Practice Problems 6

Lecture 6: The Determinant and the Inverse

Problem numbers refer to the textbook (Chiang & Wainwright, 4th edition). These are not graded, but you should attempt them all before the quiz. Try each part yourself first, then expand the Solution to check.

Exercise 5.2

1. Evaluate the following determinants:

$| \begin{array}{lll} 4 & 0 & 2 \\ 6 & 0 & 3 \\ 8 & 2 & 3 \end{array} |$

Solution

$\begin{aligned} 8 | \begin{array}{ll} 0 & 1 \\ 0 & 3 \end{array} | - 1 | \begin{array}{ll} 4 & 1 \\ 6 & 3 \end{array} | + 3 | \begin{array}{ll} 4 & 0 \\ 6 & 0 \end{array} | \\ = & 8 (0 - 0) - 1 (12 - 6) + 3 (0 - 0) \\ = & 0 - 6 + 0 \\ = & - 6 \end{aligned}$

$| \begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array} |$

Solution

$\begin{aligned} a | \begin{array}{ll} c & a \\ a & b \end{array} | - b | \begin{array}{cc} b & a \\ c & b \end{array} | + c | \begin{array}{cc} b & c \\ c & a \end{array} | \\ = & a (c b - a^{2}) - b (b^{2} - a c) + c (a b - c^{2}) \\ = & a b c - a^{3} - b^{3} + a b c + a b c - c^{3} \\ = & - a^{3} - b^{3} - c^{3} + 3 a b c \end{aligned}$

$| \begin{array}{rrr} x & 5 & 0 \\ 3 & y & 2 \\ 9 & - 1 & 8 \end{array} |$

Solution

$\begin{aligned} x | \begin{array}{cc} y & 2 \\ - 1 & 8 \end{array} | - 5 | \begin{array}{cc} 3 & 2 \\ 9 & 8 \end{array} | + 0 | \begin{array}{rr} 3 & y \\ 9 & - 1 \end{array} | \\ = & x (8 y + 2) - 5 (24 - 18) + 0 (- 3 - 9 y) \\ = & 8 x y + 2 x - 30 + 0 \\ = & 8 x y + 2 x - 30 \end{aligned}$

2. Determine the signs to be attached to the relevant minors in order to get the following cofactors of a determinant: $| C_{13} |, | C_{23} |, | C_{33} |, | C_{41} |$ , and $| C_{34} |$ .

Solution

$| C_{13} | : 1 + 3 = 4$ is even so $+$
$| C_{23} | : 2 + 3 = 5$ is odd so $-$
$| C_{33} | : 3 + 3 = 6$ is even so $+$
$| C_{41} | : 4 + 1 = 5$ is odd so $-$
$| C_{34} | : 3 + 4 = 7$ is odd so $-$

3. Given $\begin{array}{lll} a & b & c \\ d & e & f \\ g & b & i \end{array}$ , find the minors and cofactors of the elements $a, b$ and $f$ .

Solution

Minor of $a$ : $| M_{11} | = | \begin{array}{cc} e & f \\ h & i \end{array} | = e i - f h$ Cofactor of $a$ : $| C_{11} | = (- 1)^{1 + 1} | M_{11} | = | M_{11} |$

Minor of $b$ : $| M_{12} | = | \begin{array}{ll} d & f \\ g & i \end{array} | = d i - f g$ Cofactor of $b$ : $| C_{12} | = (- 1)^{3} | M_{12} | = - | M_{12} |$ Minor of $f$ : $| M_{23} | = | \begin{array}{ll} a & b \\ g & h \end{array} | = a h - b g$ Cofactor of $f$ : $| C_{23} | = (- 1)^{5} | M_{23} | = - | M_{23} |$

6. Find the minors and cofactors of the third row, given $A = [\begin{array}{rrr} 9 & 11 & 4 \\ 3 & 2 & 7 \\ 6 & 10 & 4 \end{array}]$

Solution

Minors of third row: $\begin{aligned} | M_{31} | & = | \begin{array}{cc} 11 & 4 \\ 2 & 7 \end{array} | = 69, | M_{32} | & = | \begin{array}{cc} 9 & 4 \\ 3 & 7 \end{array} | = 51, | M_{33} | & = | \begin{array}{ll} 9 & 11 \\ 3 & 2 \end{array} | = - 15 \end{aligned}$ Cofactors: $| C_{31} | = | M_{31} |, | C_{32} | = - | M_{32} |, | C_{33} | = | M_{33} |$ .

Exercise 5.3

1. Use the determinant $| \begin{array}{rrr} 4 & 0 & - 1 \\ 2 & 1 & - 7 \\ 3 & 3 & 9 \end{array} |$ to verify the first four properties of determinants.

