Question

Using properties of determinants,prove that $\left|\begin{array}{ccc}{a}^2+2a& 2a+1& 1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right|=(a-1{)}^3$.

OR find matrix A such that $\left(\begin{array}{cc}2& -1\\ 1& 0\\ -3& 4\end{array}\right)A=\left(\begin{array}{cc}-1& -8\\ 1& -2\\ 9& 22\end{array}\right).$

Answer 1

Let $△=\left|\begin{array}{ccc}{a}^2+2a& 2a+1& 1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right|$ By $R_1\to R_1-R_2,R_2\to R_2-R_3$

$\Rightarrow =\left|\begin{array}{ccc}{a}^2-1& 2a-1& 0\\ 2(a-1)& a-1& 0\\ 3& 3& 1\end{array}\right|$ $\text{Taking (a-1) common from}{R}_1\text{and}{R}_2\text{both}$

$\Rightarrow =(a-1{)}^2\left|\begin{array}{ccc}a+1& 1& 0\\ 2& 1& 0\\ 3& 3& 1\end{array}\right|$ $\text{Expanding along}{C}_3$

$\Rightarrow △=(a-1{)}^2\{0-0+1+(a+1-2\left)\right\}=(a-1{)}^3=RHS.$

OR Let A=$\left(\begin{array}{cc}m& n\\ x& y\end{array}\right)$ Now $\left(\begin{array}{ccc}2& -1& \\ 1& 0& \\ -3& 4& \end{array}\right)A=\left(\begin{array}{cc}-1& -8\\ 1& -2\\ 9& 22\end{array}\right)\Rightarrow \left(\begin{array}{ccc}2& -1& \\ 1& 0& \\ -3& 4& \end{array}\right)\left(\begin{array}{cc}m& n\\ x& y\end{array}\right)=\left(\begin{array}{cc}-1& -8\\ 1& -2\\ 9& 22\end{array}\right)$

$\Rightarrow \left(\begin{array}{cc}2m-x& 2n-y\\ m& n\\ -3m+4x& -3n+4y\end{array}\right)=\left(\begin{array}{cc}-1& -8\\ 1& -2\\ 9& 22\end{array}\right)$

By def.of equality of matrices,we get:

$2m-x=-1,2n-y=-8,m=1,n=-2,-3m+4x=9,-3n+4y=22$

So,clearly $m=1,n=-2,x=3,y=4$. Hence $A\left(\begin{array}{cc}1& -2\\ 3& 4\end{array}\right)$