Question

Solve using determinants: x + 4y − z = −14 5x + 6y + 3z = 4 −2x + 7y + 2z = −17 |a| =__________ |ax| =_________ |ay| =_________ |az| =_________

Answer 1

By using use Cramer's Rule,

`\begin{aligned}& |A|=16 \\& \left|A_x\right|=-597 \\& \left|A_y\right|=119 \\& \left|A_z\right|=494\end{aligned}`

To solve the system of linear equations using determinants, we can use Cramer's Rule. The system of equations can be written in matrix form as Ax=B, where

`A=\left[\begin{array}{ccc}1 & 4 & -1 \\5 & 6 & 3 \\-2 & 7 & 2\end{array}\right], \quad x=\left[\begin{array}{l}x \\y \\z\end{array}\right], \quad B=\left[\begin{array}{c}-14 \\4 \\-17\end{array}\right]`

The determinant of matrix A is denoted by ∣A∣. Additionally,
`A_x`,
`A_y` and
`A_z` are matrices obtained by replacing the corresponding columns of matrix A with the column vector B.

Now, let's calculate these determinants:

`\begin{aligned}& |A|=\left|\begin{array}{ccc}1 & 4 & -1 \\5 & 6 & 3 \\-2 & 7 & 2\end{array}\right| \\& \left|A_x\right|=\left|\begin{array}{ccc}-14 & 4 & -1 \\5 & 6 & 3 \\-17 & 7 & 2\end{array}\right| \\& \left|A_y\right|=\left|\begin{array}{ccc}1 & -14 & -1 \\5 & 4 & 3 \\-2 & -17 & 2\end{array}\right| \\& \left|A_z\right|=\left|\begin{array}{ccc}1 & 4 & -14 \\5 & 6 & 4 \\-2 & 7 & -17\end{array}\right|\end{aligned}`

Let's calculate these determinants:

`\begin{aligned}& |A|=1(6(2)-3(7))-4(5(2)-3(-2))-(-1)(5(7)-6(-2)) \\& \left|A_x\right|=-14(6(2)-3(7))-4(5(2)-3(-17))-(-1)(-17(7)-2(5)) \\& \left|A_y\right|=1(4(2)-3(-17))-(-14)(5(2)-3(-2))-(-1)(-2(7)-2(5)) \\& \left|A_z\right|=1(6(-17)-4(4))-4(5(-17)-6(-2))-(-14)(5(4)-6(7))\end{aligned}`

Now, let's simplify these determinants:

`\begin{aligned}& |A|=1(12+21)-4(10+6)-(-1)(35+12) \\& \left|A_x\right|=-14(12+21)-4(10+51)-(-1)(-119-10) \\& \left|A_y\right|=1(8+51)-(-14)(10+6)-(-1)(-14-10) \\& \left|A_z\right|=1(-102-16)-4(-85+12)-(-14)(20-42)\end{aligned}`

Now, perform the calculations:

`\begin{aligned}& |A|=1(33)-4(16)+47 \\& \left|A_x\right|=-14(33)-4(61)+109 \\& \left|A_y\right|=59+56+4 \\& \left|A_z\right|=-118+332+280\end{aligned}`

Let's simplify these expressions:

`\begin{aligned}&|A|=16\\&\left|A_x\right|=-597\\&\left|A_y\right|=119\\&\left|A_z\right|=494\end{aligned}`

Answer 2

The determinants are ∣A∣ = −120, ​
`A_x`, = −240,
`A_y` = 360,
`A_z` = −480

How do we find the determinants?

Let's calculate the determinant of the coefficient matrix |A|, which is:

`\left[\begin{array}{ccc}1&4&-1\\5&6&3\\-2&7&2\end{array}\right]`

The determinant of matrix A (denoted as ∣A∣) is calculated using the formula for a 3 x 3 matrix:

∣A∣ = a(ei − fh) − b(di−fg) + c(dh − eg)

∣A∣ = 1(6×2−3×7) −4(5×2−3×(−2)) + (−1)(5×7−6×(−2))

∣A∣ = - 9 - 64 - 47

∣A∣ = −120

Determinants of matrices
`A_x`,
`A_y`, and
`A_z`

`A_x`, =
`\left[\begin{array}{ccc}-14&4&-1\\4&6&3\\-17&7&2\end{array}\right]`

`A_y` =
`\left[\begin{array}{ccc}1&-14&-1\\5&4&3\\-2& -17&2\end{array}\right]`

`A_z` =
`\left[\begin{array}{ccc}1&4 & -14 \\5&6&4\\-2&7&-17\end{array}\right]`

The determinants of these matrices were calculated similarly to ∣A∣ and we obtained:

​
`A_x`, = −240

`A_y` = 360

`A_z` = −480

x = |​
`A_x`|/∣A∣ ⇒ (−240)/(−120) = 2

y = |
`A_y`|/∣A∣⇒ 360/−120 = -3

z = |
`A_z`|/ ∣A∣ ⇒−480/−120 = 4

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