Pre-requisite: Solved Examples on the Determinant of a Matrix For the first twenty questions, begin with the word problems translated here: 3 * 3 Linear Systems
You may verify your answers as applicable with: Matrices Calculators
For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.
Solve each question using the Method of Determinants
Use at least two methods whenever applicable
Show all work
Interpret your solutions as applicable.
(1.)
$
x + y = -4 ...eqn.(1) \\[3ex]
x - y = 28 ...eqn.(2) \\[3ex]
$
$ \begin{bmatrix} 1 & 1 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -4 \\[3ex] 28 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} -4 & 1 \\[3ex] 28 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{4 - 28}{-1 - 1} = \dfrac{-24}{-2} \\[5ex] x = 12 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & -4 \\[3ex] 1 & 28 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{28 + 4}{-1 - 1} = \dfrac{32}{-2} \\[5ex] y = -16 \\[3ex] $ The sum of the numbers: $12 + -16 = 12 - 16$ is $-4$
The difference between the numbers: $12 - (-16) = 12 + 16$ is $28$
$ \begin{bmatrix} 1 & 1 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -4 \\[3ex] 28 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} -4 & 1 \\[3ex] 28 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{4 - 28}{-1 - 1} = \dfrac{-24}{-2} \\[5ex] x = 12 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & -4 \\[3ex] 1 & 28 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{28 + 4}{-1 - 1} = \dfrac{32}{-2} \\[5ex] y = -16 \\[3ex] $ The sum of the numbers: $12 + -16 = 12 - 16$ is $-4$
The difference between the numbers: $12 - (-16) = 12 + 16$ is $28$
(2.)
$
x + y = 54 ...eqn.(1) \\[3ex]
y = 2 + 3x ...eqn.(2) \\[3ex]
$
Rearrange $eqn.(2)$ to "line up"
$ -2 = 3x - y \\[3ex] 3x - y = -2 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & 1 \\[3ex] 3 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 54 \\[3ex] -2 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} 54 & 1 \\[3ex] -2 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 3 & -1 \end{vmatrix}} = \dfrac{-54 + 2}{-1 - 3} = \dfrac{-52}{-4} \\[5ex] x = 13 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & 54 \\[3ex] 3 & -2 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 3 & -1 \end{vmatrix}} = \dfrac{-2 - 162}{-1 - 3} = \dfrac{-164}{-4} \\[5ex] y = 41 \\[3ex] $ There are $13$ commercial launches
There are $41$ non-commercial launches
Rearrange $eqn.(2)$ to "line up"
$ -2 = 3x - y \\[3ex] 3x - y = -2 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & 1 \\[3ex] 3 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 54 \\[3ex] -2 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} 54 & 1 \\[3ex] -2 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 3 & -1 \end{vmatrix}} = \dfrac{-54 + 2}{-1 - 3} = \dfrac{-52}{-4} \\[5ex] x = 13 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & 54 \\[3ex] 3 & -2 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 3 & -1 \end{vmatrix}} = \dfrac{-2 - 162}{-1 - 3} = \dfrac{-164}{-4} \\[5ex] y = 41 \\[3ex] $ There are $13$ commercial launches
There are $41$ non-commercial launches
(3.)
$
x = 7y - 6 ...eqn.(1) \\[3ex]
x - y = 6 ...eqn.(2) \\[3ex]
$
Rearrange $eqn.(1)$ to "line up"
$ x - 7y = -6 ...new\:\: eqn.(1) \\[3ex] \begin{bmatrix} 1 & -7 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -6 \\[3ex] 6 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} -6 & -7 \\[3ex] 6 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & -7 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{6 + 42}{-1 + 7} = \dfrac{48}{6} \\[5ex] x = 8 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & -6 \\[3ex] 1 & 6 \end{vmatrix}} {\begin{vmatrix} 1 & -7 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{6 + 6}{-1 + 7} = \dfrac{12}{6} \\[5ex] y = 2 \\[3ex] $ The two-digit number is $82$
Rearrange $eqn.(1)$ to "line up"
$ x - 7y = -6 ...new\:\: eqn.(1) \\[3ex] \begin{bmatrix} 1 & -7 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -6 \\[3ex] 6 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} -6 & -7 \\[3ex] 6 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & -7 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{6 + 42}{-1 + 7} = \dfrac{48}{6} \\[5ex] x = 8 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & -6 \\[3ex] 1 & 6 \end{vmatrix}} {\begin{vmatrix} 1 & -7 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{6 + 6}{-1 + 7} = \dfrac{12}{6} \\[5ex] y = 2 \\[3ex] $ The two-digit number is $82$
(4.)
$
x + y = 4802 ...eqn.(1) \\[3ex]
y = x - 1116 ...eqn.(2) \\[3ex]
$
Rearrange $eqn.(2)$ to "line up"
$ 1116 = x - y \\[3ex] x - y = 1116 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & 1 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 4802 \\[3ex] 1116 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} 4802 & 1 \\[3ex] 1116 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{-4802 - 1116}{-1 - 1} = \dfrac{-5918}{-2} \\[5ex] x = 2959 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & 4802 \\[3ex] 1 & 1116 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{1116 - 4802}{-1 - 1} = \dfrac{-3686}{-2} \\[5ex] y = 1843 \\[3ex] $ The average apartment rent in the City of Santa Monica is $\$2959.00$
The average apartment rent in the City of San Francisco is $\$1843.00$
Rearrange $eqn.(2)$ to "line up"
$ 1116 = x - y \\[3ex] x - y = 1116 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & 1 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 4802 \\[3ex] 1116 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} 4802 & 1 \\[3ex] 1116 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{-4802 - 1116}{-1 - 1} = \dfrac{-5918}{-2} \\[5ex] x = 2959 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & 4802 \\[3ex] 1 & 1116 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{1116 - 4802}{-1 - 1} = \dfrac{-3686}{-2} \\[5ex] y = 1843 \\[3ex] $ The average apartment rent in the City of Santa Monica is $\$2959.00$
The average apartment rent in the City of San Francisco is $\$1843.00$
(5.)
