Question

If \[AX=B\], where \[A=\left[ \begin{matrix}
   1 & 2 & 3 \\
   -1 & 1 & 2 \\
   1 & 2 & 4 \\
\end{matrix} \right]\] and \[B=\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]\], what is X?

Answer 1

Hint: In this problem, we have to find the value of X, with the given matrix A and B. We know that \[AX=B\], where we can write it as, \[X={{A}^{-1}}B\], we also know that \[{{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adjA \right)\], here we have to find the determinant and the adjoint of A, and finally multiply it with B to get the value of X.

Complete step by step solution:
We know that the given matrices are,
 \[A=\left[ \begin{matrix}
   1 & 2 & 3 \\
   -1 & 1 & 2 \\
   1 & 2 & 4 \\
\end{matrix} \right]\], \[B=\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]\].
We also given that \[AX=B\], we can now write it as,
\[X={{A}^{-1}}B\]……. (1)
Where \[{{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adjA \right),\left| A \right|\ne 0\]…… (2)
We can now find the inverse of A by finding the determinant and the adjoint of A.
 We can now find the determinant, we get
\[\Rightarrow \left| A \right|=1\left( 4-4 \right)-2\left( -4-2 \right)+3\left( -2-1 \right)=3\ne 0\]
The determinant of A is 3.
We can now find the adjoint by finding the cofactor.
We can now find the cofactor of 1 in A, we get
\[\Rightarrow {{A}_{11}}=\left| \begin{matrix}
   1 & 2 \\
   2 & 4 \\
\end{matrix} \right|=4-4=0\]
Similarly, we can find the remaining cofactors, we get
\[\begin{align}
  & \Rightarrow {{A}_{12}}=\left| \begin{matrix}
   -1 & 2 \\
   1 & 4 \\
\end{matrix} \right|=-4-2=-6 \\
 & \Rightarrow {{A}_{13}}=\left| \begin{matrix}
   -1 & 1 \\
   1 & 2 \\
\end{matrix} \right|=-2-1=-3 \\
\end{align}\]
We can now find the cofactors for the second row, we get
\[\begin{align}
  & \Rightarrow {{A}_{21}}=\left| \begin{matrix}
   2 & 3 \\
   2 & 4 \\
\end{matrix} \right|=8-6=2 \\
 & \Rightarrow {{A}_{22}}=\left| \begin{matrix}
   1 & 3 \\
   1 & 4 \\
\end{matrix} \right|=4-3=1 \\
 & \Rightarrow {{A}_{23}}=\left| \begin{matrix}
   1 & 2 \\
   1 & 2 \\
\end{matrix} \right|=2-2=0 \\
\end{align}\]
We can now find the cofactors for the third row, we get
\[\begin{align}
  & \Rightarrow {{A}_{31}}=\left| \begin{matrix}
   2 & 3 \\
   1 & 2 \\
\end{matrix} \right|=4-3=1 \\
 & \Rightarrow {{A}_{32}}=\left| \begin{matrix}
   1 & 3 \\
   -1 & 2 \\
\end{matrix} \right|=2+3=5 \\
 & \Rightarrow {{A}_{33}}=\left| \begin{matrix}
   1 & 2 \\
   -1 & 1 \\
\end{matrix} \right|=1+2=3 \\
\end{align}\]
We can now arrange the cofactors, we get
\[Cofactor=\left[ \begin{matrix}
   0 & 6 & -3 \\
   -2 & 1 & 0 \\
   1 & -5 & 3 \\
\end{matrix} \right]\]
We can now find the transpose of the cofactor which is the adjoint, we get
\[Adj=\left[ \begin{matrix}
   0 & -2 & 1 \\
   6 & 1 & -5 \\
   -3 & 0 & 3 \\
\end{matrix} \right]\]
We can now write the inverse of A, we get
\[\Rightarrow {{A}^{-1}}=\dfrac{1}{3}\times \left[ \begin{matrix}
   0 & -2 & 1 \\
   6 & 1 & -5 \\
   -3 & 0 & 3 \\
\end{matrix} \right]\]
We can now substitute the above step in (1), we get
\[\Rightarrow X=\dfrac{1}{3}\times \left[ \begin{matrix}
   0 & -2 & 1 \\
   6 & 1 & -5 \\
   -3 & 0 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]\]
We can now simplify the above step by matrix multiplication, we get
\[\Rightarrow X=\dfrac{1}{3}\left[ \begin{matrix}
   0-4+3 \\
   6+2-15 \\
   -3+0+9 \\
\end{matrix} \right]=\dfrac{1}{3}\left[ \begin{matrix}
   -1 \\
   -7 \\
   6 \\
\end{matrix} \right]\]
We can now simplify the above step, we get
\[\Rightarrow X=\dfrac{1}{3}\left[ \begin{matrix}
   -1 \\
   -7 \\
   6 \\
\end{matrix} \right]=\left[ \begin{matrix}
   -\dfrac{1}{3} \\
   -\dfrac{7}{3} \\
   2 \\
\end{matrix} \right]\]
Therefore, the answer is \[X=\left[ \begin{matrix}
   -\dfrac{1}{3} \\
   -\dfrac{7}{3} \\
   2 \\
\end{matrix} \right]\].

Note: Students make mistakes while finding the cofactor value in the symbol part. We should remember that the inverse of the matrix can be found by the formula \[{{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adjA \right),\left| A \right|\ne 0\], where, we should remember that the adjoint is the transpose of the cofactor.