- #1

E'lir Kramer

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This is problem 4.14 in Advanced Calculus of Several Variables. Two questions: Is my proof correct? And: is there a cleaner way to prove this?

Show that, if [itex] ad = bc [/itex], then the matrix [itex]\left [

\begin{array}{cc}

a & b \\

c & d \\

\end{array}

\right][/itex] has no inverse.

My attempt:

Suppose there is an inverse such that

[itex]\left [

\begin{array}{cc}

a & b \\

c & d \\

\end{array}

\right]

\times

\left [

\begin{array}{cc}

a_{i} & b_{i} \\

c_{i} & d_{i} \\

\end{array}

\right]

= I =

\left [

\begin{array}{cc}

1 & 0 \\

0 & 1 \\

\end{array}

\right]

[/itex]

By the definition of matrix multiplication, we have

[itex]aa_{i} + bc_{i} = 1 \>\> (1) \\

ab_{i} + bd_{i} = 0 \>\> (2) \\

ca_{i} + dc_{i} = 0 \>\> (3) \\

cb_{i} + dd_{i} = 1 \>\> (4)

[/itex]

Now, since [itex] aa_{i} + bc_{i} = 1 [/itex], either [itex] a_{i} [/itex] or [itex]c_{i} ≠ 0 [/itex].

First let us suppose that [itex] a_{i} ≠ 0 [/itex] and seek contradiction.

By our supposition and equations (1) and (3) above, we have

[itex] a = \frac{1-bc_{i}}{a_{i}} ≠ 0 [/itex], also, [itex] c = \frac{-dc_{i}}{a_{i}} [/itex]

To proceed further, we can also suppose that that [itex]c_{i} ≠ 0 [/itex]. Then we have

[itex] d = \frac{-ca_{i}}{c_{i}} [/itex]

so

[itex] ad = \frac{-ca_{i}}{c_{i}} * \frac{1-bc_{i}}{a_{i}} \\

= \frac{-ca_{i}}{c_{i}} + bc

[/itex]

by substitution.

Since we have by hypothesis [itex] ad = bc [/itex],

[itex] bc = \frac{-ca_{i}}{c_{i}} + bc [/itex]

so

[itex] 0 = \frac{-ca_{i}}{c_{i}} [/itex].

By supposition, [itex]a_{i}, c_{i} ≠ 0 [/itex], so, [itex] c = 0 [/itex]

But this is a contradiction, because then by (4),

[itex]dd_{i} = 1[/itex], meaning that d ≠ 0,

but we have from above that

[itex] d = \frac{-ca_{i}}{c_{i}} = 0 [/itex].

Now if we suppose that [itex] a_{i} ≠ 0 [/itex], [itex]c_{i} = 0[/itex], we are led to another contradiction.

By (3),

[itex] 0 = ca_{i} [/itex],

and by our first supposition that [itex] a_{i} ≠ 0 [/itex], [itex] c = 0 [/itex].

Since [itex] ad = bc [/itex], either [itex]a[/itex] or [itex]d[/itex] must then be 0.

If [itex] a = 0[/itex], we have by (1), and our supposition that [itex]c_{i} = 0[/itex], then

[itex] 0 * a_{i} + 0 * b_{i} = 1 [/itex], a contradiction.

If [itex] d = 0 [/itex], we have by (4) and our supposition that [itex]c_{i} = 0[/itex], then

[itex] 0 * b_{i} + 0 * d_{i} = 1 [/itex], a contradiction.

Note that this proves that [itex] c ≠ 0 [/itex] when [itex]c_{i} = 0[/itex]. (We'll use this again in a second.)

Now we have proved that if [itex] a_{i} ≠ 0, [/itex] we run into a contradiction.

The final case to consider is that [itex] a_{i} = 0 [/itex]. This is impossible. If [itex] a_{i} = 0[/itex], then [itex] c_{i} ≠ 0 [/itex] by (1) . Then we have

[itex] b = \frac{1-aa_{i}}{c_{i}} ≠ 0 \>\> (1) \\

d = \frac{-ca_{i}}{c_{i}} \>\> (3)

[/itex].

Again by hypothesis,

[itex] a * \frac{-ca_{i}}{c_{i}} = c * \frac{1-aa_{i}}{c_{i}} \\

a * -ca_{i} = c * (1-aa_{i}) \\

c = 0

[/itex]

We've already learned from above that c ≠ 0, so we're done.

