Section DM Determinant of a Matrix

Elementary matrices are very simple, as you might have suspected from their name. Their purpose is to effect row operations (Definition RO) on a matrix through matrix multiplication (Definition MM). Their definitions look much more complicated than they really are, so be sure to skip over them on your first reading and head right for the explanation that follows and the first example.

Again, these matrices are not as complicated as their definitions suggest, since they are just small perturbations of the n × n identity matrix (Definition IM). {E}_{i,j} is the identity matrix with rows (or columns) i and j trading places, {E}_{i}\left (α\right ) is the identity matrix where the diagonal entry in row i and column i has been replaced by α, and {E}_{i,j}\left (α\right ) is the identity matrix where the entry in row j and column i has been replaced by α. (Yes, those subscripts look backwards in the description of {E}_{i,j}\left (α\right )). Notice that our notation makes no reference to the size of the elementary matrix, since this will always be apparent from the context, or unimportant.

The raison d’être for elementary matrices is to “do” row operations on matrices with matrix multiplication. So here is an example where we will both see some elementary matrices and see how they accomplish row operations when used with matrix multiplication.

Example EMRO
Elementary matrices and row operations
We will perform a sequence of row operations (Definition RO) on the 3 × 4 matrix A, while also multiplying the matrix on the left by the appropriate 3 × 3 elementary matrix.

\eqalignno{ {R}_{1} ↔ {R}_{3} :\ &\left [\array{ 5&0&3&1\cr 1&3 &2 &4 \cr 2&1&3&1 } \right ] &{E}_{1,3} :\ &\left [\array{ 0&0&1\cr 0&1 &0 \cr 1&0&0} \right ]\left [\array{ 2&1&3&1\cr 1&3 &2 &4 \cr 5&0&3&1 } \right ] = \left [\array{ 5&0&3&1\cr 1&3 &2 &4 \cr 2&1&3&1 } \right ] & & & & \cr 2{R}_{2} :\ &\left [\array{ 5&0&3&1\cr 2&6 &4 &8 \cr 2&1&3&1 } \right ] &{E}_{2}\left (2\right ) :\ &\left [\array{ 1&0&0\cr 0&2 &0 \cr 0&0&1} \right ]\left [\array{ 5&0&3&1\cr 1&3 &2 &4 \cr 2&1&3&1 } \right ] = \left [\array{ 5&0&3&1\cr 2&6 &4 &8 \cr 2&1&3&1 } \right ] & & & & \cr 2{R}_{3} + {R}_{1} :\ &\left [\array{ 9&2&9&3\cr 2&6 &4 &8 \cr 2&1&3&1 } \right ] &{E}_{3,1}\left (2\right ) :\ &\left [\array{ 1&0&2\cr 0&1 &0 \cr 0&0&1} \right ]\left [\array{ 5&0&3&1\cr 2&6 &4 &8 \cr 2&1&3&1 } \right ] = \left [\array{ 9&2&9&3\cr 2&6 &4 &8 \cr 2&1&3&1 } \right ] & & & & }

The next three theorems establish that each elementary matrix effects a row operation via matrix multiplication.

Theorem EMDRO
Elementary Matrices Do Row Operations
Suppose that A is an m × n matrix, and B is a matrix of the same size that is obtained from A by a single row operation (Definition RO). Then there is an elementary matrix of size m that will convert A to B via matrix multiplication on the left. More precisely,

Proof In each of the three conclusions, performing the row operation on A will create the matrix B where only one or two rows will have changed. So we will establish the equality of the matrix entries row by row, first for the unchanged rows, then for the changed rows, showing in each case that the result of the matrix product is the same as the result of the row operation. Here we go.

Row k of the product {E}_{i,j}A, where k\mathrel{≠}i, k\mathrel{≠}j, is unchanged from A,

\eqalignno{ {\left [{E}_{i,j}A\right ]}_{kℓ} & ={ \mathop{∑ }}_{p=1}^{n}{\left [{E}_{ i,j}\right ]}_{kp}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \left [{E}_{i,j}\right ]}_{kk}{\left [A\right ]}_{kℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}k\end{array}}^{n}{\left [{E}_{ i,j}\right ]}_{kp}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)} & & & & \cr & = 1{\left [A\right ]}_{kℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}k\end{array}}^{n}0{\left [A\right ]}_{ pℓ} & &\text{@(a href="#definition.ELEM")Definition ELEM@(/a)} & & & & \cr & ={ \left [A\right ]}_{kℓ} & & & & }

\eqalignno{ {\left [{E}_{i,j}A\right ]}_{iℓ} & ={ \mathop{∑ }}_{p=1}^{n}{\left [{E}_{ i,j}\right ]}_{ip}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \left [{E}_{i,j}\right ]}_{ij}{\left [A\right ]}_{jℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}j \end{array}}^{n}{\left [{E}_{ i,j}\right ]}_{ip}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)} & & & & \cr & = 1{\left [A\right ]}_{jℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}j \end{array}}^{n}0{\left [A\right ]}_{ pℓ} & &\text{@(a href="#definition.ELEM")Definition ELEM@(/a)} & & & & \cr & ={ \left [A\right ]}_{jℓ} & & & & }

