13

Prove that each statement holds for ${\displaystyle 2\!\times \!2}$ matrices.

1. The determinant of a product is the product of the determinants ${\displaystyle \det(ST)=\det(S)\cdot \det(T)}$.
2. If ${\displaystyle T}$ is invertible then the determinant of the inverse is the inverse of the determinant ${\displaystyle \det(T^{-1})=(\,\det(T)\,)^{-1}}$.

Matrices ${\displaystyle T}$ and ${\displaystyle T^{\prime }}$ are similar if there is a nonsingular matrix ${\displaystyle P}$ such that ${\displaystyle T^{\prime }=PTP^{-1}}$. (This definition is in Chapter Five.) Show that similar ${\displaystyle 2\!\times \!2}$ matrices have the same determinant.

1. Plug and chug: the determinant of the product is this
${\displaystyle {\begin{array}{rl}\det({\begin{pmatrix}a&b\\c&d\end{pmatrix}}{\begin{pmatrix}w&x\\y&z\end{pmatrix}})&=\det({\begin{pmatrix}aw+by&ax+bz\\cw+dy&cx+dz\end{pmatrix}})\\&={\begin{array}{l}acwx+adwz+bcxy+bdyz\\\quad -acwx-bcwz-adxy-bdyz\end{array}}\end{array}}}$
while the product of the determinants is this.
${\displaystyle \det({\begin{pmatrix}a&b\\c&d\end{pmatrix}})\cdot \det({\begin{pmatrix}w&x\\y&z\end{pmatrix}})=(ad-bc)\cdot (wz-xy)}$
Verification that they are equal is easy.
2. Use the prior item.

That similar matrices have the same determinant is immediate from the above two: ${\displaystyle \det(PTP^{-1})=\det(P)\cdot \det(T)\cdot \det(P^{-1})}$.

This exercise is recommended for all readers.
Problem 14

Prove that the area of this region in the plane

is equal to the value of this determinant.

${\displaystyle \det({\begin{pmatrix}x_{1}&x_{2}\\y_{1}&y_{2}\end{pmatrix}})}$

Compare with this.

${\displaystyle \det({\begin{pmatrix}x_{2}&x_{1}\\y_{2}&y_{1}\end{pmatrix}})}$

One way is to count these areas

by taking the area of the entire rectangle and subtracting the area of ${\displaystyle A}$ the upper-left rectangle, ${\displaystyle B}$ the upper-middle triangle, ${\displaystyle D}$ the upper-right triangle, ${\displaystyle C}$ the lower-left triangle, ${\displaystyle E}$ the lower-middle triangle, and ${\displaystyle F}$ the lower-right rectangle ${\displaystyle (x_{1}+x_{2})(y_{1}+y_{2})-x_{2}y_{1}-(1/2)x_{1}y_{1}-(1/2)x_{2}y_{2}-(1/2)x_{2}y_{2}-(1/2)x_{1}y_{1}-x_{2}y_{1}}$. Simplification gives the determinant formula.

This determinant is the negative of the one above; the formula distinguishes whether the second column is counterclockwise from the first.

Problem 15

Prove that for ${\displaystyle 2\!\times \!2}$ matrices, the determinant of a matrix equals the determinant of its transpose. Does that also hold for ${\displaystyle 3\!\times \!3}$ matrices?

The computation for ${\displaystyle 2\!\times \!2}$ matrices, using the formula quoted in the preamble, is easy. It does also hold for ${\displaystyle 3\!\times \!3}$ matrices; the computation is routine.

This exercise is recommended for all readers.
Problem 16

Is the determinant function linear — is ${\displaystyle \det(x\cdot T+y\cdot S)=x\cdot \det(T)+y\cdot \det(S)}$?

No. Recall that constants come out one row at a time.

${\displaystyle \det({\begin{pmatrix}2&4\\2&6\\\end{pmatrix}})=2\cdot \det({\begin{pmatrix}1&2\\2&6\\\end{pmatrix}})=2\cdot 2\cdot \det({\begin{pmatrix}1&2\\1&3\\\end{pmatrix}})}$

This contradicts linearity (here we didn't need ${\displaystyle S}$, i.e., we can take ${\displaystyle S}$ to be the zero matrix).

Problem 17

Show that if ${\displaystyle A}$ is ${\displaystyle 3\!\times \!3}$ then ${\displaystyle \det(c\cdot A)=c^{3}\cdot \det(A)}$ for any scalar ${\displaystyle c}$.

Bring out the ${\displaystyle c}$'s one row at a time.

Problem 18

Which real numbers ${\displaystyle \theta }$ make

${\displaystyle {\begin{pmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{pmatrix}}}$

singular? Explain geometrically.

There are no real numbers ${\displaystyle \theta }$ that make the matrix singular because the determinant of the matrix ${\displaystyle \cos ^{2}\theta +\sin ^{2}\theta }$ is never ${\displaystyle 0}$, it equals ${\displaystyle 1}$ for all ${\displaystyle \theta }$. Geometrically, with respect to the standard basis, this matrix represents a rotation of the plane through an angle of ${\displaystyle \theta }$. Each such map is one-to-one — for one thing, it is invertible.

? Problem 19

If a third order determinant has elements ${\displaystyle 1}$, ${\displaystyle 2}$, ..., ${\displaystyle 9}$, what is the maximum value it may have? (Haggett & Saunders 1955)

This is how the answer was given in the cited source. Let ${\displaystyle P}$ be the sum of the three positive terms of the determinant and ${\displaystyle -N}$ the sum of the three negative terms. The maximum value of ${\displaystyle P}$ is

${\displaystyle 9\cdot 8\cdot 7+6\cdot 5\cdot 4+3\cdot 2\cdot 1=630.}$

The minimum value of ${\displaystyle N}$ consistent with ${\displaystyle P}$ is

${\displaystyle 9\cdot 6\cdot 1+8\cdot 5\cdot 2+7\cdot 4\cdot 3=218.}$

Any change in ${\displaystyle P}$ would result in lowering that sum by more than ${\displaystyle 4}$. Therefore ${\displaystyle 412}$ the maximum value for the determinant and one form for the determinant is

${\displaystyle {\begin{vmatrix}9&4&2\\3&8&6\\5&1&7\end{vmatrix}}.}$

## References

• Haggett, Vern (proposer); Saunders, F. W. (solver) (Apr. 1955), "Elementary problem 1135", American Mathematical Monthly (American Mathematical Society) 62 (5): 257