Question

Solve by matrix method 2x – y + z = 4, x + y + z = 1, x – 3y – 2z = 2.

Answer 1

Answer:

X = 9/5

Y = 1

Z  = -1/5

Explanation:

Solve by matrix method

2x – y + z = 4

x + y + z = 1

x – 3y – 2z = 2

write equations in forms AX = B

A= $\left[\begin{array}{ccc}2&-1&1\\1&1&1\\1&-1&-2\end{array}\right]$        

X= $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$

B= $\left[\begin{array}{ccc}4\\1\\2\end{array}\right]$  

Now

|A| =$\begin{vmatrix}2&-1 &1\\1 &1 &1\\1 &-1 &2\end{vmatrix}$  

= 2$\begin{vmatrix}1&1\\-1&2\end{vmatrix}$ +1$\begin{vmatrix}1&1\\1&2\end{vmatrix}$+1$\begin{vmatrix}1&1\\1&-1\end{vmatrix}$

=2(2+1) + 1(2-1) + 1 (-1-1)

= 2×3 + 1×1 + 1×(-2)

=6+1-2

=5

|A| ≠ 0

Equations has unique equation

X = $A^{-1}$B

$A^{-1}$= 1/|A| ×adj (A)

adj (A) = $\begin{vmatrix}A_{11} &A_{12} &A_{13}\\A_{21} &A_{22} &A_{23}\\A_{31} &A_{32} &A_{33}\end{vmatrix}'$=$\begin{vmatrix}A_{11} &A_{21} &A_{31}\\A_{12} &A_{22} &A_{32}\\A_{13} &A_{23} &A_{33}\end{vmatrix}$

$M_{11}$ = $\begin{vmatrix}1&1\\-1&2\end{vmatrix}$ = (2 +1) = 3

$M_{21}$ = $\begin{vmatrix}1&1\\1&2\end{vmatrix}$= (2-1) = 1

$M_{13}$=$\begin{vmatrix}1&1\\1&-1\end{vmatrix}$ = (-1-1) = -2

$M_{21}$=$\begin{vmatrix}-1&1\\-1&2\end{vmatrix}$ = (-2+1) = -1

$M_{22}$ =$\begin{vmatrix}2&1\\1&2\end{vmatrix}$ = (4-1) = 3

$M_{23}$=$\begin{vmatrix}2&-1\\1&- 1\end{vmatrix}$ = (-2+1) = -1

$M_{31}$=$\begin{vmatrix}-1&1\\1&1\end{vmatrix}$ = (-1-1) = -2

$M_{32}$= $\begin{vmatrix}2&1\\1&1\end{vmatrix}$=(2-1)= 1

$M_{33}$=$\begin{vmatrix}2&-1\\1&1\end{vmatrix}$=(2+1)=3

$A_{11}$=$(-1)^{2}$ . $M_{11}$= 3

$A_{12}=(-1)^{3}$. $M_{21}$ = 1

$A_{13}$ = $(-1)^{4}$ . $M_{13}$ = -2

$A_{21}$ = $(-1)^{3}$ . $M_{21}$ = 1

$A_{22}$ = $(-1)^{4}$ . $M_{22}$ = 3

$A_{23}$ = $(-1)^{5}$ . $M_{23}$ = 1

$A_{31}$ = $(-1)^{4}$ . $M_{31}$ = -2

$A_{32}$ = $(-1)^{5}$ . $M_{32}$ = -1

$A_{33}$ = $(-1)^{6}$ . $M_{33}$ = 3

adj(A) = $\left[\begin{array}{ccc}3&1&-2\\1&3&-1\\-2&1&3\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$=$\frac{1}{5}$$\left[\begin{array}{ccc}3&1&-2\\1&3&-1\\-2&1&3\end{array}\right]$ $\left[\begin{array}{ccc}4\\1\\2\end{array}\right]$

=$\left[\begin{array}{ccc}3/5&1/5&-2/5\\1/5&3/5&-1/5\\-2/5&1/5&3/5\end{array}\right]$ $\left[\begin{array}{ccc}4\\1\\2\end{array}\right]$

= $\left[\begin{array}{ccc}9/5\\1\\-1/5\end{array}\right]$

X = 9/5

Y = 1

Z  = -1/5