\eqalignno{ {\left [{E}_{i,j}A\right ]}_{jℓ} & ={ \mathop{∑ }}_{p=1}^{n}{\left [{E}_{ i,j}\right ]}_{jp}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \left [{E}_{i,j}\right ]}_{ji}{\left [A\right ]}_{iℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}i \end{array}}^{n}{\left [{E}_{ i,j}\right ]}_{jp}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)} & & & & \cr & = 1{\left [A\right ]}_{iℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}i \end{array}}^{n}0{\left [A\right ]}_{ pℓ} & &\text{@(a href="#definition.ELEM")Definition ELEM@(/a)} & & & & \cr & ={ \left [A\right ]}_{iℓ} & & & & }

So the matrix product {E}_{i,j}A is the same as the row operation that swaps rows i and j.

Row k of the product {E}_{i}\left (α\right )A, where k\mathrel{≠}i, is unchanged from A,

\eqalignno{ {\left [{E}_{i}\left (α\right )A\right ]}_{kℓ} & ={ \mathop{∑ }}_{p=1}^{n}{\left [{E}_{ i}\left (α\right )\right ]}_{kp}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \left [{E}_{i}\left (α\right )\right ]}_{kk}{\left [A\right ]}_{kℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}k\end{array}}^{n}{\left [{E}_{ i}\left (α\right )\right ]}_{kp}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)} & & & & \cr & = 1{\left [A\right ]}_{kℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}k\end{array}}^{n}0{\left [A\right ]}_{ pℓ} & &\text{@(a href="#definition.ELEM")Definition ELEM@(/a)} & & & & \cr & ={ \left [A\right ]}_{kℓ} & & & & }

\eqalignno{ {\left [{E}_{i}\left (α\right )A\right ]}_{iℓ} & ={ \mathop{∑ }}_{p=1}^{n}{\left [{E}_{ i}\left (α\right )\right ]}_{ip}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \left [{E}_{i}\left (α\right )\right ]}_{ii}{\left [A\right ]}_{iℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}i \end{array}}^{n}{\left [{E}_{ i}\left (α\right )\right ]}_{ip}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)} & & & & \cr & = α{\left [A\right ]}_{iℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}i \end{array}}^{n}0{\left [A\right ]}_{ pℓ} & &\text{@(a href="#definition.ELEM")Definition ELEM@(/a)} & & & & \cr & = α{\left [A\right ]}_{iℓ} & & & & }

So the matrix product {E}_{i}\left (α\right )A is the same as the row operation that swaps multiplies row i by α.

Row k of the product {E}_{i,j}\left (α\right )A, where k\mathrel{≠}j, is unchanged from A,

\eqalignno{ {\left [{E}_{i,j}\left (α\right )A\right ]}_{kℓ} & ={ \mathop{∑ }}_{p=1}^{n}{\left [{E}_{ i,j}\left (α\right )\right ]}_{kp}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \left [{E}_{i,j}\left (α\right )\right ]}_{kk}{\left [A\right ]}_{kℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}k\end{array}}^{n}{\left [{E}_{ i,j}\left (α\right )\right ]}_{kp}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)} & & & & \cr & = 1{\left [A\right ]}_{kℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}k\end{array}}^{n}0{\left [A\right ]}_{ pℓ} & &\text{@(a href="#definition.ELEM")Definition ELEM@(/a)} & & & & \cr & ={ \left [A\right ]}_{kℓ} & & & & }

Row j of the product {E}_{i,j}\left (α\right )A, is α times row i of A and then added to row j of A,

\eqalignno{ {\left [{E}_{i,j}\left (α\right )A\right ]}_{jℓ} & ={ \mathop{∑ }}_{p=1}^{n}{\left [{E}_{ i,j}\left (α\right )\right ]}_{jp}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli31.html#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \left [{E}_{i,j}\left (α\right )\right ]}_{jj}{\left [A\right ]}_{jℓ}+ & & & & \cr &\quad \quad {\left [{E}_{i,j}\left (α\right )\right ]}_{ji}{\left [A\right ]}_{iℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}j,i\end{array}}^{n}{\left [{E}_{ i,j}\left (α\right )\right ]}_{jp}{\left [A\right ]}_{pℓ} & &\text{@(a href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)} & & & & \cr & = 1{\left [A\right ]}_{jℓ} + α{\left [A\right ]}_{iℓ} +{ \mathop{∑ }}_{\begin{array}{c}p=1 \\ p\mathrel{≠}j,i\end{array}}^{n}0{\left [A\right ]}_{ pℓ} & &\text{@(a href="#definition.ELEM")Definition ELEM@(/a)} & & & & \cr & ={ \left [A\right ]}_{jℓ} + α{\left [A\right ]}_{iℓ} & & & & }

