Solving systems using matrix inversion

A linear system like $2x+3y=5,\ x-y=1$ can be packed into ONE matrix equation $AX=B$ , and solved in one shot with $X=A^{-1}B$ .

The Big Picture

WHY learn this? Because it converts "solve 3 equations by elimination" into "compute one inverse and multiply" — a mechanical, mistake-proof recipe that also tells you when a unique solution exists at all.

Step 1 — Writing a system as $AX=B$

HOW does this equal the system? Multiply row $i$ of $A$ by column $X$ : you get $a_{i1}x_1+\dots+a_{in}x_n$ , and setting it equal to entry $b_i$ of $B$ reproduces equation $i$ exactly. So $AX=B$ is the system, just repackaged.

Step 2 — Deriving the solution from scratch

We want $X$ . Start from $AX=B.$

Why can we "divide"? There is no division of matrices, but there IS an inverse. Suppose $A^{-1}$ exists (i.e. $A^{-1}A=I$ ). Multiply both sides on the LEFT by $A^{-1}$ — left, because matrix multiplication is not commutative and $A$ sits on the left of $X$ : $A^{-1}(AX)=A^{-1}B.$ Use associativity: $(A^{-1}A)X=A^{-1}B.$ Since $A^{-1}A=I$ (identity) and $IX=X$ :

WHY the $\det A\neq0$ condition? Because $A^{-1}=\dfrac{1}{\det A}\,\operatorname{adj}(A)$ , and dividing by $\det A=0$ is illegal. Geometrically $\det A=0$ means $A$ collapses space, so information is lost and you can't undo it.

Consistency: what happens when $\det A=0$ ?

WHY? Multiply $AX=B$ by $\operatorname{adj}(A)$ : $(\operatorname{adj}A)AX=(\operatorname{adj}A)B$ . But $(\operatorname{adj}A)A=(\det A)I$ . If $\det A=0$ the left side is $O$ , forcing $(\operatorname{adj}A)B=O$ ; if that's violated, no $X$ can exist.

Figure — Solving systems using matrix inversion

Worked Example 1 — a clean $2\times2$

Solve $2x+3y=5$ , $\ x-y=1$ .

$A=\begin{bmatrix}2&3\\1&-1\end{bmatrix},\quad B=\begin{bmatrix}5\\1\end{bmatrix}.$

Step A: $\det A=2(-1)-3(1)=-5$ . Why this step? Non-zero ⇒ unique solution exists, safe to invert.

Step B: $A^{-1}=\dfrac{1}{-5}\begin{bmatrix}-1&-3\\-1&2\end{bmatrix}$ . Why? $2\times2$ rule: swap diagonal, negate off-diagonal, divide by $\det$ .

Step C: $X=A^{-1}B=\dfrac{1}{-5}\begin{bmatrix}-1&-3\\-1&2\end{bmatrix}\begin{bmatrix}5\\1\end{bmatrix}=\dfrac{1}{-5}\begin{bmatrix}-5-3\\-5+2\end{bmatrix}=\dfrac{1}{-5}\begin{bmatrix}-8\\-3\end{bmatrix}=\begin{bmatrix}8/5\\3/5\end{bmatrix}.$ Why order $A^{-1}B$ ? Because $X=A^{-1}B$ — $B$ is on the right.

Check: $2(8/5)+3(3/5)=16/5+9/5=25/5=5.$ ✓

Worked Example 2 — a $3\times3$

Solve $x+y+z=6$ , $\ y+3z=11$ , $\ x+z=2z$ (i.e. $x-z=0$ ).

$A=\begin{bmatrix}1&1&1\\0&1&3\\1&0&-1\end{bmatrix},\ B=\begin{bmatrix}6\\11\\0\end{bmatrix}.$

Step A: $\det A$ . Expand along row 1: $1(1\cdot(-1)-3\cdot0) -1(0\cdot(-1)-3\cdot1)+1(0\cdot0-1\cdot1)=1(-1)-1(-3)+1(-1)=-1+3-1=1.$ Why? $\det=1\neq0$ ⇒ unique solution, and dividing by 1 keeps numbers clean.

Step B: cofactors → adj. Computing (transpose of cofactor matrix): $\operatorname{adj}A=\begin{bmatrix}-1&1&2\\3&-2&-3\\-1&1&1\end{bmatrix}.$ Why transpose? $\operatorname{adj}(A)_{ij}=C_{ji}$ ; forgetting the transpose is the #1 slip.

