If $A=\begin{pmatrix} -5 & 1 & 3 \ 7 & 1 & -5 \ 1 & -1 & 1\end{pmatrix}$ and $B=\begin{pmatrix} 1 & 1 & 2 \ 3 & 2 & 1 \ 2 & 1 & 3\end{pmatrix}$, find the products $AB$ and $BA$ and hence solve the system of equations $x+y+2z=1,\ 3x+2y+z=7,\ 2x+y+3z=2$.

Question

Accepted Answer

> [!TLDR]
>
> We first compute the two products $AB$ and $BA$ directly. Both turn out to equal $4I_3$ — this means $B\cdot\left(\dfrac{A}{4}ight)=\left(\dfrac{A}{4}ight)\cdot B=I_3$, i.e. $\dfrac{A}{4}$ is *both* a left and a right inverse of $B$, so $B^{-1}=\dfrac{A}{4}$. Since the given system's coefficients are exactly $B$'s rows, $X=B^{-1}C=\dfrac{A}{4}C$ solves it instantly.

**Step 1. Compute $AB$.** $A=\begin{pmatrix}-5&1&3\7&1&-5\1&-1&1\end{pmatrix},\ B=\begin{pmatrix}1&1&2\3&2&1\2&1&3\end{pmatrix}$.
Row 1 of $A$ times $B$: $(-5(1)+1(3)+3(2),\ -5(1)+1(2)+3(1),\ -5(2)+1(1)+3(3)) = (-5+3+6,\ -5+2+3,\ -10+1+9) = (4,0,0)$
Row 2 of $A$ times $B$: $(7(1)+1(3)+(-5)(2),\ 7(1)+1(2)+(-5)(1),\ 7(2)+1(1)+(-5)(3)) = (7+3-10,\ 7+2-5,\ 14+1-15) = (0,4,0)$
Row 3 of $A$ times $B$: $(1(1)+(-1)(3)+1(2),\ 1(1)+(-1)(2)+1(1),\ 1(2)+(-1)(1)+1(3)) = (1-3+2,\ 1-2+1,\ 2-1+3) = (0,0,4)$
$$AB=\begin{pmatrix}4&0&0\0&4&0\0&0&4\end{pmatrix}=4I_3$$

**Step 2. Compute $BA$.**
Row 1 of $B$ times $A$: $(1(-5)+1(7)+2(1),\ 1(1)+1(1)+2(-1),\ 1(3)+1(-5)+2(1)) = (-5+7+2,\ 1+1-2,\ 3-5+2) = (4,0,0)$
Row 2 of $B$ times $A$: $(3(-5)+2(7)+1(1),\ 3(1)+2(1)+1(-1),\ 3(3)+2(-5)+1(1)) = (-15+14+1,\ 3+2-1,\ 9-10+1) = (0,4,0)$
Row 3 of $B$ times $A$: $(2(-5)+1(7)+3(1),\ 2(1)+1(1)+3(-1),\ 2(3)+1(-5)+3(1)) = (-10+7+3,\ 2+1-3,\ 6-5+3) = (0,0,4)$
$$BA=\begin{pmatrix}4&0&0\0&4&0\0&0&4\end{pmatrix}=4I_3$$
So indeed $AB=BA=4I_3$.

**Step 3. Derive $B^{-1}$ from this relation.** Dividing throughout by $4$: $B\left(\dfrac{A}{4}ight)=I_3$ and $\left(\dfrac{A}{4}ight)B=I_3$. A matrix that is simultaneously a left and right inverse of $B$ *is* $B^{-1}$, so
$$B^{-1}=\frac{A}{4}=\frac14\begin{pmatrix}-5&1&3\7&1&-5\1&-1&1\end{pmatrix}.$$

**Step 4. Recognise the given system as $BX=C$.** The system $x+y+2z=1,\ 3x+2y+z=7,\ 2x+y+3z=2$ has coefficient rows $(1,1,2),\ (3,2,1),\ (2,1,3)$ — exactly the rows of $B$. So it is $BX=C$ with $C=\begin{pmatrix}1\7\2\end{pmatrix}$.

**Step 5. Solve using $X=B^{-1}C=\d

Q.If $A=\begin{pmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1\end{pmatrix}$ and $B=\begin{pmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3\end{pmatrix}$ , find the products $AB$ and $BA$ and hence solve the system of equations $x+y+2z=1,\ 3x+2y+z=7,\ 2x+y+3z=2$ .

Concept understanding — Solving Linear Systems by Matrix Inversion

🎓Unlock everything free for 14 days

Unlock everything free for 14 days