So the matrix product {E}_{i,j}\left (α\right )A is the same as the row operation that multiplies row i by α and adds the result to row j. ■

Proof We show that we can row-reduce each elementary matrix to the identity matrix. Given an elementary matrix of the form {E}_{i,j}, perform the row operation that swaps row j with row i. Given an elementary matrix of the form {E}_{i}\left (α\right ), with α\mathrel{≠}0, perform the row operation that multiplies row i by 1∕α. Given an elementary matrix of the form {E}_{i,j}\left (α\right ), with α\mathrel{≠}0, perform the row operation that multiplies row i by − α and adds it to row j. In each case, the result of the single row operation is the identity matrix. So each elementary matrix is row-equivalent to the identity matrix, and by Theorem NMRRI is nonsingular. ■

Notice that we have now made use of the nonzero restriction on α in the definition of {E}_{i}\left (α\right ). One more key property of elementary matrices.

Theorem NMPEM
Nonsingular Matrices are Products of Elementary Matrices
Suppose that A is a nonsingular matrix. Then there exists elementary matrices {E}_{1},\kern 1.95872pt {E}_{2},\kern 1.95872pt {E}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {E}_{t} so that A = {E}_{1}{E}_{2}{E}_{3}\mathop{\mathop{…}}{E}_{t}. □

Proof Since A is nonsingular, it is row-equivalent to the identity matrix by Theorem NMRRI, so there is a sequence of t row operations that converts I to A. For each of these row operations, form the associated elementary matrix from Theorem EMDRO and denote these matrices by {E}_{1},\kern 1.95872pt {E}_{2},\kern 1.95872pt {E}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {E}_{t}. Applying the first row operation to I yields the matrix {E}_{1}I. The second row operation yields {E}_{2}({E}_{1}I), and the third row operation creates {E}_{3}{E}_{2}{E}_{1}I. The result of the full sequence of t row operations will yield A, so

Other than the cosmetic matter of re-indexing these elementary matrices in the opposite order, this is the desired result. ■

Subsection DD: Definition of the Determinant

We’ll now turn to the definition of a determinant and do some sample computations. The definition of the determinant function is recursive, that is, the determinant of a large matrix is defined in terms of the determinant of smaller matrices. To this end, we will make a few definitions.

Definition SM
SubMatrix
Suppose that A is an m × n matrix. Then the submatrix A\left (i|j\right ) is the (m − 1) × (n − 1) matrix obtained from A by removing row i and column j.

\eqalignno{ A\left (2|3\right ) = \left [\array{ 1&−2&9\cr 3& 5 &1 } \right ] & &A\left (3|1\right ) = \left [\array{ −2&3&9\cr −2 &0 &1 } \right ] & & & & }

Definition DM
Determinant of a Matrix
Suppose A is a square matrix. Then its determinant, \mathop{ det} \left (A\right ) = \left \vert A\right \vert , is an element of {ℂ}^{} defined recursively by:
If A is a 1 × 1 matrix, then \mathop{ det} \left (A\right ) ={ \left [A\right ]}_{11}.
If A is a matrix of size n with n ≥ 2, then

\eqalignno{ \mathop{ det} \left (A\right ) & ={ \left [A\right ]}_{11}\mathop{ det} \left (A\left (1|1\right )\right ) −{\left [A\right ]}_{12}\mathop{ det} \left (A\left (1|2\right )\right ) +{ \left [A\right ]}_{13}\mathop{ det} \left (A\left (1|3\right )\right )− & & \cr &\quad {\left [A\right ]}_{14}\mathop{ det} \left (A\left (1|4\right )\right ) + \mathrel{⋯} + {(−1)}^{n+1}{\left [A\right ]}_{ 1n}\mathop{ det} \left (A\left (1|n\right )\right ) & & }

So to compute the determinant of a 5 × 5 matrix we must build 5 submatrices, each of size 4. To compute the determinants of each the 4 × 4 matrices we need to create 4 submatrices each, these now of size 3 and so on. To compute the determinant of a 10 × 10 matrix would require computing the determinant of 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 3, 628, 800 1 × 1 matrices. Fortunately there are better ways. However this does suggest an excellent computer programming exercise to write a recursive procedure to compute a determinant.

