DLA Examples

In most cases, you are better off using the row reduction method. However, there are situations where you might want to use the adjoint instead, as in the example below.

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Example 10.4.1. Using the adjoint to compute an inverse.

It can be tedious to row reduce a matrix with variable entries. Consider

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X = [\begin{matrix} x & 1 & - 1 \\ x - 1 & 0 & x \\ 0 & x & 1 \end{matrix}] .

To row reduce

X,

our first step would be to obtain a leading one in the first column. We might choose to perform

R_{1} \to \frac{1}{x} R_{1},

except that this operation would be invalid in the case that

x = 0 .

Or we might choose to perform

R_{2} \to \frac{1}{x - 1} R_{2},

except that this operation would be invalid in the case that

x = 1 .

So to row reduce

X

we would need to consider three different cases,

x = 0,

x = 1,

and

x \neq 0, 1,

performing different row reduction sequences in each of these cases. And when we get to the point of trying to obtain a leading one in the second column, we might discover there are even more cases to consider.

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So instead we will attempt to compute the inverse of

X

using the adjoint. First, the minors.

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\begin{aligned} M_{11} & = | \begin{array}{c} 0 & x \\ x & 1 \end{array} | = - x^{2} & M_{12} & = | \begin{array}{c} x - 1 & x \\ 0 & 1 \end{array} | = x - 1 & M_{13} & = | \begin{array}{c} x - 1 & 0 \\ 0 & x \end{array} | = x^{2} - x \\ M_{21} & = | \begin{array}{c} 1 & - 1 \\ x & 1 \end{array} | = 1 + x & M_{22} & = | \begin{array}{c} x & - 1 \\ 0 & 1 \end{array} | = x & M_{23} & = | \begin{array}{c} x & 1 \\ 0 & x \end{array} | = x^{2} \\ M_{31} & = | \begin{array}{c} 1 & - 1 \\ 0 & x \end{array} | = x & M_{32} & = | \begin{array}{c} x & - 1 \\ x - 1 & x \end{array} | = x^{2} + x - 1 & M_{33} & = | \begin{array}{c} x & 1 \\ x - 1 & 0 \end{array} | = 1 - x \end{aligned}

We obtain the matrix of cofactors by making certain minor determinants negative, according to the

3 \times 3

pattern of cofactor signs, and then the adjoint is the transpose.

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\begin{aligned} C_{X} & = [\begin{array}{c} - x^{2} & 1 - x & x^{2} - x \\ - 1 - x & x & - x^{2} \\ x & - x^{2} - x + 1 & 1 - x \end{array}] \\ adj X & = [\begin{array}{c} - x^{2} & - 1 - x & x \\ 1 - x & x & - x^{2} - x + 1 \\ x^{2} - x & - x^{2} & 1 - x \end{array}] \end{aligned}

To compute the inverse of

X,

we still need its determinant. But we already have all the cofactors, so a cofactor expansion will be easy. Let’s do a cofactor expansion of

det X

along the third row. (Remember that the cofactors already have the appropriate signs, so we are just summing “entry times cofactor” terms.)

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det X = 0 x + x (- x^{2} - x + 1) + 1 (1 - x) = 1 - x^{3} - x^{2}

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Finally, we obtain a formula for the inverse of

X

that is valid for every value of

x

for which the determinant is nonzero,

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X^{- 1} = \frac{1}{1 - x^{3} - x^{2}} [\begin{matrix} - x^{2} & - 1 - x & x \\ 1 - x & x & - x^{2} - x + 1 \\ x^{2} - x & - x^{2} & 1 - x \end{matrix}] .

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Subsection 10.4.3 Cramer’s rule

Example 10.4.2. Using Cramer’s rule to compute individual variable values in a system of equations.

Consider the system

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{\begin{array}{rcrcrcrcr} x_{1} & - & x_{2} & + & 2 x_{3} & + & x_{4} & = & 1, \\ 2 x_{1} & + & x_{3} & + & x_{4} & = & 1, \\ x_{2} & - & 3 x_{4} & = & 0, \\ x_{1} & - & 2 x_{2} & - & x_{3} & = & 1, \end{array}

with coefficient matrix and vector of constants,

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\begin{aligned} A & = [\begin{array}{rrrr} 1 & - 1 & 2 & 1 \\ 2 & 0 & 1 & 1 \\ 0 & 1 & 0 & - 3 \\ 1 & - 2 & - 1 & 0 \end{array}], & b & = [\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array}] . \end{aligned}

Conveniently, we have already computed

det A = - 27

in Example 8.4.8 (and again in Example 9.3.1). Since

det A \neq 0,

we know that

A

is invertible and so the system has one unique solution. Suppose we want to know the value of

x_{2}

in the solution. We can form the matrix

A_{2},

where the second column of

A

is replaced by

b,

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A_{2} = [\begin{array}{rrrr} 1 & 1 & 2 & 1 \\ 2 & 1 & 1 & 1 \\ 0 & 0 & 0 & - 3 \\ 1 & 1 & - 1 & 0 \end{array}],

and compute

det A_{2}

by a cofactor expansion along the third row (expanding the corresponding

3 \times 3

minor determinant along the first row),

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\begin{aligned} det A_{2} & = - (- 3) | \begin{array}{rrr} 1 & 1 & 2 \\ 2 & 1 & 1 \\ 1 & 1 & - 1 \end{array} | \\ = 3 (1 | \begin{array}{rr} 1 & 1 \\ 1 & - 1 \end{array} | - 1 | \begin{array}{rr} 2 & 1 \\ 1 & - 1 \end{array} | + 2 | \begin{array}{rr} 2 & 1 \\ 1 & 1 \end{array} |) \\ = 3 ((- 1 - 1) - (- 2 - 1) + 2 (2 - 1)) \\ = 9 . \end{aligned}

Thus, the value of

x_{2}

in the one unique solution to the system is

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x_{2} = \frac{det A_{2}}{det A} = \frac{9}{- 27} = - \frac{1}{3} .

If we also want to know the value of

x_{4},

we form the matrix

A_{4},

where the fourth column of

A

is replaced by

b,

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A_{4} = [\begin{array}{rrrr} 1 & - 1 & 2 & 1 \\ 2 & 0 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & - 2 & - 1 & 1 \end{array}],

and compute

det A_{4}

(again by a cofactor expansion along the third row, followed by an expansion along the first row of the corresponding

3 \times 3

minor determinant),

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\begin{aligned} det A_{4} & = - 1 [\begin{array}{rrr} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & - 1 & 1 \end{array}] \\ = - 1 (1 [\begin{array}{rr} 1 & 1 \\ - 1 & 1 \end{array}] - 2 [\begin{array}{rr} 2 & 1 \\ 1 & 1 \end{array}] + 1 [\begin{array}{rr} 2 & 1 \\ 1 & - 1 \end{array}]) \\ = - ((1 + 1) - 2 (2 - 1) + (- 2 - 1)) \\ = 3. \end{aligned}

Thus, the value of

x_{4}

in the one unique solution to the system is

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x_{4} = \frac{det A_{4}}{det A} = \frac{3}{- 27} = - \frac{1}{9} .

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Section 10.4 Examples

In this section.

Subsection 10.4.1 The 2×2 case

Subsection 10.4.2 Computing an inverse using the adjoint

Example 10.4.1. Using the adjoint to compute an inverse.

Subsection 10.4.3 Cramer’s rule

Example 10.4.2. Using Cramer’s rule to compute individual variable values in a system of equations.

Subsection 10.4.1 The $2 \times 2$ case