Question

1 Consider a 3x3 square matrix given as
⎡⎣⎢200020001⎤⎦⎥
. What is the element in
a22
2
0
1
0,0

2 A + B = B + A
associative
m
commutative
m-n

3 Find z by the use of determinant : 3x-4y+2z+8=0, x+5y-3z+2=0, 5x+3y-z+6=0
6
7
3
5

4 Find x by the use of determinant : 3x-4y+2z+8=0, x+5y-3z+2=0, 5x+3y-z+6+0
3
5
2
-2

5 Compute the determinant using elements in the first row:
A∣∣∣∣1540−7−8371∣∣∣∣
-7
32
13
3

Answer 1

Answer on Question #55842 – Math – Linear Algebra

1. Consider a $3 \times 3$ square matrix given as

```

200020001

```

. What is the element in

```

a22?

```

2

```

```

1

```

0,0

```

Solution

$$A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad a_{22} = 2$$

2. A + B = B + A

```

associative

```

m

```

commutative

```

m-n

```

Solution

$$A + B = B + A - \text{commutative}$$

3. Find z by the use of determinant: $3x - 4y + 2z + 8 = 0$, $x + 5y - 3z + 2 = 0$, $5x + 3y - z + 6 = 0$

6

7

3

5

Solution

$$\Delta = \left| \begin{array}{ccc} 3 & -4 & 2 \\ 1 & 5 & -3 \\ 5 & 3 & -1 \end{array} \right| = -15 + 6 + 60 - 50 - 4 + 27 = 24$$ $$\Delta_z = \left| \begin{array}{ccc} 3 & -4 & -8 \\ 1 & 5 & -2 \\ 5 & 3 & -6 \end{array} \right| = -90 - 24 + 40 + 200 + 18 - 24 = 120$$ $$z = \frac{\Delta_z}{\Delta} = \frac{120}{24} = 5$$

4. Find $x$ by the use of determinant: $3x - 4y + 2z + 8 = 0$, $x + 5y - 3z + 2 = 0$, $5x + 3y - z + 6 = 0$

3

5

2

-2

Solution

$$\Delta = \left| \begin{array}{ccc} 3 & -4 & 2 \\ 1 & 5 & -3 \\ 5 & 3 & -1 \end{array} \right| = -15 + 6 + 60 - 50 - 4 + 27 = 24$$ $$\Delta_x = \left| \begin{array}{ccc} -8 & -4 & 2 \\ -2 & 5 & -3 \\ -6 & 3 & -1 \end{array} \right| = 40 - 12 - 72 + 60 - 72 + 8 = -48$$ $$x = \frac{\Delta_x}{\Delta} = \frac{-48}{24} = -2$$

5. Compute the determinant using elements in the first row:

A||||1540-7-8371||||

-7

32

13

3

Solution

$$det(A) = \begin{vmatrix} 1 & 5 & 4 \\ 0 & -7 & -8 \\ 3 & 7 & 1 \end{vmatrix} = 1 \cdot \begin{vmatrix} -7 & -8 \\ 7 & 1 \end{vmatrix} - 5 \cdot \begin{vmatrix} 0 & -8 \\ 3 & 1 \end{vmatrix} + 4 \cdot \begin{vmatrix} 0 & -7 \\ 3 & 7 \end{vmatrix} =$$ $$= (-7 \cdot 1 - 7 \cdot (-8)) - 5(0 \cdot 1 - 3 \cdot (-8)) + 4 \cdot (0 \cdot 7 - 3 \cdot (-7)) = 49 - 120 + 84 = 13.$$

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