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Wize University Linear Algebra Textbook > Determinants

Determinants and Inverse

(Optional) Shortcut for 3x3 DeterminantsPrevious Section
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Determinants and Inverse

Cofactor Matrix

Recall that every entry in a square matrix AA has an associated cofactor called CijC_{ij}.
We can form a matrix called the matrix of cofactors of AA. For example, for A3×3A_{3 \times 3}:
CA=[C11C12C13C21C22C23C31C32C33]=[+M11−M12+M13−M21+M22−M23+M31−M32+M33]C_A =\left[ \begin{array}{c} C_{11} & C_{12} & C_{13}\\ C_{21} & C_{22} & C_{23}\\ C_{31} & C_{32} & C_{33} \end{array}\right] =\left[ \begin{array}{c} +M_{11} & -M_{12} & +M_{13}\\ -M_{21} & +M_{22} & -M_{23}\\ +M_{31} & -M_{32} & +M_{33} \end{array}\right]
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Adjoint Matrix

The adjoint matrix of AA, denoted adj(A)\text{adj}(A), is the transpose of the matrix of cofactors of AA.
adj(A)=(CA)T\boxed{\quad {\rm adj}(A)=\big(C_A\big)^T \quad}

Example
Given A=[abcd]A=\left[\begin{array}{rr}a&b\\c&d\end{array}\right], find adj(A){\rm adj}(A).
The minors of each entry are:
M11=dM12=cM21=bM22=aM_{11}=d \qquad M_{12}=c \qquad M_{21}=b \qquad M_{22}=a
Then the matrix of cofactors of AA is
CA=[C11C12C21C22]=[+M11−M12−M21+M22]=[d−c−ba]C_A= \left[\begin{array}{rr} C_{11}&C_{12}\\ C_{21}&C_{22} \end{array}\right] = \left[\begin{array}{rr} +M_{11}&-M_{12}\\ -M_{21}&+M_{22} \end{array}\right] = \left[\begin{array}{rr} d&-c\\ -b&a \end{array}\right]
Therefore, the adjoint is:
adj(A)=(CA)T=[d−b−ca] \text{adj}(A) =\big(C_A\big)^T =\left[\begin{array}{rr}d&-b\\-c&a\end{array}\right]

Wize Concept
The result is the same matrix found in the formula for finding the inverse of a 2×22\times 2 matrix.
Given A=[abcd]A=\left[\begin{array}{rr}a&b\\c&d\end{array}\right],
A−1=1ad−bc[d−b−ca]⏟adj(A)=1det(A)adj(A)A^{-1} =\dfrac{1}{ad-bc} \underbrace{\left[\begin{array}{rr}d&-b\\-c&a\end{array}\right]}_{\colorThree{\normalsize {\rm adj}(A)}} =\dfrac{1}{\text{det}(A)}\colorThree{{\rm adj}(A)}


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Matrix Inverse Formula

We can generalize this result to write a formula for the inverse of any n×nn \times n matrix AA:
A−1=1det(A) adj(A)\boxed{\quad A^{-1} = \dfrac{1}{\text{det}(A)}\ {\rm adj}(A) \quad}
Watch Out!
The notation for the determinant and adjoint are similar, but they produce very different results.
  • The determinant is a scalar: det(A)=a∈R⟵single number\det A = a \in \reals \quad \longleftarrow \text{single number}
  • The adjoint is a matrix: adj(A)=(CA)T∈Rn×n⟵n×n  matrix{\rm adj}(A) = (C_A)^T \in \reals^{n \times n} \quad \longleftarrow n \times n \ \ \text{matrix}


Wize Tip
If det(A)=0\text{det}(A)=0, then the inverse A−1A^{-1} does not exist.

So, for AA to be invertible, we must have that det(A)≠0\det{A}\neq0

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Example: Inverse and Determinants

Find the inverse of matrix A=[−20131−1047]A=\left[\begin{array}{rrr} -2&0&1\\ 3&1&-1\\ 0&4&7 \end{array}\right] using the adjoint.

