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Wize University Linear Algebra Textbook > Determinants

Determinants of Large Matrices

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Minors and Cofactors

Minor of a Matrix Entry

Let AA be a square matrix. The minor of entry aija_{ij}, denoted MijM_{ij}, is the determinant of the submatrix that remains after removing row ii and column jj from AA.
Example
Given A=[−3   1−4245156]A = \left[\begin{array}{rrr} -3&\ \ \ 1&-4\\[0.5em] 2&4&5\\[0.5em] 1&5&6 \end{array}\right], find the minors M11M_{11} and M23M_{23}:
For M11M_{11}, we need to cross out row 1 and column 1, then find the determinant of what remains:
M11=∣4556∣=4(6)−5(5)=−1M_{11}=\left|\begin{array}{rr} 4&5\\[0.5em] 5&6 \end{array}\right| = 4(6)-5(5)=-1
For M23M_{23}, we need to cross out row 2 and column 3:
M23=∣−3115∣=−3(5)−1(1)=−16M_{23}=\left|\begin{array}{rr} -3&1\\[0.5em] 1&5 \end{array}\right| = -3(5)-1(1)=-16
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Cofactor of a Matrix Entry

LetAA be a square matrix. The cofactor of entry aija_{ij}, denoted CijC_{ij}, is given as:
Cij=(−1)i+jMij\boxed{\quad C_{ij}=(-1)^{i+j}M_{ij} \quad}
Wize Tip
Imagine a checkerboard pattern over the given matrix to decide whether you should write MijM_{ij} or −Mij-M_{ij}:
[+−+⋯−+−⋯+−+⋯⋮⋮⋮⋱]\left[\begin{array}{cccc} +&-&+&\cdots\\[0.5em] -&+&-&\cdots\\[0.5em] +&-&+&\cdots\\[0.5em] \vdots&\vdots&\vdots&\ddots \end{array}\right]


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Example
Given A=[−3   1−4245156]A = \left[\begin{array}{rrr} -3&\ \ \ 1&-4\\[0.5em] 2&4&5\\[0.5em] 1&5&6 \end{array}\right] , find cofactors C11C_{11} and C23C_{23}.
Using the checkerboard pattern, we can see whether we need to negate when computing C11, C23C_{11}, \ C_{23}:
A=[+11−+−+−23+−+]A = \left[\begin{array}{rrr} \overset{11}{\boxed{+}} & - & +\\[0.3em] - & + & \overset{23}{\boxed{-}}\\[1em] + & - & + \end{array}\right]
C11=+M11=−1C_{11} = \boxed{+}M_{11} = -1
C23=−M23=−(−16)=16C_{23} = \boxed{-}M_{23} = -(-16) = 16
Note: ++ just means multiply by positive 1 (no change), it does not mean "make it positive".

Alternatively, you can use the formula:
C11=(−1)1+1M11=(−1)2(−1)=−1C_{11}=(-1)^{1+1}M_{11}=(-1)^2(-1)=-1
C23=(−1)2+3M23=−1(−16)=16C_{23}=(-1)^{2+3}M_{23}=-1(-16)=16
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Determinants of Large Matrices

Cofactor Expansion

Given a square matrix An×nA_{n \times n}, its determinant can be calculated as:
det(A)=a11C11+a12C12+a13C13+⋯+a1nC1n\text{det}(A)=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}+\cdots+a_{1n}C_{1n}
This expression is referred as the cofactor expansion along row 1.

Wize Tip
You can calculate det(A)\text{det}(A) by expanding cofactors along any row or any column.
Choose a row or column that has lots of zeros!