\eqalignno{ \mathop{ det} \left (A\right ) = \left \vert A\right \vert & = \left \vert \array{ 3 & 2 &−1\cr 4 & 1 & 6 \cr −3&−1& 2 } \right \vert & & \cr & = 3\left \vert \array{ 1 &6\cr −1 &2 } \right \vert − 2\left \vert \array{ 4 &6\cr −3 &2 } \right \vert + (−1)\left \vert \array{ 4 & 1\cr −3 &−1 } \right \vert & & \cr & = 3\left (1\left \vert \array{ 2\cr } \right \vert − 6\left \vert \array{ −1 } \right \vert \right ) − 2\left (4\left \vert \array{ 2 } \right \vert − 6\left \vert \array{ −3 } \right \vert \right ) −\left (4\left \vert \array{ −1 } \right \vert − 1\left \vert \array{ −3 } \right \vert \right ) & & \cr & = 3\left (1(2) − 6(−1)\right ) − 2\left (4(2) − 6(−3)\right ) −\left (4(−1) − 1(−3)\right ) & & \cr & = 24 − 52 + 1 & & \cr & = −27 & & }

In practice it is a bit silly to decompose a 2 × 2 matrix down into a couple of 1 × 1 matrices and then compute the exceedingly easy determinant of these puny matrices. So here is a simple theorem.

Theorem DMST
Determinant of Matrices of Size Two
Suppose that A = \left [\array{ a&b\cr c&d } \right ]. Then \mathop{ det} \left (A\right ) = ad − bc. □

Subsection CD: Computing Determinants

There are a variety of ways to compute the determinant. We will establish first that we can choose to mimic our definition of the determinant, but by using matrix entries and submatrices based on a row other than the first one.

\eqalignno{ \mathop{ det} \left (A\right )& = {(−1)}^{i+1}{\left [A\right ]}_{ i1}\mathop{ det} \left (A\left (i|1\right )\right ) + {(−1)}^{i+2}{\left [A\right ]}_{ i2}\mathop{ det} \left (A\left (i|2\right )\right ) && && \cr &\quad + {(−1)}^{i+3}{\left [A\right ]}_{ i3}\mathop{ det} \left (A\left (i|3\right )\right ) + \mathrel{⋯} + {(−1)}^{i+n}{\left [A\right ]}_{ in}\mathop{ det} \left (A\left (i|n\right )\right )&&1 ≤ i ≤ n&&&& }

Proof First, the statement of the theorem coincides with Definition DM when i = 1, so throughout, we need only consider i > 1.

Given the recursive definition of the determinant, it should be no surprise that we will use induction for this proof (Technique I). When n = 1, there is nothing to prove since there is but one row. When n = 2, we just examine expansion about the second row,

\eqalignno{ {(−1)}^{2+1}{\left [A\right ]}_{ 21} &\mathop{ det} \left (A\left (2|1\right )\right ) + {(−1)}^{2+2}{\left [A\right ]}_{ 22}\mathop{ det} \left (A\left (2|2\right )\right ) & & & & \cr & = −{\left [A\right ]}_{21}{\left [A\right ]}_{12} +{ \left [A\right ]}_{22}{\left [A\right ]}_{11} & &\text{@(a href="#definition.DM")Definition DM@(/a)} & & & & \cr & ={ \left [A\right ]}_{11}{\left [A\right ]}_{22} −{\left [A\right ]}_{12}{\left [A\right ]}_{21} & & & & \cr & =\mathop{ det} \left (A\right ) & &\text{@(a href="#theorem.DMST")Theorem DMST@(/a)} & & & & \cr & & & & }

So the theorem is true for matrices of size n = 1 and n = 2. Now assume the result is true for all matrices of size n − 1 as we derive an expression for expansion about row i for a matrix of size n. We will abuse our notation for a submatrix slightly, so A\left ({i}_{1},{i}_{2}|{j}_{1},{j}_{2}\right ) will denote the matrix formed by removing rows {i}_{1} and {i}_{2}, along with removing columns {j}_{1} and {j}_{2}. Also, as we take a determinant of a submatrix, we will need to “jump up” the index of summation partway through as we “skip over” a missing column. To do this smoothly we will set

