High School
SAT
ACT
University
MCAT
LSAT
DAT

Determinants and Inverses

Related Topics

Wize University Linear Algebra Textbook > Determinants

Determinants and Inverse

3 Activities

Wize University Linear Algebra Textbook > Determinants

Determinants of Large Matrices

5 Activities

checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.
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.
More Determinants and Inverse Questions:
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)}
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).
More Determinants of Large Matrices Questions:
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 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} .