Example
If A=[a11a12a13a21a22a23a31a32a33]A=\left[\begin{array}{rrr}a_{11} & a_{12}&a_{13}\\a_{21} & a_{22}&a_{23}\\a_{31} & a_{32}&a_{33}\end{array}\right], then by expanding cofactors along row 1, we get the result:
det(A)=a11∣a22a23a32a33∣−a12∣a21a23a31a33∣+a13∣a21a22a31a32∣\text{det}(A)= a_{11}\left|\begin{array}{rrr}a_{22}&a_{23}\\a_{32}&a_{33}\end{array}\right| - a_{12}\left|\begin{array}{rrr}a_{21}&a_{23}\\a_{31}&a_{33}\end{array}\right| + a_{13}\left|\begin{array}{rrr}a_{21}&a_{22}\\a_{31}&a_{32}\end{array}\right|
Note: Finding the determinant of a 3×33\times 3 matrix becomes finding 33 determinants 2×22\times 2 matrices.
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Example
If B=[b11b12b13b14b21b22b23b24b31b32b33b34b41b42b43b44]B=\left[\begin{array}{rrrr}b_{11} & b_{12}&b_{13}&b_{14}\\b_{21} & b_{22}&b_{23}&b_{24}\\b_{31} & b_{32}&b_{33}&b_{34}\\b_{41} & b_{42}&b_{43}&b_{44}\end{array}\right] then by expanding cofactors along row 1 we get the result:
det(B)=   b11⋅∣b22b23b24b32b33b34b42b43b44∣−b12⋅∣b21b23b24b31b33b34b41b43b44∣+b13⋅∣b21b22b24b31b32b34b41b42b44∣−b14⋅∣b21b22b23b31b32b33b41b42b43∣\begin{array}{rcl} \text{det}(B)&=& \ \ \ b_{11}\cdot\left|\begin{array}{rrr}b_{22}&b_{23}&b_{24}\\b_{32}&b_{33}&b_{34}\\b_{42}&b_{43}&b_{44}\end{array}\right|\\[2em] &&- b_{12}\cdot\left|\begin{array}{rrr}b_{21}&b_{23}&b_{24}\\b_{31}&b_{33}&b_{34}\\b_{41}&b_{43}&b_{44}\end{array}\right|\\[2em] &&+ b_{13}\cdot\left|\begin{array}{rrr}b_{21}&b_{22}&b_{24}\\b_{31}&b_{32}&b_{34}\\b_{41}&b_{42}&b_{44}\end{array}\right|\\[2em] &&- b_{14}\cdot \left|\begin{array}{rrr}b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\\b_{41}&b_{42}&b_{43}\end{array}\right| \end{array}
Exercise: To compute a 4×44 \times 4 determinant, we must compute up to
12
2×22\times 2 determinants.

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Steps

  1. Choose a row or column with lots of 0s
  2. Assign ++ or −- signs to each entry in that row or column (top left corner starts with ++, continue checkerboard pattern)
  3. Set up the determinant equation with these signs: det⁡A=+−+…\det A = + \quad - \quad + \quad\dots
  4. Write the entries of your chosen row or column: det⁡A=+entry1−entry2+entry3…\det A = +\text{entry}_1 \quad -\text{entry}_2 \quad +\text{entry}_3 \quad\dots
  5. Multiply by the determinants of the sub-matrices (matrix minors):
det⁡A=+entry1⋅Mij−entry2⋅Mkl+entry3⋅Mmn…\boxed{\quad \det A = +\text{entry}_1\cdot M_{ij} \quad -\text{entry}_2 \cdot M_{kl} \quad +\text{entry}_3\cdot M_{mn}\dots \quad}
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Example: Determinant of a 3×3\colorOne{3 \times 3} Matrix

𝐴=[1−23041−21−5]𝐴 =\begin{bmatrix} 1&-2&3\\ 0&4&1\\ -2&1&-5 \end{bmatrix}. Find det(A)\detp{A} using cofactor expansion.
  1. Choose a row or column with lots of 0s Column 1 contains a 0 (as well as Row 2). Let's choose Column 1. 𝐴=[   1−23   041−21−5]𝐴 =\begin{bmatrix} \boxed{\ \ \ 1}&-2&3\\ \boxed{\ \ \ 0}&4&1\\ \boxed{-2}&1&-5 \end{bmatrix}
  2. Assign ++ or −- signs to each entry in that row or column (top left corner starts with ++, continue checkerboard pattern) [+−+−+−+−+]\begin{bmatrix} \boxed+&-&+\\ \boxed-&+&-\\ \boxed+&-&+ \end{bmatrix}
  3. Set up the determinant equation with these signs det⁡A=+−+\det A= + \qquad - \qquad +
  4. Write the entries of your chosen row or column det⁡A=+1−0+(−2)…\det A = +1 \qquad -0 \qquad +(-2) \quad\dots
  5. Multiply by the determinants of the sub-matrices (matrix minors) det⁡A=+1 ∣411−5∣−0 ∣−231−5∣+(−2) ∣−2341∣=1[(4)(−5)−(1)(1)]−0+(−2)[(−2)(1)−(3)(4)]=−21−0+28=7\begin{array}{rclll} \det A &=& +1\ \left\vert \begin{array}{rr} 4&1\\ 1&-5 \end{array} \right\vert & -0\ \left\vert \begin{array}{rr} -2&3\\ 1&-5 \end{array} \right\vert & +(-2)\ \left\vert \begin{array}{rr} -2&3\\ 4&1 \end{array} \right\vert\\[2em] &=& 1\big[ (4)(-5) - (1)(1) \big] & -0 & +(-2)\big[ (-2)(1) - (3)(4) \big]\\[1.5em] &=& -21&-0&+28\\[1.5em] &=& \boxed{7} \end{array}
B=[40151002−2]B= \begin{bmatrix} 4&0&1\\ 5&1&0\\ 0&2&-2 \end{bmatrix}

Find det⁡B\det B.