Step C: $A^{-1}=\frac{1}{1}\operatorname{adj}A$ , so $X=A^{-1}B=\begin{bmatrix}-1&1&2\\3&-2&-3\\-1&1&1\end{bmatrix}\begin{bmatrix}6\\11\\0\end{bmatrix}=\begin{bmatrix}-6+11+0\\18-22-0\\-6+11+0\end{bmatrix}=\begin{bmatrix}5\\-4\\5\end{bmatrix}.$

Wait — check row 2: $y+3z=-4+15=11$ ✓, $x+y+z=5-4+5=6$ ✓, $x-z=5-5=0$ ✓. Solution $(5,-4,5)$ .

Common Mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine a vending machine. You press a code $X$ and it drops snack $B$ . The machine's rule is $A$ . If someone tells you the snack $B$ and you want to know which code $X$ they pressed, you need the reverse machine $A^{-1}$ — put in the snack, out pops the code. That reverse only exists if the machine never gives the same snack for two different codes. In maths, that "confusing machine" is when $\det A=0$ — then you can't figure out the unique code.

Active-Recall Flashcards

What is the matrix form of a linear system?

AX=B

, where

A

=coefficient matrix,

X

=variables,

B

=constants.

Derive the inversion solution from

AX=B

Left-multiply by

A^{-1}

A^{-1}AX=A^{-1}B\Rightarrow IX=A^{-1}B\Rightarrow X=A^{-1}B

Why left-multiply, not right?

Matrices don't commute;

A

is on the left of

X

, so only left-multiplying by

A^{-1}

cancels it.

Condition for a unique solution by inversion?

\det A\neq0

(so

A^{-1}

exists).

Formula for

A^{-1}

A^{-1}=\frac{1}{\det A}\operatorname{adj}(A)

2\times2

inverse of

\begin{bmatrix}a&b\\c&d\end{bmatrix}

\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}

\det A=0

and

(\operatorname{adj}A)B\neq O

No solution — system is inconsistent.

\det A=0

and

(\operatorname{adj}A)B=O

Infinitely many solutions (or none) — check by substitution.

What is

\operatorname{adj}(A)

Transpose of the cofactor matrix.

Key identity linking adj and det?

A\,\operatorname{adj}(A)=\operatorname{adj}(A)\,A=(\det A)I

Connections

Determinants — the $\det A\neq0$ gate that decides solvability.
Adjoint and Inverse of a Matrix — builds the $A^{-1}$ we use here.
Cramer's Rule — alternative solver, same $\det A\neq0$ condition.
Cofactors and Minors — ingredients of $\operatorname{adj}(A)$ .
Consistency of Linear Systems — the $\det=0$ branch.
Gaussian Elimination — the practical rival for large systems.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab tumhare paas do-teen linear equations hote hain, unhe alag-alag solve karne ke bajaye hum unhe ek hi matrix equation mein pack kar dete hain: $AX=B$ . Yahan $A$ coefficient matrix hai, $X$ variables ka column, aur $B$ constants ka column. Yeh $AX=B$ actually wahi equations hain, bas ek neat form mein.

Ab solve kaise? Normal algebra mein $ax=b$ ho to $x=b/a$ . Matrices mein "divide" nahi hota, lekin inverse hota hai. Dono taraf ko LEFT se $A^{-1}$ se multiply karo: $A^{-1}AX=A^{-1}B$ , aur kyunki $A^{-1}A=I$ , mil jaata hai $X=A^{-1}B$ . Yaad rakho — inverse ko LEFT mein rakhna hai, kyunki matrices ka multiplication order matter karta hai. $BA^{-1}$ likhna galat hai.

Sabse important cheez: yeh trick tabhi chalega jab $\det A\neq0$ , kyunki $A^{-1}=\frac{1}{\det A}\operatorname{adj}(A)$ — aur zero se divide nahi kar sakte. Agar $\det A=0$ ho, to ya system ka koi solution nahi (inconsistent), ya infinite solutions. Toh hamesha pehle $\det A$ nikalo, warna time waste.

Intuition: socho $A$ ek machine hai jo code $X$ ko snack $B$ mein badalti hai. $A^{-1}$ ulti machine hai — snack daalo, code wapas. Yeh ulti machine tabhi banti hai jab machine confuse na karti ho, yaani $\det A\neq0$ . Bas itna samajh lo, exam mein har system solve kar loge.

The Big Picture

Step 1 — Writing a system as $AX=B$

Step 2 — Deriving the solution from scratch

Consistency: what happens when $\det A=0$ ?

Worked Example 1 — a clean $2\times2$

Worked Example 2 — a $3\times3$

Common Mistakes (Steel-manned)

Active-Recall Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Matrices & Determinants — Introduction

Connections