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Cramer's Rule for Solving Linear Systems: Determinants and Inverses

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Wize University Linear Algebra Textbook > Determinants

Cramer's Rule for Solving Linear Systems

3 Activities

Wize University Linear Algebra Textbook > Determinants

Determinants and Inverse

3 Activities

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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}
More Cramer's Rule for Solving Linear Systems Questions:
Solving System of Linear Equations using Cramer's Rule
Solve for zz if [4−11023201][xyz]=[123]\begin{bmatrix}4&-1&1\\0&2&3\\2&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}
Cramer Rule to solve Linear System of equations
Cramer’s rule
If AA is an n×nn\times n invertible matrix and we wish to solve AX=BAX=B for X=[x1, …, xn]TX=\left[x_1,\ \ldots,\ x_n\right]^T, then Cramer's Rule gives
xi=det(Ai)det(A)x_i=\frac{det\left(A_i\right)}{det\left(A\right)}
Cramer's Rule
Find yy given the system of linear equations:
[0−1142021−2][xyz]=[214]\left[ \begin{array}{rrr} 0&-1&1\\ 4&2&0\\ 2&1&-2 \end{array} \right] \begin{bmatrix} x\\y\\z\\ \end{bmatrix} = \begin{bmatrix} 2\\1\\4\\ \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}
Cramer's Rule: Linear Systems
Given that det[a3−12−bc1−1d]=10,  det[p3−1q−bc5−1d]=3,  det[ap−12qc15d]=7det \begin{bmatrix} a&3&-1\\ 2&-b&c\\ 1&-1&d \end{bmatrix}=10,\ \ det \begin{bmatrix} p&3&-1\\ q&-b&c\\ 5&-1&d \end{bmatrix}=3,\ \ det \begin{bmatrix} a&p&-1\\ 2&q&c\\ 1&5&d \end{bmatrix}=7 , find the value of yy in the unique solution to the following system of linear equations"
ax+3y−z=p2x−by+cz=qx−y+dz=5\begin{array}{c} ax+3y-z=p\\ 2x-by+cz=q\\ x-y+dz=5 \end{array}
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)}
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 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).