Compute the determinant of A=[−214−110−125−121003−1]A=\left[\begin{array}{rrrr} -2&1&4&-1\\ 1&0&-1&2\\ 5&-1&2&1\\ 0&0&3&-1 \end{array}\right].
Extra Practice
133 - FML 3 - 18.1W e.g. 8
Find the determinant of the matrix A=  [102−321−21−101−10−21−1]\bcb{\boldsymbol{ A =  \begin{bmatrix} 1 & 0 & 2 & -3 \\ 2 & 1 & -2 & 1 \\ -1 & 0 & 1 & -1 \\ 0 & -2 & 1 & - 1\end{bmatrix} }}
133 - FML 3 - 18.1W e.g. 77
Find the determinant of the matrix A=[102−3021−210−101100−21−11001−10]\bcb{\boldsymbol{A = \begin{bmatrix} 1 & 0 & 2 & -3 & 0 \\ 2 & 1 & -2 & 1 & 0 \\ -1 & 0 & 1 & 1 & 0 \\ 0 & -2 & 1 & -1 & 1 \\ 0 & 0 & 1 & -1 & 0 \end{bmatrix} }}
Determinants of Large Matrices
Compute the determinant of A=[−214−110−125−121003−1]A=\left[\begin{array}{rrrr} -2&1&4&-1\\ 1&0&-1&2\\ 5&-1&2&1\\ 0&0&3&-1 \end{array}\right].
Determinants of Large Matrices
B=[40151002−2]B= \begin{bmatrix} 4&0&1\\ 5&1&0\\ 0&2&-2 \end{bmatrix}
Find det⁡B\det B.
Determinants and Inverse
Let B=[1−102111000−11−21−2−1]B=\left[\begin{array}{rrrr} 1&-1&0&2\\ 1&1&1&0\\ 0&0&-1&1\\ -2&1&-2&-1 \end{array}\right], and note that det(B)=−1\text{det}(B)=-1.
Use the inverse formula to find the (4,3)(4,3)-entry of B−1B^{-1}.
Practice: Determinant
Practice Question: Determinant
Find the determinant of [40151002−2]\begin{bmatrix} 4&0&1\\ 5&1&0\\ 0&2&-2 \end{bmatrix}.
Practice: Determinant
Practice: Determinant
Find the determinant of [40151002−2]\begin{bmatrix} 4&0&1\\ 5&1&0\\ 0&2&-2 \end{bmatrix}.
Cramer's Rule: Linear Systems
Let A=[−1−2430−215−1]A=\begin{bmatrix} -1&-2&4\\ 3&0&-2\\ 1&5&-1 \end{bmatrix}.
(a) Find detA\text{det}A.
(b) User Cramer's Rule to find the value of xx in the unique solution to Ax=bA\bm{x}=\bm{b}, where x=[xyz]\bm{x}=\begin{bmatrix} x\\y\\z \end{bmatrix} and b=[00−2]\bm{b}=\begin{bmatrix} 0\\0\\-2 \end{bmatrix}
Determinants and Inverses
Let A=[abcdef12−3]A=\begin{bmatrix} a&b&c\\ d&e&f\\ 1&2&-3 \end{bmatrix}, where it is known that det[abde]=5\text{det}\begin{bmatrix} a&b\\d&e \end{bmatrix}=5, det[acdf]=−2\text{det}\begin{bmatrix} a&c\\d&f \end{bmatrix}=-2 and det[bcef]=3\text{det}\begin{bmatrix} b&c\\e&f \end{bmatrix}=3.
(a) Find detA\text{det}A.
(b) Find the (2,3)-entry of AdjA\text{Adj}A.
Determinants of Large Matrices
Given A=[459093−20−1]A= \begin{bmatrix} 4&5&9\\ 0&9&3\\ -2&0&-1\\ \end{bmatrix},
a.) Find the (2, 1)-minor of A.
b.) Find the (1, 3)-cofactor of A.
Determinants of Large Matrices
Compute the determinant of B=[021−4−1321001−2−1322]B= \begin{bmatrix} 0&2&1&-4\\ -1&3&2&1\\ 0&0&1&-2\\ -1&3&2&2 \end{bmatrix} .