Worked examples — Determinant of 2×2 matrix

This page is a practice arena for Determinant of 2×2 matrix. Before we compute anything, let us recall the one rule everything rests on.

Every symbol here means a plain thing:

• $a,b,c,d$ are just the four numbers written inside the box.
• $\det (A)$ (say "det of A") is one single number you get out.
• The sign of that number tells you whether the shape got flipped; its size tells you the area stretch. Look at the picture below before reading on.

The red parallelogram is what the unit square becomes after the matrix acts. Its area equals $|\det (A)|$, and if the corner-labels go the "wrong way round" (clockwise instead of counterclockwise), $\det (A)$ is negative.

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The scenario matrix

Every determinant problem you will ever meet falls into one of these cells. Our job below is to hit all of them.

Cell	What makes it special	Example that hits it
C1 — plain positive	$ad-bc>0$, ordinary numbers	Ex 1
C2 — negative (flip)	$ad-bc<0$, orientation reverses	Ex 2
C3 — zero / degenerate	columns parallel, area collapses	Ex 3
C4 — one zero column	a whole column is $\vec{0}$	Ex 4
C5 — negatives inside	some of $a,b,c,d$ are negative	Ex 5
C6 — fractions / decimals	non-integer entries	Ex 6
C7 — used in inverse	feed $\det$ into $A^{-1}$	Ex 7
C8 — real-world word problem	area of a land plot	Ex 8
C9 — exam twist (unknown)	solve for $k$ so that $\det =0$	Ex 9
C10 — limiting behaviour	let an entry $\to$ a value, watch $\det$	Ex 10

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Worked examples

Forecast: guess — will the answer be positive, negative, or zero? (The rows don't look like multiples of each other, so probably non-zero.)

1. Falling diagonal: $a\cdot d=4\cdot 3=12$. Why this step? The formula's first piece is always the top-left times bottom-right — that is the "$ad$" in $ad-bc$.
2. Rising diagonal: $b\cdot c=1\cdot 2=2$. Why this step? This is the part we subtract; it corrects for how much the two vectors "lean" together.
3. Subtract: $12-2=10$.

$\det (A)=10$

Verify: Area of the parallelogram with sides $(4,2)$ and $(1,3)$ using the shoelace idea $|ad-bc|=|10|=10$. Positive $\Rightarrow$ orientation kept. ✔

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Forecast: notice this is Example 1 with the two columns swapped. Swapping columns flips the sign, so predict $-10$.

1. $ad=1\cdot 2=2$.
2. $bc=4\cdot 3=12$.
3. $ad-bc=2-12=-10$.

$\det (A)=-10$

Why negative matters: the minus sign says the transformed square is turned inside-out (like flipping a pancake). Look at the two figures — the corner arrows spin opposite ways.

Verify: Column swap should multiply the Example-1 determinant by $-1$: $-1\times 10=-10$. ✔ Size $|-10|=10$ (same area stretch as Ex 1, as expected). ✔

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Forecast: the second column $(6,9)$ looks like $3\times (2,3)$ — the first column. Parallel columns $\Rightarrow$ predict $0$.

1. $ad=2\cdot 9=18$.
2. $bc=6\cdot 3=18$.
3. $18-18=0$.

$\det (A)=0$

What it looks like: the two vectors point the same way, so the "parallelogram" has zero width — it is squashed onto a single line. No area, no inverse.

Verify: Is column 2 a multiple of column 1? $(6,9)=3\cdot (2,3)$. Yes $\Rightarrow$ dependent columns $\Rightarrow$ determinant must be $0$. ✔

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Forecast: the whole first column is the zero vector. A zero-length side can't enclose area, so predict $0$.

1. $ad=0\cdot 7=0$.
2. $bc=5\cdot 0=0$.
3. $0-0=0$.

$\det (A)=0$

Why this is still "degenerate": a zero column is the extreme case of dependency — one side of the parallelogram has collapsed to a point. Everything gets mapped onto the vertical line $x=0$.

Verify: Any matrix with a zero column has determinant $0$; here both products vanish. ✔

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Forecast: negatives inside are fine — just plug in carefully. Two negatives multiply to a positive.

1. $ad=(-3)\cdot (-5)=15$. Why this step? Sign rule: negative $\times$ negative $=$ positive. This is where careless students slip.
2. $bc=2\cdot 4=8$.
3. $15-8=7$.

$\det (A)=7$

Verify: Recompute with sign care: $(-3)(-5)-(2)(4)=15-8=7>0$, so orientation is preserved despite the minus signs in the entries. ✔

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Forecast: fractions don't change the method; expect a smallish number.