Find the cofactor of each entry, keeping in mind the "checkerboard pattern" of when to change the sign of the minor:
C11=∣1−147∣=11C12=−∣3−107∣=−21C13=∣3104∣=12C21=−∣0147∣=4C22=∣−2107∣=−14C23=−∣−2004∣=8C31=∣011−1∣=−1C32=−∣−213−1∣=1C33=∣−2031∣=−2\begin{array}{lll} C_{11}=\left|\begin{array}{rr}1&-1\\4&7\end{array}\right|=11 \qquad& C_{12}=-\left|\begin{array}{rr}3&-1\\0&7\end{array}\right|=-21 \qquad& C_{13}=\left|\begin{array}{rr}3&1\\0&4\end{array}\right|=12\\[1.5em] C_{21}=-\left|\begin{array}{rr}0&1\\4&7\end{array}\right|=4 & C_{22}=\left|\begin{array}{rr}-2&1\\0&7\end{array}\right|=-14 & C_{23}=-\left|\begin{array}{rr}-2&0\\0&4\end{array}\right|=8\\[1.5em] C_{31}=\left|\begin{array}{rr}0&1\\1&-1\end{array}\right|=-1 & C_{32}=-\left|\begin{array}{rr}-2&1\\3&-1\end{array}\right|=1 & C_{33}=\left|\begin{array}{rr}-2&0\\3&1\end{array}\right|=-2 \end{array}
Arranging these into a matrix we get the matrix of cofactors of AA:
CA=[11−21124−148−11−2]C_A=\left[\begin{array}{rrr} 11&-21&12\\ 4&-14&8\\ -1&1&-2 \end{array}\right]
Then the adjoint of AA is the transpose of this matrix:
adj(A)=[114−1−21−141128−2]\text{adj}(A)= \left[\begin{array}{rrr} 11&4&-1\\ -21&-14&1\\ 12&8&-2 \end{array}\right]
The inverse formula requires us to find det(A)\det A.
Since we already have the cofactors, we can easily write the cofactor expansion along Row 1 of AA:
det(A)=a11C11+a12C12+a13C13=(−2)(11)+(0)(−21)+(1)(12)=−10\begin{aligned} \det A &= a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}\\ &=(-2)(11) + (0)(-21) + (1)(12)\\ &=-10 \end{aligned}
∴ A−1=1det(A) adj(A)=−110[114−1−21−141128−2]\therefore \ A^{-1} = \dfrac{1}{\det A}\ {\rm adj}(A) =-\dfrac{1}{10} \left[\begin{array}{rrr} 11&4&-1\\ -21&-14&1\\ 12&8&-2 \end{array}\right]
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}.
Extra Practice
Practice Question: Determinant Properties
True or False?
If 𝐴 is a 4 × 4 matrix and det(𝐴) = −3, then the homogeneous system of linear equations 𝐴𝑥⃗=0⃗𝐴𝑥⃗=\vec{0} has at least one non-trivial solution.
For what value of d, the following linear system has at least one solution:
{𝑥+𝑦+𝑧=0−𝑥+𝑧=1𝑥+3𝑧=𝑑\begin{cases} 𝑥 + 𝑦 + 𝑧 = 0\\ −𝑥 + 𝑧 = 1\\ 𝑥 + 3𝑧 = 𝑑 \end{cases}
133 - FML 3 - 18.1W e.g. 17.2
If   ⁣A‾\bcb{\boldsymbol{ \ul{A} }} is a 3×3\bcb{\boldsymbol{ 3 \times 3}} matrix with det(  ⁣A‾)=−4\bcb{\boldsymbol{ \det{\ul{A}} = -4}}, find:
det((2A)−1)\boldsymbol{ \det{(2A)^{-1}}}
133 - FML 3 - 18.1W e.g. 15
  ⁣A‾2=4  ⁣A‾ and   ⁣A‾\bcb{\boldsymbol{ \ul{A}^2 = 4\ul{A} \text{ and } \ul{A} }} is an invertible matrix, then a possible value for det(  ⁣A‾)\bcb{\boldsymbol{\det{\ul{A}}}} is:
133 - FML 3 - 18.1W e.g. 69
Given   ⁣A‾ and   ⁣B‾ 3×3\bcb{\boldsymbol{ \ul{A} \text{ and } \ul{B} ~ 3 \times 3}} matrices with det(  ⁣A‾)=−4 and det(BT)=3\bcb{\boldsymbol{\det{\ul{A}} = -4 \text{ and } \det{ \mtran{B} } = 3}} find det(2  ⁣A‾ BT   ⁣A‾−2  ⁣B‾ AT B−1)\bcb{\boldsymbol{ \det { 2\ul{A} \, \mtran{B} \, \ul{A}^{-2} \ul{B} \, \mtran{A} \, \minv{B} } }}
.
133 - FML 3 - 18.1W e.g. 63
  ⁣A‾\bcb{\boldsymbol{ \ul{A} }} is a 3×3\bcb{\boldsymbol{ 3 \times 3}} matrix such that det(−2(AT)−1)=8 then det(  ⁣A‾)\bcb{\boldsymbol{ \det{-2(\mtran{A})^{-1} }= 8 \text{ then } \det{\ul{A}}}} is:
Shortcut for 3x3 Determinants
Let A=[−61aa−2001−2]A=\left[\begin{array}{rrr} -6&1& a\\ a&-2&0\\ 0&1& -2 \end{array}\right]. Find the value(s) of aa such that AA is not invertible.
Practice: Determinant of $3\times3$ Matrix
Let A=[61aa−20012]A=\left[\begin{array}{rrr} 6&1&a\\ a&-2&0\\ 0&1&2 \end{array}\right] using cofactor expansion along any row or column
Find the value(s) of aa that will make AA invertible
Let B=[𝑎𝑏𝑐𝑑𝑒𝑓𝑔h𝑖]B=\begin{bmatrix} 𝑎&𝑏&𝑐\\ 𝑑&𝑒&𝑓\\ 𝑔&ℎ&𝑖 \end{bmatrix} be a 3×33\times3 matrix. Assume that det(B)=20\det B = 20. Determine if the matrix A that appears below is invertible by computing its determinant.
A=[5𝑎5𝑏05𝑐0010−𝑔−h0−i𝑑−10𝑎𝑒−10𝑏0𝑓−10𝑐]A=\begin{bmatrix} 5𝑎&5𝑏&0&5𝑐\\ 0&0&1&0\\ −𝑔&-h&0&-i\\ 𝑑 − 10𝑎&𝑒 − 10𝑏&0&𝑓 − 10𝑐 \end{bmatrix}
Let B=[𝑎𝑏𝑐𝑑𝑒𝑓𝑔h𝑖]be a 3X3 matrix. Assume that det(B)=20.Let\ B=\begin{bmatrix} 𝑎&𝑏&𝑐\\ 𝑑&𝑒&𝑓\\ 𝑔&ℎ&𝑖 \end{bmatrix}be\ a\ 3X3\ matrix.\ Assume\ that\ det(B) = 20.
Determine if the matrix A that appears below is invertible by computing its determinant.
Practice: Determinant Properties
Practice: Determinant Properties
True or False?
If 𝐴 is a 4 × 4 matrix and det(𝐴) = −3, then det(AT)=1det(A−1)det\left(A^T\right)=\frac{1}{det\left(A^{-1}\right)}
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.
Cramer's Rule for Solving Linear Systems: Determinants and Inverses
Let AA be a 3 x 3 matrix with detA=−18\text{det}A=-18 and Adj A=[−9340−6−2006]\text{Adj}\ A= \begin{bmatrix} -9&3&4\\ 0&-6&-2\\ 0&0&6 \end{bmatrix}
(a) Find A−1A^{-1}.
(b) Find the solution to Ax=bA\bm{x}=\bm{b}, where x=[xyz]\bm{x}=\begin{bmatrix} x\\y\\z \end{bmatrix} and b=[−120]\bm{b}=\begin{bmatrix} -1\\2\\0 \end{bmatrix}
Determinants and Inverses
Find the value(s) of kk such that the matrix [10−2123k04]\begin{bmatrix} 1&0&-2\\ 1&2&3\\ k&0&4 \end{bmatrix} has no inverse.
Determinants: Basics
If A=[1234021−500−30000−1]A= \begin{bmatrix} 1&2&3&4\\ 0&2&1&-5\\ 0&0&-3&0\\ 0&0&0&-1 \end{bmatrix}, calculate det(Adj A)det(Adj\ A).