Show that, if [itex] ad = bc [/itex], then the matrix [itex]\left [

\begin{array}{cc}

a & b \\

c & d \\

\end{array}

\right][/itex] has no inverse.

My attempt:

Suppose there is an inverse such that

[itex]\left [

\begin{array}{cc}

a & b \\

c & d \\

\end{array}

\right]

\times

\left [

\begin{array}{cc}

a_{i} & b_{i} \\

c_{i} & d_{i} \\

\end{array}

\right]

= I =

\left [

\begin{array}{cc}

1 & 0 \\

0 & 1 \\

\end{array}

\right]

[/itex]

By the definition of matrix multiplication, we have

[itex]aa_{i} + bc_{i} = 1 \>\> (1) \\

ab_{i} + bd_{i} = 0 \>\> (2) \\

ca_{i} + dc_{i} = 0 \>\> (3) \\

cb_{i} + dd_{i} = 1 \>\> (4)

[/itex]

Now, since [itex] aa_{i} + bc_{i} = 1 [/itex], either [itex] a_{i} [/itex] or [itex]c_{i} ≠ 0 [/itex].

First let us suppose that [itex] a_{i} ≠ 0 [/itex] and seek contradiction.

By our supposition and equations (1) and (3) above, we have

[itex] a = \frac{1-bc_{i}}{a_{i}} ≠ 0 [/itex], also, [itex] c = \frac{-dc_{i}}{a_{i}} [/itex]

To proceed further, we can also suppose that that [itex]c_{i} ≠ 0 [/itex]. Then we have

[itex] d = \frac{-ca_{i}}{c_{i}} [/itex]

so

[itex] ad = \frac{-ca_{i}}{c_{i}} * \frac{1-bc_{i}}{a_{i}} \\

= \frac{-ca_{i}}{c_{i}} + bc

[/itex]

by substitution.

Since we have by hypothesis [itex] ad = bc [/itex],

[itex] bc = \frac{-ca_{i}}{c_{i}} + bc [/itex]

so

[itex] 0 = \frac{-ca_{i}}{c_{i}} [/itex].

By supposition, [itex]a_{i}, c_{i} ≠ 0 [/itex], so, [itex] c = 0 [/itex]

But this is a contradiction, because then by (4),

[itex]dd_{i} = 1[/itex], meaning that d ≠ 0,

but we have from above that

[itex] d = \frac{-ca_{i}}{c_{i}} = 0 [/itex].

Now if we suppose that [itex] a_{i} ≠ 0 [/itex], [itex]c_{i} = 0[/itex], we are led to another contradiction.

By (3),

[itex] 0 = ca_{i} [/itex],

and by our first supposition that [itex] a_{i} ≠ 0 [/itex], [itex] c = 0 [/itex].

Since [itex] ad = bc [/itex], either [itex]a[/itex] or [itex]d[/itex] must then be 0.

If [itex] a = 0[/itex], we have by (1), and our supposition that [itex]c_{i} = 0[/itex], then

[itex] 0 * a_{i} + 0 * b_{i} = 1 [/itex], a contradiction.

If [itex] d = 0 [/itex], we have by (4) and our supposition that [itex]c_{i} = 0[/itex], then

[itex] 0 * b_{i} + 0 * d_{i} = 1 [/itex], a contradiction.

Note that this proves that [itex] c ≠ 0 [/itex] when [itex]c_{i} = 0[/itex]. (We'll use this again in a second.)

Now we have proved that if [itex] a_{i} ≠ 0, [/itex] we run into a contradiction.

The final case to consider is that [itex] a_{i} = 0 [/itex]. This is impossible. If [itex] a_{i} = 0[/itex], then [itex] c_{i} ≠ 0 [/itex] by (1) . Then we have

[itex] b = \frac{1-aa_{i}}{c_{i}} ≠ 0 \>\> (1) \\

d = \frac{-ca_{i}}{c_{i}} \>\> (3)

[/itex].

Again by hypothesis,

[itex] a * \frac{-ca_{i}}{c_{i}} = c * \frac{1-aa_{i}}{c_{i}} \\

a * -ca_{i} = c * (1-aa_{i}) \\

c = 0

[/itex]

We've already learned from above that c ≠ 0, so we're done.

Last edited: