Question

Senior Secondary Question
Using matrices, solve the following system of equations:
x+y+z=2 , 2x-y=3 , 2y+z=0

Answer 1

Given:

3 equations

• x+y+z=2
• 2x-y=3
• 2y+z=0

To Find:

• Solution of given equations or value of x, y and z

Solution:

Given equations can be written as

x+y+z=2

2x-y+(0)z=3

(0)x+2y+z=0

We can also write above equations as

✏ A=XB

where,

$A=\left[\begin{array}{ccc}1&1&1\\2&-1&0\\0&2&1\end{array}\right]$

$B=\left[\begin{array}{c}2\\3\\0\end{array}\right]$

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

We know that,

$\longrightarrow\boxed{\pink{X=A^{-1}B}}$

$\longrightarrow\sf{A^{-1}=\dfrac{adj(A)}{\mid A\mid}}$

• Firstly we have to find adjoint of matrix A

For that we have to find cofactor matrix of A

Let cofactor matrix of A be C

So,

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

Now, transpose of C is equal to adjoint of matrix A

i.e. $\sf{adj(A)=C^{T}}$

$\sf{adj(A)=\left[\begin{array}{ccc}-1&1&1\\-2&1&2\\4&-2&-3\end{array}\right]}$

Also, here |A|= 1

Now,

$\rightarrow\sf{A^{-1}=\dfrac{adj(A)}{\mid A\mid}}$

$\rightarrow\sf{A^{-1}=\dfrac{1}{1} \left[\begin{array}{ccc}-1&1&1\\-2&1&2\\4&-2&-3\end{array}\right]}$

$\rightarrow\sf{A^{-1}=\left[\begin{array}{ccc}-1&1&1\\-2&1&2\\4&-2&-3\end{array}\right]}$

Now,

$\longrightarrow\sf{\pink{X=A^{-1}B}}$

$\longrightarrow\sf{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}-1&1&1\\-2&1&2\\4&-2&-3\end{array}\right]\left[\begin{array}{c}2\\3\\0\end{array}\right]}$

$\longrightarrow\sf{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\-1\\2\end{array}\right]}$

So,

x=1, y=-1, z=2

Hence, solution is (1,-1,2).