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Cramer’s Rule and Inverse Matrices

Suppose that we need to solve the $n \times n$ $n \times n$ linear system

A x = B,

$A x = B,$ (18)

where

\begin{array}{l} A = [a_{i j}], & x = [\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{array}] & and & B = [\begin{array}{c} b_{1} \\ b_{2} \\ ⋮ \\ b_{n} \end{array}] \end{array} .

$\begin{array}{l} A = [a_{i j}], & x = [\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{array}] & and & B = [\begin{array}{c} b_{1} \\ b_{2} \\ ⋮ \\ b_{n} \end{array}] \end{array} .$ (19)

We assume that the coefficient matrix A is invertible, so we know in advance that a unique solution x of (18) exists. The question is how to write x explicitly in terms of the coefficients $a_{i j}$ $a_{i j}$ and the constants $b_{i} .$ $b_{i} .$ In the following discussion, we think of x as a fixed (though as yet unknown) vector.

If we denote by $a_{1}, a_{2}, \dots, a_{n}$ $a_{1}, a_{2}, \dots, a_{n}$ the column vectors of the $n \times n$ $n \times n$ matrix A, then

A = [\begin{array}{l} a_{1} & a_{2} & \dots & a_{n} \end{array}] .

$A = [\begin{array}{l} a_{1} & a_{2} & \dots & a_{n} \end{array}] .$ (20)

By Fact 1 in Section 3.5, we can rewrite Eq. (18) as

x_{1} a_{1} + x_{2} a_{2} + \dots x_{n} a_{n} = b .

$x_{1} a_{1} + x_{2} a_{2} + \dots x_{n} a_{n} = b .$

Thus the constant vector b is expressed in terms of the entries $x_{1}, x_{2}, \dots, x_{n}$ $x_{1}, x_{2}, \dots, x_{n}$ of the solution vector x and the column vectors of A by

b = \sum_{j = 1}^{n} x_{j} a_{j} .

$b = \sum_{j = 1}^{n} x_{j} a_{j} .$ (21)

The trick for finding the ith unknown $x_{i}$ $x_{i}$ is to compute the determinant of the matrix

[\begin{array}{c} a_{1} & \dots & b & \dots & a_{n} \end{array}] = [\begin{array}{c} a_{11} & \dots & b_{1} & \dots & a_{1 n} \\ a_{21} & \dots & b_{2} & \dots & a_{2 n} \\ ⋮ & ⋱ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & \dots & b_{n} & \dots & a_{n n} \end{array}]

$[\begin{array}{c} a_{1} & \dots & b & \dots & a_{n} \end{array}] = [\begin{array}{c} a_{11} & \dots & b_{1} & \dots & a_{1 n} \\ a_{21} & \dots & b_{2} & \dots & a_{2 n} \\ ⋮ & ⋱ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & \dots & b_{n} & \dots & a_{n n} \end{array}]$ (22)

that we obtain by replacing the ith column $a_{i}$ $a_{i}$ of A with the constant vector b. Using

Eq. (21) to substitute for b, we find that

\begin{array}{rcl} | \begin{array}{r} a_{1} & \dots & b & \dots & a_{n} \end{array} | & = & | \begin{array}{c} a_{1} & \dots & \sum_{j = 1}^{n} x_{j} a_{j} & \dots & a_{n} \end{array} | \\ = & \sum_{j = 1}^{n} | \begin{array}{c} a_{1} & \dots & x_{j} a_{j} & \dots & a_{n} \end{array} | & \begin{array}{l} (by Property 4 of \\ determinants) \end{array} \\ = & \sum_{j = 1}^{n} x_{j} | \begin{array}{c} a_{1} & \dots & a_{j} & \dots & a_{n} \end{array} | & \begin{array}{l} (by Property 1 of \\ determinants) . \end{array} \end{array}

$\begin{array}{rcl} | \begin{array}{r} a_{1} & \dots & b & \dots & a_{n} \end{array} | & = & | \begin{array}{c} a_{1} & \dots & \sum_{j = 1}^{n} x_{j} a_{j} & \dots & a_{n} \end{array} | \\ = & \sum_{j = 1}^{n} | \begin{array}{c} a_{1} & \dots & x_{j} a_{j} & \dots & a_{n} \end{array} | & \begin{array}{l} (by Property 4 of \\ determinants) \end{array} \\ = & \sum_{j = 1}^{n} x_{j} | \begin{array}{c} a_{1} & \dots & a_{j} & \dots & a_{n} \end{array} | & \begin{array}{l} (by Property 1 of \\ determinants) . \end{array} \end{array}$

Note that, in the jth term of this summation, the vector $a_{j}$ $a_{j}$ appears in the ith position. Thus we have found that

\begin{aligned} | \begin{array}{c} a_{1} & \dots & b & \dots & a_{n} \end{array} | & = x_{1} | \begin{array}{c} a_{1} & a_{2} & \dots & a_{1} & \dots & a_{n} \end{array} | \\ + x_{2} | \begin{array}{c} a_{1} & a_{2} & \dots & a_{2} & \dots & a_{n} \end{array} | \\ ⋮ \\ + x_{i} | \begin{array}{c} a_{1} & a_{2} & \dots & a_{i} & \dots & a_{n} \end{array} | \\ ⋮ \\ + x_{n} | \begin{array}{c} a_{1} & a_{2} & \dots & a_{n} & \dots & a_{n} \end{array} | . \end{aligned}

$\begin{aligned} | \begin{array}{c} a_{1} & \dots & b & \dots & a_{n} \end{array} | & = x_{1} | \begin{array}{c} a_{1} & a_{2} & \dots & a_{1} & \dots & a_{n} \end{array} | \\ + x_{2} | \begin{array}{c} a_{1} & a_{2} & \dots & a_{2} & \dots & a_{n} \end{array} | \\ ⋮ \\ + x_{i} | \begin{array}{c} a_{1} & a_{2} & \dots & a_{i} & \dots & a_{n} \end{array} | \\ ⋮ \\ + x_{n} | \begin{array}{c} a_{1} & a_{2} & \dots & a_{n} & \dots & a_{n} \end{array} | . \end{aligned}$

Of the n determinants on the right-hand side here, all but the ith one have two identical columns and therefore are equal to zero. The coefficient of $x_{i}$ $x_{i}$ in the ith term is simply

| A | = | \begin{array}{c} a_{1} & a_{2} & \dots & a_{i} & \dots & a_{n} \end{array} | .

$| A | = | \begin{array}{c} a_{1} & a_{2} & \dots & a_{i} & \dots & a_{n} \end{array} | .$

Consequently, a result of our computation is that

| \begin{array}{c} a_{1} & \dots & b & \dots & a_{n} \end{array} | = x_{i} | A | .

$| \begin{array}{c} a_{1} & \dots & b & \dots & a_{n} \end{array} | = x_{i} | A | .$ (23)

We get the desired simple formula for $x_{i}$ $x_{i}$ after we divide each side by $| A | \neq 0.$ $| A | \neq 0.$

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Table of Contents for Cramer’s Rule and Inverse Matrices

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Table of Contents for
Cramer’s Rule and Inverse Matrices