MATH 240

Mon. October 7th, 2019


Applications of the Determinant


Cramer’s Rule

Given an n×nn\times n matrix AA, by Ai(b)A_{i}(b) we denote the matrix obtained by replacing the iith column of AA with bb.

Cramer’s Rule: Let AA be an n×nn\times n invertible matrix. For any bb in Rn\R^n, the solution to Ax=bAx=b is given by xi=detAi(b)detA.x_i=\dfrac{\text{det}A_i(b)}{\text{det}A}.

This is not very useful for calculations, but makes some theoretical proofs simpler.

Proof. Consider Ii(x)I_i(x).
Ii(x)=(10...x1...001...x2...0...00...xi...0...00...xn...1)detIi(x)=10...x1...001...x2...0...00...xi...0...00...xn...1=xi10...x1...001...x2...0...00...1...0...00...xn...1=xi10...0...001...0...0...00...1...0...00...0...1=xi detIn=xiAIi(x)=A[e1e2...xei+1...en]=[Ae1Ae2...AxAei+1...Aen]=[Ae1Ae2...bAei+1...Aen]=Ai(b)detAi(b)=detAIi(x)=detA detIi(x)=xidetAxi=detAi(b)detA.\begin{aligned} I_i(x)&=\begin{pmatrix} 1&0&...&x_1&...&0\\ 0&1&...&x_2&...&0\\ ...\\ 0&0&...&x_i&...&0\\ ...\\ 0&0&...&x_n&...&1\\ \end{pmatrix}\\ \\ \text{det}I_i(x)&=\begin{vmatrix} 1&0&...&x_1&...&0\\ 0&1&...&x_2&...&0\\ ...\\ 0&0&...&x_i&...&0\\ ...\\ 0&0&...&x_n&...&1\\ \end{vmatrix}\\ &=x_i\begin{vmatrix} 1&0&...&x_1&...&0\\ 0&1&...&x_2&...&0\\ ...\\ 0&0&...&1&...&0\\ ...\\ 0&0&...&x_n&...&1\\ \end{vmatrix}\\ &=x_i\begin{vmatrix} 1&0&...&0&...&0\\ 0&1&...&0&...&0\\ ...\\ 0&0&...&1&...&0\\ ...\\ 0&0&...&0&...&1\\ \end{vmatrix}\\ &=x_i\ \text{det}I_n\\ &=x_i\\ \\ AI_i(x)&=A\begin{bmatrix} e_1&e_2&...&x&e_{i+1}&...&e_n \end{bmatrix}\\ &=\begin{bmatrix} Ae_1&Ae_2&...&Ax&Ae_{i+1}&...&Ae_n \end{bmatrix}\\ &=\begin{bmatrix} Ae_1&Ae_2&...&b&Ae_{i+1}&...&Ae_n \end{bmatrix}\\ &=A_i(b)\\ \\ \text{det}A_i(b)&=\text{det}AI_i(x)\\ &=\text{det}A\ \text{det}I_i(x)\\ &=x_i\text{det}A\\ \\ x_i&=\dfrac{\text{det}A_i(b)}{\text{det}A}. \end{aligned}


Ex. Let ss be a parameter
{3sx12x2=46x1+sx2=1\begin{cases} 3sx_1-2x_2=4\\-6x_1+sx_2=1 \end{cases}
and assume that the system has a unique solution.
A=(3s26s)detA=3s212=3(s24)=3(s2)(s+2)s2,2A1(b)=(421s)detA1(b)=4s+2A2(b)=(3s461)detA2(b)=3s+24\begin{aligned} A&=\begin{pmatrix} 3s&-2\\-6&s \end{pmatrix}\\ \text{det}A&=3s^2-12\\ &=3(s^2-4)\\ &=3(s-2)(s+2)\\ s&\neq 2,-2\\ \\ A_1(b)&=\begin{pmatrix} 4&-2\\1&s \end{pmatrix}\\ \text{det}A_1(b)&=4s+2\\ \\ A_2(b)&=\begin{pmatrix} 3s&4\\-6&1 \end{pmatrix}\\ \text{det}A_2(b)&=3s+24\\ \end{aligned}

By Cramer’s Rule:
x1=4s+23(s2)(s+2)x2=3s+243(s2)(s+2)\begin{aligned} x_1&=\dfrac{4s+2}{3(s-2)(s+2)}\\ x_2&=\dfrac{3s+24}{3(s-2)(s+2)} \end{aligned}


Cramer’s Rule also provides a formula to find the inverse of matrices. (You really shouldn’t use this for computations because it’s harder than other methods we’ve learned.)
A1=1detA(C11C21...Cn1C12C22...Cn2...C1nC2n...Cnn)A^{-1}=\dfrac{1}{\text{det}A}\begin{pmatrix} C_11&C_21&...&C_n1\\ C_12&C_22&...&C_n2\\ ...\\ C_1n&C_2n&...&Cnn \end{pmatrix}
where CijC_ij are cofactors that were defined previously. (The matrix on the right here is called the classical adjoint of AA.)


Geometric Applications

If AA is a 2×22\times 2 matrix, the area of the parallelogram determined by the columns of AA is detA\mid\text{det}A\mid. This also applies to the volume of parallelopipeds for 3×33\times 3 matrices, and for higher-dimensional equivalents of parallelograms and volume for larger square matrices.

By determined by the columns of AA, we mean the parallelogram that has (0,0)(0,0) and each column of AA as vertices. The remaning vertices can be determined from this information.


Let T:R2R2T: \R^2\rightarrow\R^2 be a linear transformation determined by the matrix AA. If SS is an object in R2\R^2 with finite area, then the area of T(S)T(S) is the product of detA\mid\text{det}A\mid and the area of SS. (T(S)T(S) means applying TT to every point in SS.)