1. $ad=0.5\cdot 3=1.5$.
2. $bc=2\cdot 1=2$.
3. $1.5-2=-0.5$.

$\det (A)=-0.5$

Verify: $|-0.5|=0.5$ means the shape's area is halved, and the negative sign flips orientation. Convert to a fraction: $-\frac{1}{2}$. ✔

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Forecast: the inverse formula divides by $\det (A)$, so first we need that number to be non-zero.

1. $\det (A)=(2)(3)-(5)(1)=6-5=1$. Why this step? From Inverse of a 2×2 matrix, the inverse only exists if $\det \ne 0$. Here $\det =1$, so we're safe.
2. Build the adjugate: swap $a\leftrightarrow d$, negate $b$ and $c$: $\left[\begin{array}{cc}3& -5\\ -1& 2\end{array}\right]$
3. Divide by $\det =1$: $A^{-1}=\frac{1}{1}\left[\begin{array}{cc}3& -5\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}3& -5\\ -1& 2\end{array}\right]$

Verify: Multiply $A{A}^{-1}$ (see Matrix multiplication): should give the identity $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$. ✔

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Forecast: two edge-vectors from $O$ span a parallelogram; the triangle is half of it. Guess area is a few square metres.

1. Edge vectors: $\overrightarrow{OP}=(6,1)$, $\overrightarrow{OQ}=(2,5)$. Put them as columns: $M=\left[\begin{array}{cc}6& 2\\ 1& 5\end{array}\right]$ Why this step? Area of parallelogram equals $|\det |$ of the matrix whose columns are the two spanning vectors.
2. $\det (M)=(6)(5)-(2)(1)=30-2=28$.
3. Parallelogram area $=|28|=28{\text{m}}^2$; triangle is half: $\text{Area}=\frac{1}{2}\cdot 28=14{\text{m}}^2$

Verify: Shoelace on the triangle $O,P,Q$: $\frac{1}{2}|x_P{y}_Q-x_Q{y}_P|=\frac{1}{2}|6\cdot 5-2\cdot 1|=\frac{1}{2}(28)=14{\text{m}}^2$. Units are area (m²). ✔

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Forecast: singular means $\det =0$. That gives an equation in $k$ — probably two answers because $k^2$ appears.

1. Write the determinant: $\det =k\cdot k-4\cdot 2=k^2-8$.
2. Set it to zero: $k^2-8=0$. Why this step? "Singular" is exactly the statement $\det =0$ (no inverse, columns dependent).
3. Solve: $k^2=8\Rightarrow k=\pm 2\sqrt{2}$.

$k=2\sqrt{2}\text{or}k=-2\sqrt{2}$

Verify: Plug $k=2\sqrt{2}$: $\det =(2\sqrt{2}{)}^2-8=8-8=0$. ✔ Same for $-2\sqrt{2}$. ✔

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Forecast: at $t=0$ it's the identity ($\det =1$); as the off-diagonal grows, the columns lean together, so $\det$ should shrink, hit zero somewhere, then go negative.

1. Determinant: $\det =(1)(1)-(t)(t)=1-t^2$.
2. Evaluate the boundary/limiting cases:
   • $t=0$: $1-0=1$ (identity, no distortion).
   • $t=1$: $1-1=0$ (columns become identical $\Rightarrow$ collapse to a line).
   • $t=2$: $1-4=-3$ (columns crossed over $\Rightarrow$ orientation flipped). Why this step? Sweeping one entry lets us watch the determinant pass through positive → zero → negative — all three cases in one family.

$\det (A(t))=1-t^2$

Verify: Values at $t=0,1,2$ are $1,0,-3$ respectively, matching the sign story. ✔

Recall Quick self-test

Why is Example 2 the negative of Example 1? ::: They are the same matrix with columns swapped, and a column swap multiplies the determinant by $-1$. In Example 8, why divide by 2? ::: The determinant gives the parallelogram area; a triangle is exactly half of it. In Example 10, at what $t$ does the matrix become non-invertible for $t\ge 0$? ::: At $t=1$, where $\det =0$.

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Connections

• Inverse of a 2×2 matrix — Example 7 uses $\det$ before inverting.
• Area of parallelogram — Example 8's word problem.
• Linear independence — Examples 3, 4, 9 all hinge on dependent columns.
• Cramer's rule — needs $\det \ne 0$, the same test as Example 9.
• Transformations and scaling — the flip in Example 2 and the collapse in Example 10.