$
x - y = 141 ...eqn.(1) \\[3ex]
x + y = 183 ...eqn.(2) \\[3ex]
$
$ \begin{bmatrix} 1 & -1 \\[3ex] 1 & 1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 141 \\[3ex] 183 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} 141 & -1 \\[3ex] 183 & 1 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 1 & 1 \end{vmatrix}} = \dfrac{141 + 183}{1 + 1} = \dfrac{324}{2} \\[5ex] x = 162 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & 141 \\[3ex] 1 & 183 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 1 & 1 \end{vmatrix}} = \dfrac{183 - 141}{1 + 1} = \dfrac{42}{2} \\[5ex] y = 21 \\[3ex] $ The speed of the plane is $162\: mph$
The speed of the wind is $21\: mph$
$ \begin{bmatrix} 1 & -1 \\[3ex] 1 & 1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 141 \\[3ex] 183 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} 141 & -1 \\[3ex] 183 & 1 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 1 & 1 \end{vmatrix}} = \dfrac{141 + 183}{1 + 1} = \dfrac{324}{2} \\[5ex] x = 162 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & 141 \\[3ex] 1 & 183 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 1 & 1 \end{vmatrix}} = \dfrac{183 - 141}{1 + 1} = \dfrac{42}{2} \\[5ex] y = 21 \\[3ex] $ The speed of the plane is $162\: mph$
The speed of the wind is $21\: mph$
(6.)
$
10 + x = y ...eqn.(1) \\[3ex]
4y = 6x + 25 ...eqn.(2) \\[3ex]
$
Rearrange $eqns. (1.) \:\:and\:\: (2)$ to "line up"
$ x - y = -10 ...new\:\: eqn.(1) \\[3ex] -25 = 6x - 4y \\[3ex] 6x - 4y = -25 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & -1 \\[3ex] 6 & -4 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -10 \\[3ex] -25 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} -10 & -1 \\[3ex] -25 & -4 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 6 & -4 \end{vmatrix}} = \dfrac{40 - 25}{-4 + 6} = \dfrac{15}{2} \\[5ex] x = 7.5 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & -10 \\[3ex] 6 & -25 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 6 & -4 \end{vmatrix}} = \dfrac{-25 + 60}{-4 + 6} = \dfrac{35}{2} \\[5ex] y = 17.5 \\[3ex] $ $7.5$ pounds of cashews needs to be mixed with $10$ pounds of peanuts to give $17.5$ pounds of cashews and peanuts so that the mixture will produce the same revenue as would selling the nuts separately.
Rearrange $eqns. (1.) \:\:and\:\: (2)$ to "line up"
$ x - y = -10 ...new\:\: eqn.(1) \\[3ex] -25 = 6x - 4y \\[3ex] 6x - 4y = -25 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & -1 \\[3ex] 6 & -4 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -10 \\[3ex] -25 \end{bmatrix} \\[3ex] $
$ x = \dfrac{ \begin{vmatrix} -10 & -1 \\[3ex] -25 & -4 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 6 & -4 \end{vmatrix}} = \dfrac{40 - 25}{-4 + 6} = \dfrac{15}{2} \\[5ex] x = 7.5 \\[3ex] $
$ y = \dfrac{ \begin{vmatrix} 1 & -10 \\[3ex] 6 & -25 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 6 & -4 \end{vmatrix}} = \dfrac{-25 + 60}{-4 + 6} = \dfrac{35}{2} \\[5ex] y = 17.5 \\[3ex] $ $7.5$ pounds of cashews needs to be mixed with $10$ pounds of peanuts to give $17.5$ pounds of cashews and peanuts so that the mixture will produce the same revenue as would selling the nuts separately.
(21.) WASSCE:FM (a) Show that the determinant
$ \begin{vmatrix} x + y & x & x \\[3ex] x & x + y & x \\[3ex] x & x & x + y \end{vmatrix} = (3x + y)y^2 \\[3ex] $
(b) Evaluate
$ \begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix} \\[3ex] $
Hence, solve the following equations:
$ 3x + 4y + 2z = 4 \\[3ex] x - 5y + 3z = -1 \\[3ex] 2x + 3y + z = 3 \\[3ex] $
We shall find the determinant in two ways.
Use any method you wish.
