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Cramer's Rule: Linear Systems

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Wize University Linear Algebra Textbook > Determinants

Cramer's Rule for Solving Linear Systems

3 Activities

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 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 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}