Question

If three numbers are added, their sum is ‘2’. If two times the second number is subtracted from the sum of the first and third numbers, we get ‘8’, and if three times the first number is added to the sum of the second and third numbers we get ‘4’. Find the numbers using matrices.

Answer 1

Hint: Here, first we will assume the three variables as x, y, z and then $AX=B$ this equation will be formed From this making X as subject we will get $X=\dfrac{B}{A}={{A}^{-1}}B$ . To find inverse of matrix formula required is ${{A}^{-1}}=\dfrac{1}{\left| A \right|}Adj\left( A \right)$ . To find Adjoint of A, the formula will be $Adj\left( A \right)={{\left( \text{cofactor} \right)}^{T}}$. So, on solving the equation $X=\dfrac{B}{A}={{A}^{-1}}B$ by putting all the values, we will get the answer.

Complete step-by-step solution -
Here, we are given three statements in form of sentences, and three numbers are taken. So, we will assume the three numbers as x equals the first number, y equals the second number, z equals the third number.
Now, it is given that if three numbers are added, their sum is ‘2’. So, writing this in mathematical form, we get as
$x+y+z=2$ …………………………(1)
Now, if two times the second number is subtracted from the sum of the first and third numbers, we get ‘8’. In mathematical form, we get
$x+z-2y=8$ On rearranging terms, we get $\Rightarrow x-2y+z=8$ ………………………….(2)
Next, if three times the first number is added to the sum of the second and third numbers we get ‘4’. So, writing this in mathematical form, we get as
$3x+y+z=4$ ……………………….(3)
Now, all the three equations can be written in matrix form as,
$\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} {{c}_{1}} \\\ {{c}_{2}} \\\ {{c}_{3}} \\\ \end{matrix} \right]$
Where ${{a}_{11}}=1,{{a}_{12}}=1,{{a}_{13}}=1,{{a}_{21}}=1,{{a}_{22}}=-2,{{a}_{23}}=1,{{a}_{31}}=3,{{a}_{32}}=1,{{a}_{33}}-1$, and ${{c}_{1}},{{c}_{2}},{{c}_{3}}$ are the constants of the 3 equations. So, putting the values we get,
$\left[ \begin{matrix} 1 & 1 & 1 \\\ 1 & -2 & 1 \\\ 3 & 1 & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 2 \\\ 8 \\\ 4 \\\ \end{matrix} \right]$
Here, we will assume as $AX=B$ . we need to find value of X.
So, on dividing both side by A, we get $X=\dfrac{B}{A}={{A}^{-1}}B$ . ……………………(4)
To find ${{A}^{-1}}$ of the matrix, formula we will used is ${{A}^{-1}}=\dfrac{1}{\left| A \right|}Adj\left( A \right)$ …………………………….(5)
Now, to find $\left| A \right|$ from the matrix $\left[ \begin{matrix} 1 & 1 & 1 \\\ 1 & -2 & 1 \\\ 3 & 1 & 1 \\\ \end{matrix} \right]$ we get as,
$\left| A \right|=1\left( -2\cdot 1-1\cdot 1 \right)-1\left( 1\cdot 1-3\cdot 1 \right)+1\left( 1\cdot 1-3\cdot \left( -2 \right) \right)$
$\left| A \right|=\left( -2-1 \right)-\left( 1-3 \right)+\left( 1+6 \right)$
$\left| A \right|=-3+2+7=6$
Now to find $Adj\left( A \right)$, we need cofactors as shown below:
${{B}_{11}}=\left| \begin{matrix} -2 & 1 \\\ 1 & 1 \\\ \end{matrix} \right|=\left( -2\cdot 1-1\cdot 1 \right)=-2-1=-3$
Cofactor will be ${{\left( -1 \right)}^{\left( i+j \right)}}={{\left( -1 \right)}^{\left( 1+1 \right)}}=-{{1}^{2}}=1$ . So, we will multiply $1\times \left( -3 \right)=-3$ . Thus, cofactor of ${{B}_{11}}=-3$ .
${{B}_{12}}=\left| \begin{matrix} 1 & 1 \\\ 3 & 1 \\\ \end{matrix} \right|=\left( 1\cdot 1-3\cdot 1 \right)=1-3=-2$
Cofactor will be ${{\left( -1 \right)}^{\left( i+j \right)}}={{\left( -1 \right)}^{\left( 1+2 \right)}}=-{{1}^{3}}=-1$ . So, we will multiply $-1\times \left( -2 \right)=2$ . Thus, cofactor of ${{B}_{12}}=2$ .
${{B}_{13}}=\left| \begin{matrix} 1 & -2 \\\ 3 & 1 \\\ \end{matrix} \right|=\left( 1\cdot 1-3\cdot \left( -2 \right) \right)=1+6=7$