\eqalignno{ \mathop{ det} \left (A\right )& ={ \mathop{∑ }}_{j=1}^{n}{(−1)}^{1+j}{\left [A\right ]}_{ 1j}\mathop{ det} \left (A\left (1|j\right )\right ) &&\text{@(a href="#definition.DM")Definition DM@(/a)} &&&& \cr & ={ \mathop{∑ }}_{j=1}^{n}{(−1)}^{1+j}{\left [A\right ]}_{ 1j}{ \mathop{∑ }}_{\begin{array}{c}1≤ℓ≤n \\ ℓ\mathrel{≠}j \end{array}}{(−1)}^{i−1+ℓ−{ϵ}_{ℓj} }{\left [A\right ]}_{iℓ}\mathop{ det} \left (A\left (1,i|j,ℓ\right )\right )&&\text{Induction Hypothesis}&&&& \cr & ={ \mathop{∑ }}_{j=1}^{n}{ \mathop{∑ }}_{\begin{array}{c}1≤ℓ≤n \\ ℓ\mathrel{≠}j \end{array}}{(−1)}^{j+i+ℓ−{ϵ}_{ℓj} }{\left [A\right ]}_{1j}{\left [A\right ]}_{iℓ}\mathop{ det} \left (A\left (1,i|j,ℓ\right )\right ) &&\text{@(a href="fcla-jsmath-latestli69.html#property.DCN")Property DCN@(/a)} &&&& \cr & ={ \mathop{∑ }}_{ℓ=1}^{n}{ \mathop{∑ }}_{\begin{array}{c}1≤j≤n \\ j\mathrel{≠}ℓ \end{array}}{(−1)}^{j+i+ℓ−{ϵ}_{ℓj} }{\left [A\right ]}_{1j}{\left [A\right ]}_{iℓ}\mathop{ det} \left (A\left (1,i|j,ℓ\right )\right ) &&\text{@(a href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)} &&&& \cr & ={ \mathop{∑ }}_{ℓ=1}^{n}{(−1)}^{i+ℓ}{\left [A\right ]}_{ iℓ}{ \mathop{∑ }}_{\begin{array}{c}1≤j≤n \\ j\mathrel{≠}ℓ \end{array}}{(−1)}^{j−{ϵ}_{ℓj} }{\left [A\right ]}_{1j}\mathop{ det} \left (A\left (1,i|j,ℓ\right )\right ) &&\text{@(a href="fcla-jsmath-latestli69.html#property.DCN")Property DCN@(/a)} &&&& \cr & ={ \mathop{∑ }}_{ℓ=1}^{n}{(−1)}^{i+ℓ}{\left [A\right ]}_{ iℓ}{ \mathop{∑ }}_{\begin{array}{c}1≤j≤n \\ j\mathrel{≠}ℓ \end{array}}{(−1)}^{{ϵ}_{ℓj}+j}{\left [A\right ]}_{ 1j}\mathop{ det} \left (A\left (i, 1|ℓ,j\right )\right ) &&\text{$2{ϵ}_{ℓj}$ is even} &&&& \cr & ={ \mathop{∑ }}_{ℓ=1}^{n}{(−1)}^{i+ℓ}{\left [A\right ]}_{ iℓ}\mathop{ det} \left (A\left (i|ℓ\right )\right ) &&\text{@(a href="#definition.DM")Definition DM@(/a)} &&&& }

We can also obtain a formula that computes a determinant by expansion about a column, but this will be simpler if we first prove a result about the interplay of determinants and transposes. Notice how the following proof makes use of the ability to compute a determinant by expanding about any row.

Theorem DT
Determinant of the Transpose
Suppose that A is a square matrix. Then \mathop{ det} \left ({A}^{t}\right ) =\mathop{ det} \left (A\right ). □

Proof With our definition of the determinant (Definition DM) and theorems like Theorem DER, using induction (Technique I) is a natural approach to proving properties of determinants. And so it is here. Let n be the size of the matrix A, and we will use induction on n.

For n = 1, the transpose of a matrix is identical to the original matrix, so vacuously, the determinants are equal.

\eqalignno{ \mathop{ det} \left ({A}^{t}\right ) & = {1\over n}{\mathop{∑ }}_{i=1}^{n}\mathop{ det} \left ({A}^{t}\right ) & & & & \cr & = {1\over n}{\mathop{∑ }}_{i=1}^{n}{ \mathop{∑ }}_{j=1}^{n}{(−1)}^{i+j}{\left [{A}^{t}\right ]}_{ ij}\mathop{ det} \left ({A}^{t}\left (i|j\right )\right ) & &\text{@(a href="#theorem.DER")Theorem DER@(/a)} & & & & \cr & = {1\over n}{\mathop{∑ }}_{i=1}^{n}{ \mathop{∑ }}_{j=1}^{n}{(−1)}^{i+j}{\left [A\right ]}_{ ji}\mathop{ det} \left ({A}^{t}\left (i|j\right )\right ) & &\text{@(a href="fcla-jsmath-latestli30.html#definition.TM")Definition TM@(/a)} & & & & \cr & = {1\over n}{\mathop{∑ }}_{i=1}^{n}{ \mathop{∑ }}_{j=1}^{n}{(−1)}^{i+j}{\left [A\right ]}_{ ji}\mathop{ det} \left ({\left (A\left (j|i\right )\right )}^{t}\right ) & &\text{@(a href="fcla-jsmath-latestli30.html#definition.TM")Definition TM@(/a)} & & & & \cr & = {1\over n}{\mathop{∑ }}_{i=1}^{n}{ \mathop{∑ }}_{j=1}^{n}{(−1)}^{i+j}{\left [A\right ]}_{ ji}\mathop{ det} \left (A\left (j|i\right )\right ) & &\text{Induction Hypothesis} & & & & \cr & = {1\over n}{\mathop{∑ }}_{j=1}^{n}{ \mathop{∑ }}_{i=1}^{n}{(−1)}^{j+i}{\left [A\right ]}_{ ji}\mathop{ det} \left (A\left (j|i\right )\right ) & &\text{@(a href="fcla-jsmath-latestli69.html#property.CACN")Property CACN@(/a)} & & & & \cr & = {1\over n}{\mathop{∑ }}_{j=1}^{n}\mathop{ det} \left (A\right ) & &\text{@(a href="#theorem.DER")Theorem DER@(/a)} & & & & \cr & =\mathop{ det} \left (A\right ) & & & & }

Now we can easily get the result that a determinant can be computed by expansion about any column as well.