Solution

$\begin{aligned} | \begin{array}{ccc} 4 & 0 & - 1 \\ 2 & 1 & - 7 \\ 3 & 3 & 9 \end{array} | & = 4 | \begin{array}{rr} 1 & - 7 \\ 3 & 9 \end{array} | - 0 | \begin{array}{cc} 2 & - 7 \\ 3 & 9 \end{array} | - 1 | \begin{array}{ll} 2 & 1 \\ 3 & 3 \end{array} | \\ = 4 (9 + 21) - 0 (18 + 21) - 1 (6 - 3) \\ = 120 - 0 - 3 = 117 \end{aligned}$

Interchanging rows and columns:
$\begin{aligned} | \begin{array}{ccc} 4 & 2 & 3 \\ 0 & 1 & 3 \\ - 1 & - 7 & 9 \end{array} | & = 4 | \begin{array}{cc} 1 & 3 \\ - 7 & 9 \end{array} | - 2 | \begin{array}{cc} 0 & 3 \\ - 1 & 9 \end{array} | + 3 | \begin{array}{cc} 0 & 1 \\ - 1 & - 7 \end{array} | \\ = 4 (9 + 21) - 2 (0 + 3) + 3 (0 + 1) \\ = 120 - 6 + 3 = 117 \end{aligned}$
Interchange row 1 and 2: $\begin{aligned} | \begin{array}{rrr} 2 & 1 & - 7 \\ 4 & 0 & - 1 \\ 3 & 3 & 9 \end{array} | & = 2 | \begin{array}{rr} 0 & - 1 \\ 3 & 9 \end{array} | - 1 | \begin{array}{cc} 4 & - 1 \\ 3 & 9 \end{array} | - 7 | \begin{array}{ll} 4 & 0 \\ 3 & 3 \end{array} | \\ = 2 (0 + 3) - 1 (36 + 3) - 7 (12 - 0) \\ = 6 - 39 - 84 = - 117 \end{aligned}$ You can verify the other two properties in the same way.

4. Show that when all the elements of an $n$ th-order determinant |A| are multiplied by a number $k$ , the result will be $k^{n} | A |$ .

Solution

Multiplying every element of an $n$ th-order determinant by $k$ multiplies each of its $n$ rows by $k$ . Multiplying a single row by $k$ multiplies the determinant by $k$ , so doing it to all $n$ rows gives $k^{n} | A |$ .

5. Calculate the determinant for the following matrices. Comment on whether the matrices are nonsingular.

$[\begin{array}{rrr} 4 & 0 & 1 \\ 19 & 1 & - 3 \\ 7 & 1 & 0 \end{array}]$

Solution

$\begin{array}{l} 4 | \begin{array}{cc} 1 & - 3 \\ 1 & 0 \end{array} | + 1 | \begin{array}{cc} 19 & 1 \\ 7 & 1 \end{array} | \\ = 4 (0 + 3) + 1 (19 - 7) \\ = 12 + 12 = 24 \end{array}$

Non-singular.\

$[\begin{array}{rrr} 4 & - 2 & 1 \\ - 5 & 6 & 0 \\ 7 & 0 & 3 \end{array}]$

Solution

$\begin{aligned} 5 | \begin{array}{cc} - 2 & 1 \\ 0 & 3 \end{array} | + 6 | \begin{array}{ll} 4 & 1 \\ 7 & 3 \end{array} | \\ = & 5 (- 6 - 0) + 6 (12 - 7) \\ = & - 30 + 30 = 0 \end{aligned}$

Singular.

$[\begin{array}{rrr} 7 & - 1 & 0 \\ 1 & 1 & 4 \\ 13 & - 3 & - 4 \end{array}]$

Solution

$\begin{aligned} 7 | \begin{array}{cc} 1 & 4 \\ - 3 & - 4 \end{array} | + 1 | \begin{array}{cc} 1 & 4 \\ 13 & - 4 \end{array} | \\ = & 7 (- 4 + 12) + 1 (- 4 - 52) \\ = & 56 - 56 = 0 \end{aligned}$

Singular.

$[\begin{array}{rrr} - 4 & 9 & 5 \\ 3 & 0 & 1 \\ 10 & 8 & 6 \end{array}]$

Solution

$\begin{aligned} - 3 | \begin{array}{ll} 9 & 5 \\ 8 & 6 \end{array} | - 1 | \begin{array}{cc} - 4 & 9 \\ 10 & 8 \end{array} | \\ = & - 3 (54 - 40) - 1 (- 32 - 90) \\ = & - 42 + 122 = 80 \end{aligned}$ Non-singular.

8. Comment on the validity of the following statements:

Given any matrix A, we can always derive from it a transpose and a determinant.

Solution

False. While we can find the transpose of any matrix, the determinant is only defined for square matrices.

Multiplying each element of an $n \times n$ determinant by 2 will double the value of that determinant.

Solution

False. Multiplying each element of an $n \times n$ by 2 will increase the determinant by $2^{n}$ times.

If a square matrix $A$ vanishes, then we can be sure that the equation system $A x = d$ is nonsingular.

Solution

$A$ cannot be 0, but the determinant of $A$ can. In any case, if $| A |$ =0, $A$ is singular.

