Question

solve the system of equations using the matrix tool.(explanation please)
4x+5y=13
8x-2y=14

Answer 1

\[write \in the form AX=B\]
\[Find \left| A \right|,then adjoint A\]
\[inverseA=\frac{ adj A }{ \left| A \right|}\]
X=inverse of A* B

Answer 2

\[A=\left[\begin{matrix}4 &5 \\ 8 & -2\end{matrix}\right]\]
\[X=\left(\begin{matrix}x \\ y\end{matrix}\right)\]
\[B=\left(\begin{matrix}13 \\ 14\end{matrix}\right)\]\[\left| A \right|=\left| \begin{matrix}4 & 5 \\8 & -2\end{matrix} \right|=4\left( -2 \right) -5\left( 8 \right) =-8-40=-48\neq 0 \]
hence inverse of A exists.

Answer 3

Find matrix of of co factors of each element of each row
then adj. of A is the transpose of above matrix.
i think now you can solve it.

Answer 4

that was so confusing

Answer 5

cofactors of elements of first row of \left| A \right|are -2,-8
cofactors of elements of second row of \left| A \right| are -5,4
\[adj. A=\left[\begin{matrix}-2 & -8 \\ -5 & 4\end{matrix}\right]\prime \]
\[adj.A=\left[\begin{matrix}-2 & -5 \\ -8 & 4\end{matrix}\right]\]
\[A ^{-1}=\frac{ 1 }{ -48 }\left[\begin{matrix}-2 &-5 \\ -8 & 4\end{matrix}\right]\]
\[X=A ^{-1}B\]
Now try to find X

Answer 6

Let me know if you can solve.

Answer 7

cofactors of elements of first row of | A| are-2,-8

Answer 8

\[X=\frac{ -1 }{48 }\left[\begin{matrix}-2 & -5\\ -8 & 4\end{matrix}\right]\left(\begin{matrix}13 \\14\end{matrix}\right)\]
\[X=\frac{ -1 }{ 48 }\left(\begin{matrix}-2*13+(-5)*14 \\ -8*13+4*14\end{matrix}\right)\]
\[X=\frac{ -1 }{ 48 }\left(\begin{matrix}-26-70 \\ -104+56\end{matrix}\right)\]
\[X=\frac{ -1 }{48 }\left(\begin{matrix}-96 \\-48\end{matrix}\right)=\left(\begin{matrix}-96/-48 \\ -48/-48\end{matrix}\right)\]
\[\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}2 \\ 1\end{matrix}\right)\]
x=2,y=1

Answer 9

\[\left| A \right|=\left| \begin{matrix}a & b \\ c &d\end{matrix} \right|\]
\[co factor of a=\left( -1 \right)^{ri+j}d=\left( -1 \right)^{1+1}d=\left( -1 \right)^{2}d=d\]
i=no. of row
j=no. of column
\[similarly cofactor of b=\left( -1 \right)^{i+j}c=\left( -1 \right)^{i+j}c=\left( -1 \right)^{1+2 }c=-c\]
\[cofactor of c=\left( -1 \right)^{i+j}b=\left( -1 \right)^{2+1}c\]
\[and cofactor of d=\left( -1 \right)^{i+j}a=\left( -1 \right)^{2+2}a=a\]