\eqalignno{ \mathop{ det} \left (A\right )& = {(−1)}^{1+j}{\left [A\right ]}_{ 1j}\mathop{ det} \left (A\left (1|j\right )\right ) + {(−1)}^{2+j}{\left [A\right ]}_{ 2j}\mathop{ det} \left (A\left (2|j\right )\right ) && && \cr &\quad + {(−1)}^{3+j}{\left [A\right ]}_{ 3j}\mathop{ det} \left (A\left (3|j\right )\right ) + \mathrel{⋯} + {(−1)}^{n+j}{\left [A\right ]}_{ nj}\mathop{ det} \left (A\left (n|j\right )\right )&&1 ≤ j ≤ n&&&& }

\eqalignno{ \mathop{ det} \left (A\right ) & =\mathop{ det} \left ({A}^{t}\right ) & &\text{@(a href="#theorem.DT")Theorem DT@(/a)} & & & & \cr & ={ \mathop{∑ }}_{i=1}^{n}{(−1)}^{j+i}{\left [{A}^{t}\right ]}_{ ji}\mathop{ det} \left ({A}^{t}\left (j|i\right )\right ) & &\text{@(a href="#theorem.DER")Theorem DER@(/a)} & & & & \cr & ={ \mathop{∑ }}_{i=1}^{n}{(−1)}^{j+i}{\left [{A}^{t}\right ]}_{ ji}\mathop{ det} \left ({\left (A\left (i|j\right )\right )}^{t}\right ) & &\text{@(a href="fcla-jsmath-latestli30.html#definition.TM")Definition TM@(/a)} & & & & \cr & ={ \mathop{∑ }}_{i=1}^{n}{(−1)}^{j+i}{\left [{A}^{t}\right ]}_{ ji}\mathop{ det} \left (A\left (i|j\right )\right ) & &\text{@(a href="#theorem.DT")Theorem DT@(/a)} & & & & \cr & ={ \mathop{∑ }}_{i=1}^{n}{(−1)}^{i+j}{\left [A\right ]}_{ ij}\mathop{ det} \left (A\left (i|j\right )\right ) & &\text{@(a href="fcla-jsmath-latestli30.html#definition.TM")Definition TM@(/a)} & & & & }

That the determinant of an n × n matrix can be computed in 2n different (albeit similar) ways is nothing short of remarkable. For the doubters among us, we will do an example, computing a 4 × 4 matrix in two different ways.

\eqalignno{ \left \vert A\right \vert & = (4){(−1)}^{4+1}\left \vert \array{ 3 & 0 & 1\cr −2 & 0 & 1 \cr 3 &−2&−1 } \right \vert + (1){(−1)}^{4+2}\left \vert \array{ −2& 0 & 1\cr 9 & 0 & 1 \cr 1 &−2&−1 } \right \vert & & \cr &\quad \quad + (2){(−1)}^{4+3}\left \vert \array{ −2& 3 & 1\cr 9 &−2 & 1 \cr 1 & 3 &−1 } \right \vert + (6){(−1)}^{4+4}\left \vert \array{ −2& 3 & 0\cr 9 &−2 & 0 \cr 1 & 3 &−2 } \right \vert & & \cr & = (−4)(10) + (1)(−22) + (−2)(61) + 6(46) = 92 & & }

\eqalignno{ \left \vert A\right \vert & = (0){(−1)}^{1+3}\left \vert \array{ 9&−2& 1\cr 1& 3 &−1 \cr 4& 1 & 6 } \right \vert + (0){(−1)}^{2+3}\left \vert \array{ −2&3& 1\cr 1 &3 &−1 \cr 4 &1& 6 } \right \vert + & & \cr &\quad \quad (−2){(−1)}^{3+3}\left \vert \array{ −2& 3 &1\cr 9 &−2 &1 \cr 4 & 1 &6 } \right \vert + (2){(−1)}^{4+3}\left \vert \array{ −2& 3 & 1\cr 9 &−2 & 1 \cr 1 & 3 &−1 } \right \vert & & \cr & = 0 + 0 + (−2)(−107) + (−2)(61) = 92 & & }

Notice how much easier the second computation was. By choosing to expand about the third column, we have two entries that are zero, so two 3 × 3 determinants need not be computed at all! ⊠

When a matrix has all zeros above (or below) the diagonal, exploiting the zeros by expanding about the proper row or column makes computing a determinant insanely easy.

We will compute the determinant of this 5 × 5 matrix by consistently expanding about the first column for each submatrix that arises and does not have a zero entry multiplying it.