$ (a)\:\: From\:\:LHS \\[3ex] \underline{First\:\:Method} \\[3ex] \begin{vmatrix} + & - & + \\[3ex] - & + & - \\[3ex] + & - & + \end{vmatrix} \\[10ex] \underline{First\:\:Method} \\[3ex] 1st\:\:Row \\[3ex] \begin{vmatrix} x + y & x & x \\[3ex] x & x + y & x \\[3ex] x & x & x + y \end{vmatrix} = (x + y)\begin{vmatrix} x + y & x \\[3ex] x & x + y \end{vmatrix} - x\begin{vmatrix} x & x \\[3ex] x & x + y \end{vmatrix} + x\begin{vmatrix} x & x + y \\[3ex] x & x \end{vmatrix} \\[10ex] = (x + y)[(x + y)(x + y) - x^2] - x[x(x + y) - x^2] + x[x^2 - x(x + y)] \\[3ex] = (x + y)[x^2 + xy + xy + y^2 - x^2] - x[x^2 + xy - x^2] + x[x^2 - x^2 - xy] \\[3ex] = (x + y)(2xy + y^2) - x(xy) + x(-xy) \\[3ex] = 2x^2y + xy^2 + 2xy^2 + y^3 - x^2y - x^2y \\[3ex] = 3xy^2 + y^3 \\[3ex] = y^2(3x + y) \\[3ex] = RHS \\[3ex] \underline{Second\:\:Method:\;\; Diagonal\:\:Method} \\[3ex] \begin{vmatrix} x + y & x & x & x + y & x \\[3ex] x & x + y & x & x & x + y \\[3ex] x & x & x + y & x & x \end{vmatrix} \\[10ex] \underline{UpDown} \\[3ex] 1st:\:\: (x + y)(x + y)(x + y) = (x + y)^3 = x^3 + 3x^2 + 3xy^2 + y^3 ...Pascal's\:\:Triangle \\[3ex] 2nd:\:\: x(x)(x) = x^3 \\[3ex] 3rd:\:\: x(x)(x) = x^3 \\[3ex] UpDown\:\:Sum: \\[3ex] = (x^3 + 3x^2 + 3xy^2 + y^3) + x^3 + x^3 = x^3 + 3x^2 + 3xy^2 + y^3 + x^3 + x^3 \\[3ex] = 3x^3 + 3x^2 + 3xy^2 + y^3 \\[3ex] \underline{DownUp} \\[3ex] 1st:\:\: x(x + y)(x) = x^2(x + y) = x^3 + x^2y \\[3ex] 2nd:\:\: x(x)(x + y) = x^2(x + y) = x^3 + x^2y \\[3ex] 3rd:\:\: (x + y)(x)(x) = x^2(x + y) = x^3 + x^2y \\[3ex] DownUp\:\:Sum: \\[3ex] = x^3 + x^2y + x^3 + x^2y + x^3 + x^2y \\[3ex] = 3(x^3 + x^2y) \\[3ex] = 3x^3 + 3x^2y \\[3ex] det = UpDown\:\:Sum - DownUp\:\:Sum \\[3ex] det = (3x^3 + 3x^2 + 3xy^2 + y^3) - (3x^3 + 3x^2y) \\[3ex] det = 3x^3 + 3x^2 + 3xy^2 + y^3 - 3x^3 - 3x^2y \\[3ex] det = 3xy^2 + y^3 \\[3ex] det = y^2(3x + y) \\[3ex] = RHS \\[5ex] (b) \\[3ex] \begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix} \\[10ex] \underline{First\:\:Method} \\[3ex] 3rd\:\:Row \\[3ex] = 2\begin{vmatrix} 4 & 2 \\[3ex] -5 & 3 \end{vmatrix} - 3\begin{vmatrix} 3 & 2 \\[3ex] 1 & 3 \end{vmatrix} + 1\begin{vmatrix} 3 & 4 \\[3ex] 1 & -5 \end{vmatrix} \\[10ex] = 2[4(3) - -5(2)] - 3[3(3) - 1(2)] + 1[3(-5) - 1(4)] \\[3ex] = 2(12 - - 10) - 3(9 - 2) + 1(-15 - 4) \\[3ex] = 2(12 + 10) - 3(7) + 1(-19) \\[3ex] = 2(22) - 21 - 19 \\[3ex] = 44 - 21 - 19 \\[3ex] = 4 \\[3ex] $ For the solution of the Linear System, we shall use Cramer's method because it appears they want us to make use of the determinant we found.