Exercise 5.4

2. Find the inverse of each of the following matrices:

$A = [\begin{array}{ll} 5 & 2 \\ 0 & 1 \end{array}]$

Solution

Note that $A^{- 1} = \frac{1}{| A |} adj A$

Here, $adj A = C^{T}$ where $C = [| C_{i j} |]$ .

$| A | = 5 - 0 = 5$ $adj A = [\begin{array}{ll} | C_{11} | & | C_{21} | \\ | C_{12} | & | C_{22} | \end{array}] = [\begin{array}{cc} 1 & - 2 \\ 0 & 5 \end{array}]$ $A^{- 1} = \frac{1}{| A |} adj A = \frac{1}{5} [\begin{array}{cc} 1 & - 2 \\ 0 & 5 \end{array}] = [\begin{array}{cc} 1 / 5 & - 2 / 5 \\ 0 & 1 \end{array}]$

$B = [\begin{array}{rr} - 1 & 0 \\ 9 & 2 \end{array}]$

Solution

$\begin{array}{l} | B | = - 2 - 0 = - 2 \\ B^{- 1} = \frac{- 1}{2} [\begin{array}{rr} 2 & 0 \\ - 9 & - 1 \end{array}] \end{array}$

$C = [\begin{array}{rr} 3 & 7 \\ 3 & - 1 \end{array}]$

Solution

$\begin{array}{l} | C | = - 3 - 21 = - 24 \\ C^{- 1} = \frac{- 1}{24} [\begin{array}{cc} - 1 & - 7 \\ - 3 & 3 \end{array}] \end{array}$

$D = [\begin{array}{ll} 7 & 6 \\ 0 & 3 \end{array}]$

Solution

$\begin{array}{l} | D | = 21 - 0 = 21 \\ D^{- 1} = \frac{1}{21} [\begin{array}{rr} 3 & - 6 \\ 0 & 7 \end{array}] \end{array}$

Drawing on your answers to Prob. 2, formulate a two-step rule for finding the adjoint of a given $2 \times 2$ matrix $A$ : In the first step, indicate what should be done to the two diagonal elements of $A$ in order to get the diagonal elements of $a d j A$ ; in the second step, indicate what should be done to the two off-diagonal elements of $A$ . (Warning: This rule applies only to $2 \times 2$ matrices.)

Solution

Step 1: Exchange the two diagonal elements.
Step 2: Multiply both the off-diagonal elements by -1.

Add a third step which, in conjunction with the previous two steps, yields the $2 \times 2$ inverse matrix $A^{- 1}$ .

Solution

Step 3: Multiply the resulting matrix from steps 1 and 2 by $1 / | A |$ .

4. Find the inverse of each of the following matrices:

$E = [\begin{array}{rrr} 4 & - 2 & 1 \\ 7 & 3 & 0 \\ 2 & 0 & 1 \end{array}]$

Solution

$\begin{aligned} | E | & = 1 | \begin{array}{ll} 7 & 3 \\ 2 & 0 \end{array} | + 1 | \begin{array}{rr} 4 & - 2 \\ 7 & 3 \end{array} | \\ = 1 (0 - 6) + 1 (12 + 14) \\ = - 6 + 26 = 20 \end{aligned}$

$\begin{aligned} adj E & = [\begin{array}{ccc} | C_{11} | & | C_{21} | & | C_{31} | \\ | C_{12} | & | C_{22} | & | C_{32} | \\ | C_{13} | & | C_{23} | & | C_{33} | \end{array}] \\ = [\begin{array}{ccc} 3 & 2 & - 3 \\ - 7 & 2 & 7 \\ - 6 & - 4 & 26 \end{array}] \end{aligned}$ $E^{- 1} = \frac{1}{| E |} a d j E$

$F = [\begin{array}{rrr} 1 & - 1 & 2 \\ 1 & 0 & 3 \\ 4 & 0 & 2 \end{array}]$

Solution

$F^{- 1} = \frac{- 1}{10} [\begin{array}{ccc} 0 & 2 & - 3 \\ 10 & - 6 & - 1 \\ 0 & - 4 & 1 \end{array}]$

$G = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}]$

Solution

$| C_{11} | = | \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array} | = - 1 | C_{12} | = - | \begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array} | = 0 | C_{13} | = | \begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array} | = 0$ $| C_{21} | = - | \begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array} | = 0 | C_{22} | = | \begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array} | = 0 | C_{23} | = - | \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array} | = - 1$ $| C_{31} | = | \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} | = 0 | C_{32} | = - | \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} | = - 1 | C_{33} | = | \begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array} | = 0$ $G^{- 1} = - 1 [\begin{array}{ccc} - 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & - 1 & 0 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}]$

$H = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Solution

$H^{- 1} = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

7. Is it possible for a matrix to be its own inverse?

Solution

If a matrix is its own inverse, we would need that $A^{2} = I$ . This is true for the identity matrix. There are other possibilities such as matrix $G$ in exercise 4.