\eqalignno{ \mathop{ det} \left (T\right ) & = \left \vert \array{ 2& 3 &−1& 3 & 3\cr 0&−1 & 5 & 2 &−1 \cr 0& 0 & 3 & 9 & 2\cr 0& 0 & 0 &−1 & 3 \cr 0& 0 & 0 & 0 & 5 } \right \vert & & \cr & = 2{(−1)}^{1+1}\left \vert \array{ −1&5& 2 &−1\cr 0 &3 & 9 & 2 \cr 0 &0&−1& 3\cr 0 &0 & 0 & 5 } \right \vert & & \cr & = 2(−1){(−1)}^{1+1}\left \vert \array{ 3& 9 &2\cr 0&−1 &3 \cr 0& 0 &5 } \right \vert & & \cr & = 2(−1)(3){(−1)}^{1+1}\left \vert \array{ −1&3 \cr 0 &5 } \right \vert & & \cr & = 2(−1)(3)(−1){(−1)}^{1+1}\left \vert \array{ 5 } \right \vert & & \cr & = 2(−1)(3)(−1)(5) = 30 & & }

If you consult other texts in your study of determinants, you may run into the terms “minor” and “cofactor,” especially in a discussion centered on expansion about rows and columns. We’ve chosen not to make these definitions formally since we’ve been able to get along without them. However, informally, a minor is a determinant of a submatrix, specifically \mathop{ det} \left (A\left (i|j\right )\right ) and is usually referenced as the minor of {\left [A\right ]}_{ij}. A cofactor is a signed minor, specifically the cofactor of {\left [A\right ]}_{ij} is {(−1)}^{i+j}\mathop{ det} \left (A\left (i|j\right )\right ).

Subsection READ: Reading Questions

Subsection EXC: Exercises

M10 Find a value of k so that the matrix A = \left [\array{ 2&4\cr 3&k } \right ] has \mathop{ det}(A) = 0, or explain why it is not possible.
Contributed by Chris Black Solution [1192]

M11 Find a value of k so that the matrix A = \left [\array{ 1&2&1\cr 2&0 &1 \cr 2&3&k } \right ] has \mathop{ det}(A) = 0, or explain why it is not possible.
Contributed by Chris Black Solution [1193]

M15 Given the matrix B = \left [\array{ 2 − x& 1\cr 4 &2 − x } \right ], find all values of x that are solutions of \mathop{ det}(B) = 0.
Contributed by Chris Black Solution [1193]

M16 Given the matrix B = \left [\array{ 4 − x& −4 & −4\cr 2 &−2 − x & −4 \cr 3 & −3 &−4 − x } \right ], find all values of x that are solutions of \mathop{ det}(B) = 0.
Contributed by Chris Black Solution [1194]

Subsection SOL: Solutions

\eqalignno{ \left \vert \array{ 1&3\cr 6&2 } \right \vert & = 1 ⋅ 2 − 6 ⋅ 3 = 2 − 18 = −16 & & }

\eqalignno{ \left \vert \array{ 1&3\cr 2&6 } \right \vert & = 1 ⋅ 6 − 2 ⋅ 3 = 6 − 6 = 0 & & }

C23 Contributed by Chris Black Statement [1181]
We can compute the determinant by expanding about any row or column; the most efficient ones to choose are either the second column or the third row. In any case, the determinant will be − 4.

C24 Contributed by Robert Beezer Statement [1182]
We’ll expand about the first row since there are no zeros to exploit,

\eqalignno{ \left \vert \array{ −2& 3 &−2\cr −4&−2&1 \cr 2 & 4 & 2} \right \vert & = (−2)\left \vert \array{ −2&1\cr 4 &2 } \right \vert + (−1)(3)\left \vert \array{ −4&1\cr 2 &2 } \right \vert + (−2)\left \vert \array{ −4&−2\cr 2 & 4 } \right \vert && \cr & = (−2)((−2)(2) − 1(4)) + (−3)((−4)(2) − 1(2)) + (−2)((−4)(4) − (−2)(2)) && \cr & = (−2)(−8) + (−3)(−10) + (−2)(−12) = 70 && }

C25 Contributed by Robert Beezer Statement [1182]
We can expand about any row or column, so the zero entry in the middle of the last row is attractive. Let’s expand about column 2. By Theorem DER and Theorem DEC you will get the same result by expanding about a different row or column. We will use Theorem DMST twice.