$ \underline{Cramer's\:\:Rule} \\[3ex] \begin{bmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \\[3ex] z \end{bmatrix} = \begin{bmatrix} 4 \\[3ex] -1 \\[3ex] 3 \end{bmatrix} \\[10ex] x = \dfrac{\begin{vmatrix} 4 & 4 & 2 \\[3ex] -1 & -5 & 3 \\[3ex] 3 & 3 & 1 \end{vmatrix}}{\begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix}} \:\: y = \dfrac{\begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -1 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix}}{\begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix}} \:\: z = \dfrac{\begin{vmatrix} 3 & 4 & 4 \\[3ex] 1 & -5 & -1 \\[3ex] 2 & 3 & 3 \end{vmatrix}}{\begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix}} \\[3ex] $
$ \underline{Numerator\:\:for\:\:x} \\[3ex] \underline{Second\:\:Method:\;\; Diagonal\:\:Method} \\[3ex] \begin{vmatrix} 4 & 4 & 2 & 4 & 4 \\[3ex] -1 & -5 & 3 & -1 & -5 \\[3ex] 3 & 3 & 1 & 3 & 3 \end{vmatrix} \\[10ex] \underline{Up-down} \\[3ex] 1st:\:\: 4(-5)(1) = -20 \\[3ex] 2nd:\:\: 4(3)(3) = 36 \\[3ex] 3rd:\:\: 2(-1)(3) = -6 \\[3ex] UpDown\:\:Sum: \\[3ex] = -20 + 36 + - 6 \\[3ex] = 10 \\[5ex] \underline{DownUp} \\[3ex] 1st:\:\: 3(-5)(2) = -30 \\[3ex] 2nd:\:\: 3(3)(4) = 36 \\[3ex] 3rd:\:\: 1(-1)(4) = -4 \\[3ex] DownUp\:\:Sum: \\[3ex] = -30 + 36 + -4 \\[3ex] = 2 \\[5ex] det = UpDown\:\:Sum - DownUp\:\:Sum \\[3ex] det = 10 - 2 \\[3ex] det = 8 \\[3ex] \underline{Numerator\:\:for\:\:y} \\[3ex] \underline{First\:\:Method} \\[3ex] 2nd\:\:Row \\[3ex] det = -1\begin{vmatrix} 4 & 2 \\[3ex] 3 & 1 \end{vmatrix} + -1\begin{vmatrix} 3 & 2 \\[3ex] 2 & 1 \end{vmatrix} - 3\begin{vmatrix} 3 & 4 \\[3ex] 2 & 3 \end{vmatrix} \\[10ex] det = -1[4(1) - 3(2)] - 1[3(1) - 2(2)] - 3[3(3) - 2(4)] \\[3ex] = -1(4 - 6) - 1(3 - 4) - 3(9 - 8) \\[3ex] = -1(-2) - 1(-1) - 3(1) \\[3ex] = 2 + 1 - 3 \\[3ex] = 0 \\[3ex] \underline{Numerator\:\:for\:\:z} \\[3ex] \underline{Second\:\:Method:\;\; Diagonal\:\:Method} \\[3ex] \begin{vmatrix} 3 & 4 & 4 & 3 & 4 \\[3ex] 1 & -5 & -1 & 1 & -5 \\[3ex] 2 & 3 & 3 & 2 & 3 \end{vmatrix} \\[10ex] \underline{UpDown} \\[3ex] 1st:\:\: 3(-5)(3) = -45 \\[3ex] 2nd:\:\: 4(-1)(2) = -8 \\[3ex] 3rd:\:\: 4(1)(3) = 12 \\[3ex] UpDown\:\:Sum: \\[3ex] = -45 + -8 + 12 \\[3ex] = -41 \\[5ex] \underline{DownUp} \\[3ex] 1st:\:\: 2(-5)(4) = -40 \\[3ex] 2nd:\:\: 3(-1)(3) = -9 \\[3ex] 3rd:\:\: 3(1)(4) = 12 \\[3ex] DownUp\:\:Sum: \\[3ex] = -40 + -9 + 12 \\[3ex] = -37 \\[5ex] det = UpDown\:\:Sum - DownUp\:\:Sum \\[3ex] det = -41 - - 37 \\[3ex] det = -41 + 37 \\[3ex] det = -4 \\[3ex] \therefore \\[3ex] x = \dfrac{8}{4} \\[5ex] x = 2 \\[3ex] y = \dfrac{0}{4} \\[5ex] y = 0 \\[3ex] z = -\dfrac{4}{4} \\[5ex] z = -1 \\[3ex] $ Check
$x = 2,\:y = 0,\:z = -1$
$ \begin{vmatrix} x + y & x & x \\[3ex] x & x + y & x \\[3ex] x & x & x + y \end{vmatrix} = (3x + y)y^2 \\[3ex] $
(b) Evaluate
$ \begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix} \\[3ex] $
Hence, solve the following equations:
$ 3x + 4y + 2z = 4 \\[3ex] x - 5y + 3z = -1 \\[3ex] 2x + 3y + z = 3 \\[3ex] $
We shall find the determinant in two ways.
Use any method you wish.
