To every square matrix $A$ we attach a single number, its determinant, written $\det A$ or $|A|$. It is defined only for square matrices and answers one decisive question: is $A$ invertible? ($|A|\ne 0$ means yes.) Geometrically it measures how the linear map $A$ scales area (in $2\times2$) or volume (in $3\times3$), and its sign records whether orientation is preserved.
Determinant of order 1, 2 and 3
For a $1\times1$ matrix $A=[a]$, $|A|=a$. For a $2\times2$ matrix,
$$\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}.$$
The number is the product of the leading (main) diagonal minus the product of the off-diagonal. For a $3\times3$ matrix we expand along any one row or column, multiplying each entry by its signed minor (cofactor). Expanding along the first row:
$$\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}=a_{1}\begin{vmatrix}b_{2}&c_{2}\\b_{3}&c_{3}\end{vmatrix}-b_{1}\begin{vmatrix}a_{2}&c_{2}\\a_{3}&c_{3}\end{vmatrix}+c_{1}\begin{vmatrix}a_{2}&b_{2}\\a_{3}&b_{3}\end{vmatrix}.$$
The alternating $+,-,+$ signs follow the checkerboard pattern $(-1)^{i+j}$. A vital practical point: the value does not depend on which row or column you expand along, so always choose the line with the most zeros to save arithmetic.
Properties of determinants
These properties let you simplify a determinant before expanding — often reducing a messy $3\times3$ to almost nothing. Let $A$ be $n\times n$.
- Property 1 (Reflection): $|A^{T}|=|A|$. Rows and columns play identical roles, so any property true for rows holds for columns.
- Property 2 (Interchange): swapping two rows (or two columns) multiplies the determinant by $-1$.
- Property 3 (All-zero line): if every entry of some row or column is $0$, then $|A|=0$.
- Property 4 (Identical/proportional lines): if two rows (or columns) are equal, or one is a scalar multiple of another, then $|A|=0$.
- Property 5 (Scalar / factor): if every entry of one row (or column) carries a common factor $k$, that $k$ can be taken outside the determinant. Hence for $n\times n$, $|kA|=k^{n}|A|$.
- Property 6 (Sum-splitting): if each entry of one row is written as a sum, the determinant splits into a sum of two determinants, e.g. $\begin{vmatrix}a_{1}+x&b_{1}&c_{1}\\a_{2}+y&b_{2}&c_{2}\\a_{3}+z&b_{3}&c_{3}\end{vmatrix}=\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}+\begin{vmatrix}x&b_{1}&c_{1}\\y&b_{2}&c_{2}\\z&b_{3}&c_{3}\end{vmatrix}.$
- Property 7 (Row-operation invariance): adding to one row a multiple of another row leaves the determinant unchanged, i.e. $R_{i}\to R_{i}+kR_{j}$ (or $C_{i}\to C_{i}+kC_{j}$) does not alter $|A|$. This is the single most useful tool for creating zeros and for proving identities.
- Multiplicative law: $|AB|=|A|\,|B|$, and a triangular matrix has determinant equal to the product of its diagonal entries.
A determinant in which every entry of a row or column is the same as another after an operation collapses to $0$, which is why row operations are the engine behind every "prove that the determinant equals …" question.
Deeper Insight — why the row operation $R_i\to R_i+kR_j$ changes nothing: By sum-splitting (Property 6), $|A \text{ with } R_i\to R_i+kR_j|$ equals the original determinant plus a second determinant whose row $i$ is $k$ times row $j$. That second determinant has two proportional rows, so by Property 4 it is $0$ — leaving the original value untouched. Understanding this one chain (split $\to$ proportional $\to$ zero) explains all the simplification you will do in this chapter: you are free to add multiples of rows to one another to manufacture zeros, and the answer never moves.
Evaluate $\begin{vmatrix}3&1\\2&4\end{vmatrix}$.
Solution$=3\cdot4-1\cdot2=12-2=10.$
Answer: $10$
Evaluate $\begin{vmatrix}1&2&3\\0&1&4\\0&0&5\end{vmatrix}$.
SolutionThe matrix is upper-triangular, so the determinant is the product of the diagonal: $1\cdot1\cdot5=5$. (You can confirm by expanding along the first column.)
Answer: $5$
Without expanding, show $\begin{vmatrix}2&4\\3&6\end{vmatrix}=0$.
SolutionColumn $2$ is $2\times$ column $1$ (proportional columns), so by the proportionality property the determinant is $0$. (Check: $2\cdot6-4\cdot3=0$.)
Answer: The determinant is $0$ (columns are proportional).
If $|A|=5$ for a $3\times3$ matrix, find $|2A|$.