Cofactor will be ${{\left( -1 \right)}^{\left( i+j \right)}}={{\left( -1 \right)}^{\left( 1+3 \right)}}=-{{1}^{4}}=1$ . So, we will multiply $1\times 7=7$ . Thus, cofactor of ${{B}_{13}}=7$ .
${{B}_{21}}=\left| \begin{matrix} 1 & 1 \\\ 1 & 1 \\\ \end{matrix} \right|=\left( 1\cdot 1-1\cdot 1 \right)=1-1=0$
Cofactor will be ${{\left( -1 \right)}^{\left( i+j \right)}}={{\left( -1 \right)}^{\left( 2+1 \right)}}=-{{1}^{3}}=-1$ . So, we will multiply $-1\times 0=0$ . Thus, cofactor of ${{B}_{21}}=0$ .
${{B}_{22}}=\left| \begin{matrix} 1 & 1 \\\ 3 & 1 \\\ \end{matrix} \right|=\left( 1\cdot 1-3\cdot 1 \right)=1-3=-2$
Cofactor will be ${{\left( -1 \right)}^{\left( i+j \right)}}={{\left( -1 \right)}^{\left( 2+2 \right)}}=-{{1}^{4}}=1$ . So, we will multiply $1\times \left( -2 \right)=-2$ . Thus, cofactor of ${{B}_{22}}=-2$ .
${{B}_{23}}=\left| \begin{matrix} 1 & 1 \\\ 3 & 1 \\\ \end{matrix} \right|=\left( 1\cdot 1-3\cdot 1 \right)=1-3=-2$
Cofactor will be ${{\left( -1 \right)}^{\left( i+j \right)}}={{\left( -1 \right)}^{\left( 2+3 \right)}}=-{{1}^{5}}=-1$ . So, we will multiply $-1\times \left( -2 \right)=2$ . Thus, cofactor of ${{B}_{23}}=2$ .
${{B}_{31}}=\left| \begin{matrix} 1 & 1 \\\ -2 & 1 \\\ \end{matrix} \right|=\left( 1\cdot 1-\left( -2 \right)\cdot 1 \right)=1+2=3$
Cofactor will be ${{\left( -1 \right)}^{\left( i+j \right)}}={{\left( -1 \right)}^{\left( 3+1 \right)}}=-{{1}^{4}}=1$ . So, we will multiply $1\times 3=3$ . Thus, cofactor of ${{B}_{31}}=3$ .
${{B}_{32}}=\left| \begin{matrix} 1 & 1 \\\ 1 & 1 \\\ \end{matrix} \right|=\left( 1\cdot 1-1\cdot 1 \right)=1-1=0$
Cofactor will be ${{\left( -1 \right)}^{\left( i+j \right)}}={{\left( -1 \right)}^{\left( 3+2 \right)}}=-{{1}^{5}}=-1$ . So, we will multiply $-1\times 0=0$ . Thus, cofactor of ${{B}_{32}}=0$ .
${{B}_{33}}=\left| \begin{matrix} 1 & 1 \\\ 1 & -2 \\\ \end{matrix} \right|=\left( 1\cdot \left( -2 \right)-1\cdot 1 \right)=-2-1=-3$
Cofactor will be ${{\left( -1 \right)}^{\left( i+j \right)}}={{\left( -1 \right)}^{\left( 3+3 \right)}}=-{{1}^{6}}=1$ . So, we will multiply $1\times -3=-3$ . Thus, cofactor of ${{B}_{33}}=-3$ .
$Adj\left( A \right)={{\left( \text{cofactors} \right)}^{T}}=\left[ \begin{matrix} {{B}_{11}} & {{B}_{21}} & {{B}_{31}} \\\ {{B}_{12}} & {{B}_{22}} & {{B}_{32}} \\\ {{B}_{13}} & {{B}_{23}} & {{B}_{33}} \\\ \end{matrix} \right]=\left[ \begin{matrix} -3 & 0 & 3 \\\ 2 & -2 & 0 \\\ 7 & 2 & -3 \\\ \end{matrix} \right]$ (T is known as transpose)
Now, substituting all the values in equation (5), we get as
${{A}^{-1}}=\dfrac{1}{6}\left[ \begin{matrix} -3 & 0 & 3 \\\ 2 & -2 & 0 \\\ 7 & 2 & -3 \\\ \end{matrix} \right]$
So, now writing equation (4) we get as,
$X=\dfrac{1}{6}\left[ \begin{matrix} -3 & 0 & 3 \\\ 2 & -2 & 0 \\\ 7 & 2 & -3 \\\ \end{matrix} \right]\left[ \begin{matrix} 2 \\\ 8 \\\ 4 \\\ \end{matrix} \right]$

-3\cdot 2+0\cdot 8+3\cdot 4 \\\ 2\cdot 2+\left( -2\cdot 8 \right)+0\cdot 4 \\\ 7\cdot 2+2\cdot 8+\left( -3\cdot 4 \right) \\\ \end{matrix} \right]$$ On solving, we get $$X=\dfrac{1}{6}\left[ \begin{matrix} 6 \\\ -12 \\\ 18 \\\ \end{matrix} \right]$$ $$X=\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ -2 \\\ 3 \\\ \end{matrix} \right]$$ **Thus, three numbers as 1, -2, 3** Note: Generally, students make mistakes in multiplying the matrix $3\times 3$ with $3\times 1$ and calculation will be wrong. So, just remember this rule that to multiply $m\times n$ matrix by $n\times p$ matrix, the ns must be the same and the result obtained will be $m\times p$ . Therefore, $\left( m\times n \right)\times \left( n\times p \right)=m\times p$ matrix. Also, don’t forget to transpose the cofactor we find for adjoint matrix A i.e. $Adj\left( A \right)={{\left( \text{cofactor} \right)}^{T}}$ otherwise, the answer will be wrong. So, be careful with it.