$ (a)\:\: From\:\:LHS \\[3ex] \underline{First\:\:Method} \\[3ex] \begin{vmatrix} + & - & + \\[3ex] - & + & - \\[3ex] + & - & + \end{vmatrix} \\[10ex] \underline{First\:\:Method} \\[3ex] 1st\:\:Row \\[3ex] \begin{vmatrix} x + y & x & x \\[3ex] x & x + y & x \\[3ex] x & x & x + y \end{vmatrix} = (x + y)\begin{vmatrix} x + y & x \\[3ex] x & x + y \end{vmatrix} - x\begin{vmatrix} x & x \\[3ex] x & x + y \end{vmatrix} + x\begin{vmatrix} x & x + y \\[3ex] x & x \end{vmatrix} \\[10ex] = (x + y)[(x + y)(x + y) - x^2] - x[x(x + y) - x^2] + x[x^2 - x(x + y)] \\[3ex] = (x + y)[x^2 + xy + xy + y^2 - x^2] - x[x^2 + xy - x^2] + x[x^2 - x^2 - xy] \\[3ex] = (x + y)(2xy + y^2) - x(xy) + x(-xy) \\[3ex] = 2x^2y + xy^2 + 2xy^2 + y^3 - x^2y - x^2y \\[3ex] = 3xy^2 + y^3 \\[3ex] = y^2(3x + y) \\[3ex] = RHS \\[3ex] \underline{Second\:\:Method:\;\; Diagonal\:\:Method} \\[3ex] \begin{vmatrix} x + y & x & x & x + y & x \\[3ex] x & x + y & x & x & x + y \\[3ex] x & x & x + y & x & x \end{vmatrix} \\[10ex] \underline{UpDown} \\[3ex] 1st:\:\: (x + y)(x + y)(x + y) = (x + y)^3 = x^3 + 3x^2 + 3xy^2 + y^3 ...Pascal's\:\:Triangle \\[3ex] 2nd:\:\: x(x)(x) = x^3 \\[3ex] 3rd:\:\: x(x)(x) = x^3 \\[3ex] UpDown\:\:Sum: \\[3ex] = (x^3 + 3x^2 + 3xy^2 + y^3) + x^3 + x^3 = x^3 + 3x^2 + 3xy^2 + y^3 + x^3 + x^3 \\[3ex] = 3x^3 + 3x^2 + 3xy^2 + y^3 \\[3ex] \underline{DownUp} \\[3ex] 1st:\:\: x(x + y)(x) = x^2(x + y) = x^3 + x^2y \\[3ex] 2nd:\:\: x(x)(x + y) = x^2(x + y) = x^3 + x^2y \\[3ex] 3rd:\:\: (x + y)(x)(x) = x^2(x + y) = x^3 + x^2y \\[3ex] DownUp\:\:Sum: \\[3ex] = x^3 + x^2y + x^3 + x^2y + x^3 + x^2y \\[3ex] = 3(x^3 + x^2y) \\[3ex] = 3x^3 + 3x^2y \\[3ex] det = UpDown\:\:Sum - DownUp\:\:Sum \\[3ex] det = (3x^3 + 3x^2 + 3xy^2 + y^3) - (3x^3 + 3x^2y) \\[3ex] det = 3x^3 + 3x^2 + 3xy^2 + y^3 - 3x^3 - 3x^2y \\[3ex] det = 3xy^2 + y^3 \\[3ex] det = y^2(3x + y) \\[3ex] = RHS \\[5ex] (b) \\[3ex] \begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix} \\[10ex] \underline{First\:\:Method} \\[3ex] 3rd\:\:Row \\[3ex] = 2\begin{vmatrix} 4 & 2 \\[3ex] -5 & 3 \end{vmatrix} - 3\begin{vmatrix} 3 & 2 \\[3ex] 1 & 3 \end{vmatrix} + 1\begin{vmatrix} 3 & 4 \\[3ex] 1 & -5 \end{vmatrix} \\[10ex] = 2[4(3) - -5(2)] - 3[3(3) - 1(2)] + 1[3(-5) - 1(4)] \\[3ex] = 2(12 - - 10) - 3(9 - 2) + 1(-15 - 4) \\[3ex] = 2(12 + 10) - 3(7) + 1(-19) \\[3ex] = 2(22) - 21 - 19 \\[3ex] = 44 - 21 - 19 \\[3ex] = 4 \\[3ex] $ For the solution of the Linear System, we shall use Cramer's method because it appears they want us to make use of the determinant we found.
$ \underline{Cramer's\:\:Rule} \\[3ex] \begin{bmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \\[3ex] z \end{bmatrix} = \begin{bmatrix} 4 \\[3ex] -1 \\[3ex] 3 \end{bmatrix} \\[10ex] x = \dfrac{\begin{vmatrix} 4 & 4 & 2 \\[3ex] -1 & -5 & 3 \\[3ex] 3 & 3 & 1 \end{vmatrix}}{\begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix}} \:\: y = \dfrac{\begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -1 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix}}{\begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix}} \:\: z = \dfrac{\begin{vmatrix} 3 & 4 & 4 \\[3ex] 1 & -5 & -1 \\[3ex] 2 & 3 & 3 \end{vmatrix}}{\begin{vmatrix} 3 & 4 & 2 \\[3ex] 1 & -5 & 3 \\[3ex] 2 & 3 & 1 \end{vmatrix}} \\[3ex] $