Solution$|kA|=k^{n}|A|$ with $n=3$, $k=2$: $|2A|=2^{3}\cdot5=8\cdot5=40.$
Answer: $|2A|=40$
Evaluate $\begin{vmatrix}2&-1&0\\1&3&2\\0&1&1\end{vmatrix}$ by expanding along the first column.
SolutionOnly the first two entries of column $1$ are non-zero. $=2\begin{vmatrix}3&2\\1&1\end{vmatrix}-1\begin{vmatrix}-1&0\\1&1\end{vmatrix}+0$. Now $\begin{vmatrix}3&2\\1&1\end{vmatrix}=3-2=1$ and $\begin{vmatrix}-1&0\\1&1\end{vmatrix}=-1-0=-1$. So $=2(1)-1(-1)=2+1=3.$
Answer: $3$
Find $x$ if $\begin{vmatrix}x&2\\3&x\end{vmatrix}=0$.
SolutionExpand: $x\cdot x-2\cdot3=x^{2}-6$. Set $x^{2}-6=0$, so $x^{2}=6$ and $x=\pm\sqrt6.$
Answer: $x=\pm\sqrt6$
Using row operations, evaluate $\begin{vmatrix}1&1&1\\1&2&3\\1&3&6\end{vmatrix}$.
SolutionApply $R_{2}\to R_{2}-R_{1}$ and $R_{3}\to R_{3}-R_{1}$ (these do not change the value): $\begin{vmatrix}1&1&1\\0&1&2\\0&2&5\end{vmatrix}$. Expand along column $1$: $=1\begin{vmatrix}1&2\\2&5\end{vmatrix}=1\cdot(5-4)=1.$
Answer: $1$
Prove that $\begin{vmatrix}a&b&c\\a&b&c\\x&y&z\end{vmatrix}=0$.
SolutionRows $R_{1}$ and $R_{2}$ are identical. By the identical-rows property the determinant is $0$ — no expansion needed.
Answer: The determinant is $0$ (two identical rows).
Show that $\begin{vmatrix}1&a&a^{2}\\1&b&b^{2}\\1&c&c^{2}\end{vmatrix}=(a-b)(b-c)(c-a)$ (the Vandermonde determinant).
SolutionApply $R_{1}\to R_{1}-R_{2}$ and $R_{2}\to R_{2}-R_{3}$: $\begin{vmatrix}0&a-b&a^{2}-b^{2}\\0&b-c&b^{2}-c^{2}\\1&c&c^{2}\end{vmatrix}$. Take out $(a-b)$ from $R_{1}$ and $(b-c)$ from $R_{2}$ (using $p^{2}-q^{2}=(p-q)(p+q)$): $=(a-b)(b-c)\begin{vmatrix}0&1&a+b\\0&1&b+c\\1&c&c^{2}\end{vmatrix}$. Expand along column $1$: $=(a-b)(b-c)\cdot 1\cdot\begin{vmatrix}1&a+b\\1&b+c\end{vmatrix}=(a-b)(b-c)\big[(b+c)-(a+b)\big]=(a-b)(b-c)(c-a).$
Answer: $(a-b)(b-c)(c-a)$
Prove that $\begin{vmatrix}a+b&b+c&c+a\\b+c&c+a&a+b\\c+a&a+b&b+c\end{vmatrix}=0$.
SolutionApply $C_{1}\to C_{1}+C_{2}+C_{3}$. Each entry of the new first column becomes $2(a+b+c)$, so factor it out: $=2(a+b+c)\begin{vmatrix}1&b+c&c+a\\1&c+a&a+b\\1&a+b&b+c\end{vmatrix}$. Now $R_{2}\to R_{2}-R_{1}$ and $R_{3}\to R_{3}-R_{1}$ make the first column $(1,0,0)^{T}$ and the remaining $2\times2$ block has rows that sum the same way, giving two proportional rows. The determinant is $0$.
Answer: The determinant is $0$.
If $A$ is a $3\times3$ matrix with $|A|=4$ and $|B|=2$, find $|AB|$ and $|3A|$.
SolutionBy the multiplicative law $|AB|=|A||B|=4\cdot2=8$. And $|3A|=3^{3}|A|=27\cdot4=108.$
Answer: $|AB|=8$ and $|3A|=108$.
Evaluate $\begin{vmatrix}102&18&36\\1&3&4\\17&3&6\end{vmatrix}$.
SolutionNotice $R_{1}=6\,R_{3}$? Check: $6\times(17,3,6)=(102,18,36)$. Yes — $R_{1}$ is proportional to $R_{3}$, so the determinant is $0.$
Answer: The determinant is $0$ ($R_{1}=6\,R_{3}$).