\eqalignno{ \left \vert \array{ 3&−1&4\cr 2& 5 &1 \cr 2& 0 &6 } \right \vert & = (−1){(−1)}^{1+2}\left \vert \array{ 2&1 \cr 2&6 } \right \vert + (5){(−1)}^{2+2}\left \vert \array{ 3&4 \cr 2&6 } \right \vert + (0){(−1)}^{3+2}\left \vert \array{ 3&4 \cr 2&1 } \right \vert & & \cr & = (1)(10) + (5)(10) + 0 = 60 & & }

\eqalignno{ \mathop{ det} \left (A\right ) & = \left \vert \array{ 2&0&3&2\cr 5&1 &2 &4 \cr 3&0&1&2\cr 5&3 &2 &1 } \right \vert & & \cr & = 0(−1)\left \vert \array{ 5&2&4\cr 3&1 &2 \cr 5&2&1} \right \vert + 1(1)\left \vert \array{ 2&3&2\cr 3&1 &2 \cr 5&2&1} \right \vert + 0(−1)\left \vert \array{ 2&3&2\cr 5&2 &4 \cr 5&2&1} \right \vert + 3(1)\left \vert \array{ 2&3&2\cr 5&2 &4 \cr 3&1&2} \right \vert & & \cr & = (1)\left (2(1(1) − 2(2)) − 3(3(1) − 5(2)) + 2(3(2) − 5(1))\right )+ & & \cr &\quad \quad (3)\left (2(2(2) − 4(1)) − 3(5(2) − 4(3)) + 2(5(1) − 3(2))\right ) & & \cr & = (−6 + 21 + 2) + (3)(0 + 6 − 2) = 29 & & }

\eqalignno{ \left \vert \array{ 1&0& 1 &1\cr 2&2 &−1 &1 \cr 2&1& 3 &0\cr 1&1 & 0 &1 } \right \vert & = \left \vert \array{ 2&−1&1\cr 1& 3 &0 \cr 1& 0 &1 } \right \vert − 0 + \left \vert \array{ 2&2&1\cr 2&1 &0 \cr 1&1&1} \right \vert −\left \vert \array{ 2&2&−1\cr 2&1 & 3 \cr 1&1& 0 } \right \vert & & \cr & = 4 + (−1) − (−1) = 4 & & }

\eqalignno{ \left \vert \array{ 1& 0 & 1 &1\cr 2&−1 &−1 &1 \cr 2& 5 & 3 &0\cr 1&−1 & 0 &1 } \right \vert & = \left \vert \array{ −1&−1&1\cr 5 & 3 &0 \cr −1& 0 &1 } \right \vert − 0 + \left \vert \array{ 2&−1&1\cr 2& 5 &0 \cr 1&−1&1 } \right \vert −\left \vert \array{ 2&−1&−1\cr 2& 5 & 3 \cr 1&−1& 0 } \right \vert & & \cr & = 5 − 0 + 5 − 10 = 0. & & }

\eqalignno{ \left \vert \array{ 2&3&0&2&1\cr 0&1 &1 &1 &2 \cr 0&0&1&2&3\cr 0&1 &2 &1 &0 \cr 0&0&0&1&2 } \right \vert & = 2\left \vert \array{ 1&1&1&2\cr 0&1 &2 &3 \cr 1&2&1&0\cr 0&0 &1 &2 } \right \vert + 0 + 0 + 0 + 0 & & \text{Now, expanding along the first column again, we have} \cr & = 2\left (\left \vert \array{ 1&2&3\cr 2&1 &0 \cr 0&1&2} \right \vert − 0 + \left \vert \array{ 1&1&2\cr 1&2 &3 \cr 0&1&2} \right \vert − 0\right ) & & \cr & = 2([2 + 0 + 6 − 0 − 0 − 8] + [4 + 0 + 2 − 0 − 3 − 2]) & & \cr & = 2 & & }

C30 Contributed by Chris Black Statement [1185]
In order to exploit the zeros, let’s expand along row 3. We then have

\eqalignno{ \left \vert \array{ 2&3&0&2&1\cr 0&1 &1 &1 &2 \cr 0&0&1&2&0\cr 1&0 &3 &1 &1 \cr 2&1&1&2&1 } \right \vert & = {(−1)}^{6}\left \vert \array{ 2&1& 0 &1\cr 2&1 &−1 &1 \cr 1&0& 1 &1\cr 2&1 & 2 &1 } \right \vert + {(−1)}^{7} ⋅ 2\left \vert \array{ 2&1&1&1\cr 2&1 &2 &1 \cr 1&0&3&1\cr 2&1 &1 &1 } \right \vert & & \text{Notice that the second matrix here is singular since two rows are identical and thus it cannot row-reduce to an identity matrix. We now have} \cr & = \left \vert \array{ 2&1& 0 &1\cr 2&1 &−1 &1 \cr 1&0& 1 &1\cr 2&1 & 2 &1 } \right \vert + 0 & & \text{and now we expand on the first row of the first matrix:} \cr & = 2\left \vert \array{ 1&−1&1\cr 0& 1 &1 \cr 1& 2 &1 } \right \vert −\left \vert \array{ 2&−1&1\cr 1& 1 &1 \cr 2& 2 &1 } \right \vert + 0 −\left \vert \array{ 2&1&−1\cr 1&0 & 1 \cr 2&1& 2 } \right \vert & & \cr & = 2(−3) − (−3) − (−3) = 0 & & }

M10 Contributed by Chris Black Statement [1185]
There is only one value of k that will make this matrix have a zero determinant.