$ \underline{Numerator\:\:for\:\:x} \\[3ex] \underline{Second\:\:Method:\;\; Diagonal\:\:Method} \\[3ex] \begin{vmatrix} 4 & 4 & 2 & 4 & 4 \\[3ex] -1 & -5 & 3 & -1 & -5 \\[3ex] 3 & 3 & 1 & 3 & 3 \end{vmatrix} \\[10ex] \underline{Up-down} \\[3ex] 1st:\:\: 4(-5)(1) = -20 \\[3ex] 2nd:\:\: 4(3)(3) = 36 \\[3ex] 3rd:\:\: 2(-1)(3) = -6 \\[3ex] UpDown\:\:Sum: \\[3ex] = -20 + 36 + - 6 \\[3ex] = 10 \\[5ex] \underline{DownUp} \\[3ex] 1st:\:\: 3(-5)(2) = -30 \\[3ex] 2nd:\:\: 3(3)(4) = 36 \\[3ex] 3rd:\:\: 1(-1)(4) = -4 \\[3ex] DownUp\:\:Sum: \\[3ex] = -30 + 36 + -4 \\[3ex] = 2 \\[5ex] det = UpDown\:\:Sum - DownUp\:\:Sum \\[3ex] det = 10 - 2 \\[3ex] det = 8 \\[3ex] \underline{Numerator\:\:for\:\:y} \\[3ex] \underline{First\:\:Method} \\[3ex] 2nd\:\:Row \\[3ex] det = -1\begin{vmatrix} 4 & 2 \\[3ex] 3 & 1 \end{vmatrix} + -1\begin{vmatrix} 3 & 2 \\[3ex] 2 & 1 \end{vmatrix} - 3\begin{vmatrix} 3 & 4 \\[3ex] 2 & 3 \end{vmatrix} \\[10ex] det = -1[4(1) - 3(2)] - 1[3(1) - 2(2)] - 3[3(3) - 2(4)] \\[3ex] = -1(4 - 6) - 1(3 - 4) - 3(9 - 8) \\[3ex] = -1(-2) - 1(-1) - 3(1) \\[3ex] = 2 + 1 - 3 \\[3ex] = 0 \\[3ex] \underline{Numerator\:\:for\:\:z} \\[3ex] \underline{Second\:\:Method:\;\; Diagonal\:\:Method} \\[3ex] \begin{vmatrix} 3 & 4 & 4 & 3 & 4 \\[3ex] 1 & -5 & -1 & 1 & -5 \\[3ex] 2 & 3 & 3 & 2 & 3 \end{vmatrix} \\[10ex] \underline{UpDown} \\[3ex] 1st:\:\: 3(-5)(3) = -45 \\[3ex] 2nd:\:\: 4(-1)(2) = -8 \\[3ex] 3rd:\:\: 4(1)(3) = 12 \\[3ex] UpDown\:\:Sum: \\[3ex] = -45 + -8 + 12 \\[3ex] = -41 \\[5ex] \underline{DownUp} \\[3ex] 1st:\:\: 2(-5)(4) = -40 \\[3ex] 2nd:\:\: 3(-1)(3) = -9 \\[3ex] 3rd:\:\: 3(1)(4) = 12 \\[3ex] DownUp\:\:Sum: \\[3ex] = -40 + -9 + 12 \\[3ex] = -37 \\[5ex] det = UpDown\:\:Sum - DownUp\:\:Sum \\[3ex] det = -41 - - 37 \\[3ex] det = -41 + 37 \\[3ex] det = -4 \\[3ex] \therefore \\[3ex] x = \dfrac{8}{4} \\[5ex] x = 2 \\[3ex] y = \dfrac{0}{4} \\[5ex] y = 0 \\[3ex] z = -\dfrac{4}{4} \\[5ex] z = -1 \\[3ex] $ Check
$x = 2,\:y = 0,\:z = -1$
| $LHS$ | $RHS$ |
|---|---|
| $ 3x + 4y + 2z \\[3ex] 3(2) + 4(0) + 2(-1) \\[3ex] 6 + 0 -2 \\[3ex] 4 $ | $4$ |
| $ x - 5y + 3z \\[3ex] 2 - 5(0) + 3(-1) \\[3ex] 2 - 0 - 3 \\[3ex] -1 $ | $-1$ |
| $ 2x + 3y + z \\[3ex] 2(2) + 3(0) + (-1) \\[3ex] 4 + 0 - 1 \\[3ex] 3 $ | $3$ |
(22.) WASSCE:FM Using determinants, solve the following equations simultaneously.
$ 5x - 6y + 4z = 15 \\[3ex] 7x + 4y - 3z = 19 \\[3ex] 2x + y + 6z = 46 \\[3ex] $
$ \boldsymbol{First\;\;Method} \\[3ex] \begin{vmatrix} + & - & + \\[3ex] - & + & - \\[3ex] + & - & + \end{vmatrix} \\[10ex] \begin{vmatrix} 5 & -6 & 4 \\[3ex] 7 & 4 & -3 \\[3ex] 2 & 1 & 6 \end{vmatrix} \\[10ex] \underline{First\:\:Method} \\[3ex] 3rd\:\:Row \\[3ex] = 2\begin{vmatrix} -6 & 4 \\[3ex] 4 & -3 \end{vmatrix} - 1\begin{vmatrix} 5 & 4 \\[3ex] 7 & -3 \end{vmatrix} + 6\begin{vmatrix} 5 & -6 \\[3ex] 7 & 4 \end{vmatrix} \\[10ex] = 2[-6(-3) - 4(4)] - 1[5(-3) - 7(4)] + 6[5(4) - 7(-6)] \\[3ex] = 2(18 - 16) - 1(-15 - 28) + 6(20 + 42) \\[3ex] = 2(2) - 1(-43) + 6(62) \\[3ex] = 4 + 43 + 372 \\[3ex] = 419 \\[5ex] \underline{Cramer's Rule} \\[3ex] \begin{bmatrix} 5 & -6 & 4 \\[3ex] 7 & 4 & -3 \\[3ex] 2 & 1 & 6 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \\[3ex] z \end{bmatrix} = \begin{bmatrix} 15 \\[3ex] 19 \\[3ex] 46 \end{bmatrix} \\[10ex] x = \dfrac{\begin{vmatrix} 15 & -6 & 4 \\[3ex] 19 & 4 & -3 \\[3ex] 46 & 1 & 6 \end{vmatrix}}{\begin{vmatrix} 5 & -6 & 4 \\[3ex] 7 & 4 & -3 \\[3ex] 2 & 1 & 6 \end{vmatrix}} \:\: y = \dfrac{\begin{vmatrix} 5 & 15 & 4 \\[3ex] 7 & 19 & -3 \\[3ex] 2 & 46 & 6 \end{vmatrix}}{\begin{vmatrix} 5 & -6 & 4 \\[3ex] 7 & 4 & -3 \\[3ex] 2 & 1 & 6 \end{vmatrix}} \:\: z = \dfrac{\begin{vmatrix} 5 & -6 & 15 \\[3ex] 7 & 4 & 19 \\[3ex] 2 & 1 & 46 \end{vmatrix}}{\begin{vmatrix} 5 & -6 & 4 \\[3ex] 7 & 4 & -3 \\[3ex] 2 & 1 & 6 \end{vmatrix}} \\[3ex] $
$ \underline{Numerator\;\;for\;\;x} \\[3ex] \boldsymbol{First\;\;Method} \\[3ex] 3rd\;\;Row \\[3ex] det = 46\begin{vmatrix} -6 & 4 \\[3ex] 4 & -3 \end{vmatrix} - 1\begin{vmatrix} 15 & 4 \\[3ex] 19 & -3 \end{vmatrix} + 6\begin{vmatrix} 15 & -6 \\[3ex] 19 & 4 \end{vmatrix} \\[10ex] = 46[-6(-3) - 4(4)] - 1[15(-3) - 19(4)] + 6[15(4) - 19(-6)] \\[3ex] = 46(18 - 16) - 1(-45 - 76) + 6(60 + 114) \\[3ex] = 46(2) - 1(-121) + 6(174) \\[3ex] = 92 + 121 + 1044 \\[3ex] = 1257 \\[5ex] \underline{Numerator\;\;for\;\;y} \\[3ex] \boldsymbol{First\;\;Method} \\[3ex] 1st\;\;Row \\[3ex] det = 5\begin{vmatrix} 19 & -3 \\[3ex] 46 & 6 \end{vmatrix} - 15\begin{vmatrix} 7 & -3 \\[3ex] 2 & 6 \end{vmatrix} + 4\begin{vmatrix} 7 & 19 \\[3ex] 2 & 46 \end{vmatrix} \\[10ex] = 5[19(6) - 46(-3)] - 15[7(6) - 2(-3)] + 4[7(46) - 2(19)] \\[3ex] = 5(114 + 138) - 15(42 + 6) + 4(322 - 38) \\[3ex] = 5(252) - 15(48) + 4(284) \\[3ex] = 1260 - 720 + 1136 \\[3ex] = 1676 \\[5ex] \underline{Numerator\;\;for\;\;z} \\[3ex] \underline{Second\:\:Method:\;\; Diagonal\:\:Method} \\[3ex] \begin{vmatrix} 5 & -6 & 15 & 5 & -6 \\[3ex] 7 & 4 & 19 & 7 & 4 \\[3ex] 2 & 1 & 46 & 2 & 1 \end{vmatrix} \\[10ex] \underline{UpDown} \\[3ex] 1st:\:\: 5(4)(46) = 920 \\[3ex] 2nd:\:\: -6(19)(2) = -228 \\[3ex] 3rd:\:\: 15(7)(1) = 105 \\[3ex] UpDown\:\:Sum: \\[3ex] = 920 + - 228 + 105 \\[3ex] = 797 \\[5ex] \underline{DownUp} \\[3ex] 1st:\:\: 2(4)(15) = 120 \\[3ex] 2nd:\:\: 1(19)(5) = 95 \\[3ex] 3rd:\:\: 46(7)(-6) = -1932 \\[3ex] DownUp\:\:Sum: \\[3ex] = 120 + 95 + -1932 \\[3ex] = -1717 \\[5ex] det = UpDown\:\:Sum - DownUp\:\:Sum \\[3ex] det = 797 - (-1717) \\[3ex] det = 797 + 1717 \\[3ex] det = 2514 \\[3ex] \therefore \\[3ex] x = \dfrac{1257}{419} \\[5ex] x = 3 \\[3ex] y = \dfrac{1676}{419} \\[5ex] y = 4 \\[3ex] z = \dfrac{2514}{419} \\[5ex] z = 6 \\[3ex] $ Check
$x = 3,\:y = 4,\:z = 6$
$ 5x - 6y + 4z = 15 \\[3ex] 7x + 4y - 3z = 19 \\[3ex] 2x + y + 6z = 46 \\[3ex] $
$ \boldsymbol{First\;\;Method} \\[3ex] \begin{vmatrix} + & - & + \\[3ex] - & + & - \\[3ex] + & - & + \end{vmatrix} \\[10ex] \begin{vmatrix} 5 & -6 & 4 \\[3ex] 7 & 4 & -3 \\[3ex] 2 & 1 & 6 \end{vmatrix} \\[10ex] \underline{First\:\:Method} \\[3ex] 3rd\:\:Row \\[3ex] = 2\begin{vmatrix} -6 & 4 \\[3ex] 4 & -3 \end{vmatrix} - 1\begin{vmatrix} 5 & 4 \\[3ex] 7 & -3 \end{vmatrix} + 6\begin{vmatrix} 5 & -6 \\[3ex] 7 & 4 \end{vmatrix} \\[10ex] = 2[-6(-3) - 4(4)] - 1[5(-3) - 7(4)] + 6[5(4) - 7(-6)] \\[3ex] = 2(18 - 16) - 1(-15 - 28) + 6(20 + 42) \\[3ex] = 2(2) - 1(-43) + 6(62) \\[3ex] = 4 + 43 + 372 \\[3ex] = 419 \\[5ex] \underline{Cramer's Rule} \\[3ex] \begin{bmatrix} 5 & -6 & 4 \\[3ex] 7 & 4 & -3 \\[3ex] 2 & 1 & 6 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \\[3ex] z \end{bmatrix} = \begin{bmatrix} 15 \\[3ex] 19 \\[3ex] 46 \end{bmatrix} \\[10ex] x = \dfrac{\begin{vmatrix} 15 & -6 & 4 \\[3ex] 19 & 4 & -3 \\[3ex] 46 & 1 & 6 \end{vmatrix}}{\begin{vmatrix} 5 & -6 & 4 \\[3ex] 7 & 4 & -3 \\[3ex] 2 & 1 & 6 \end{vmatrix}} \:\: y = \dfrac{\begin{vmatrix} 5 & 15 & 4 \\[3ex] 7 & 19 & -3 \\[3ex] 2 & 46 & 6 \end{vmatrix}}{\begin{vmatrix} 5 & -6 & 4 \\[3ex] 7 & 4 & -3 \\[3ex] 2 & 1 & 6 \end{vmatrix}} \:\: z = \dfrac{\begin{vmatrix} 5 & -6 & 15 \\[3ex] 7 & 4 & 19 \\[3ex] 2 & 1 & 46 \end{vmatrix}}{\begin{vmatrix} 5 & -6 & 4 \\[3ex] 7 & 4 & -3 \\[3ex] 2 & 1 & 6 \end{vmatrix}} \\[3ex] $
$ \underline{Numerator\;\;for\;\;x} \\[3ex] \boldsymbol{First\;\;Method} \\[3ex] 3rd\;\;Row \\[3ex] det = 46\begin{vmatrix} -6 & 4 \\[3ex] 4 & -3 \end{vmatrix} - 1\begin{vmatrix} 15 & 4 \\[3ex] 19 & -3 \end{vmatrix} + 6\begin{vmatrix} 15 & -6 \\[3ex] 19 & 4 \end{vmatrix} \\[10ex] = 46[-6(-3) - 4(4)] - 1[15(-3) - 19(4)] + 6[15(4) - 19(-6)] \\[3ex] = 46(18 - 16) - 1(-45 - 76) + 6(60 + 114) \\[3ex] = 46(2) - 1(-121) + 6(174) \\[3ex] = 92 + 121 + 1044 \\[3ex] = 1257 \\[5ex] \underline{Numerator\;\;for\;\;y} \\[3ex] \boldsymbol{First\;\;Method} \\[3ex] 1st\;\;Row \\[3ex] det = 5\begin{vmatrix} 19 & -3 \\[3ex] 46 & 6 \end{vmatrix} - 15\begin{vmatrix} 7 & -3 \\[3ex] 2 & 6 \end{vmatrix} + 4\begin{vmatrix} 7 & 19 \\[3ex] 2 & 46 \end{vmatrix} \\[10ex] = 5[19(6) - 46(-3)] - 15[7(6) - 2(-3)] + 4[7(46) - 2(19)] \\[3ex] = 5(114 + 138) - 15(42 + 6) + 4(322 - 38) \\[3ex] = 5(252) - 15(48) + 4(284) \\[3ex] = 1260 - 720 + 1136 \\[3ex] = 1676 \\[5ex] \underline{Numerator\;\;for\;\;z} \\[3ex] \underline{Second\:\:Method:\;\; Diagonal\:\:Method} \\[3ex] \begin{vmatrix} 5 & -6 & 15 & 5 & -6 \\[3ex] 7 & 4 & 19 & 7 & 4 \\[3ex] 2 & 1 & 46 & 2 & 1 \end{vmatrix} \\[10ex] \underline{UpDown} \\[3ex] 1st:\:\: 5(4)(46) = 920 \\[3ex] 2nd:\:\: -6(19)(2) = -228 \\[3ex] 3rd:\:\: 15(7)(1) = 105 \\[3ex] UpDown\:\:Sum: \\[3ex] = 920 + - 228 + 105 \\[3ex] = 797 \\[5ex] \underline{DownUp} \\[3ex] 1st:\:\: 2(4)(15) = 120 \\[3ex] 2nd:\:\: 1(19)(5) = 95 \\[3ex] 3rd:\:\: 46(7)(-6) = -1932 \\[3ex] DownUp\:\:Sum: \\[3ex] = 120 + 95 + -1932 \\[3ex] = -1717 \\[5ex] det = UpDown\:\:Sum - DownUp\:\:Sum \\[3ex] det = 797 - (-1717) \\[3ex] det = 797 + 1717 \\[3ex] det = 2514 \\[3ex] \therefore \\[3ex] x = \dfrac{1257}{419} \\[5ex] x = 3 \\[3ex] y = \dfrac{1676}{419} \\[5ex] y = 4 \\[3ex] z = \dfrac{2514}{419} \\[5ex] z = 6 \\[3ex] $ Check
$x = 3,\:y = 4,\:z = 6$
| $LHS$ | $RHS$ |
|---|---|
| $ 5x - 6y + 4z \\[3ex] 5(3) - 6(4) + 4(6) \\[3ex] 15 - 24 + 24 \\[3ex] 15 $ | $15$ |
| $ 7x + 4y - 3z \\[3ex] 7(3) + 4(4) - 3(6) \\[3ex] 21 + 16 - 18 \\[3ex] 19 $ | $19$ |
| $ 2x + y + 6z \\[3ex] 2(3) + 4 + 6(6) \\[3ex] 6 + 4 + 36 \\[3ex] 46 $ | $46$ |