Inverse of a matrix using adjoint

A number like $5$ has a reciprocal, $\tfrac15$ , satisfying $5 \times \tfrac15 = 1$ . A square matrix $A$ has an analogous partner, its inverse $A^{-1}$ , satisfying $A \cdot A^{-1} = I$ , the identity matrix. But not every matrix has one — only those with $|A| \neq 0$ — and CBSE Class 12 builds $A^{-1}$ by a specific, examinable route: the adjoint method, $A^{-1} = \frac{1}{|A|}\operatorname{adj}(A)$ . This topic sits in the Determinants chapter, right after the adjoint of a matrix, and it is the final piece needed before solving systems of linear equations in matrix form. Get comfortable here — checking the determinant, building the adjoint, then dividing — and that next application follows almost for free.

The core insight

For a non-singular square matrix ( $|A| \neq 0$ ), the inverse is $A^{-1} = \frac{1}{|A|}\operatorname{adj}(A)$ , and it exists if and only if $|A| \neq 0$ .

The core idea

Where it fits. The syllabus's chosen path to $A^{-1}$ is the adjoint method, not elementary row or column operations — those are not examined here. You are expected to compute $A^{-1}$ for a $2\times2$ or $3\times3$ non-singular matrix using the formula below, and to know instantly, from $|A| = 0$ , that no inverse exists.

The defining formula. For a square matrix $A$ with $|A| \neq 0$ ,

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

where: $|A|$ is the determinant of $A$ (a scalar), and $\operatorname{adj}(A)$ is the adjoint of $A$ (the transpose of the cofactor matrix, built in the previous topic). Every entry of $\operatorname{adj}(A)$ gets divided by the single number $|A|$ .

Why this formula, and not another. It follows directly from the key identity carried over from the adjoint topic, $A \cdot \operatorname{adj}(A) = |A|\,I$ . Dividing both sides by the scalar $|A|$ (legal exactly when $|A| \neq 0$ ) gives $A \cdot \left(\frac{1}{|A|}\operatorname{adj}(A)\right) = I$ — which is precisely the defining property of an inverse, $A \cdot A^{-1} = I$ . So the formula is not a separate fact to memorise in isolation; it falls straight out of the identity you already proved for the adjoint.

Existence condition. $A^{-1}$ exists if and only if $|A| \neq 0$ (a non-singular matrix). If $|A| = 0$ (a singular matrix), $A^{-1}$ simply does not exist — checking the determinant is always the first move, before any cofactor work is attempted.

The arithmetic parallel. For a non-zero real number $a$ , the inverse is $\frac1a$ and $a \cdot \frac1a = 1$ ; for a non-singular matrix $A$ , the inverse is $\frac{1}{|A|}\operatorname{adj}(A)$ and $A \cdot A^{-1} = I$ , with the identity matrix playing the role that the number $1$ plays for ordinary numbers. Just as $0$ has no reciprocal, a matrix with $|A| = 0$ has no inverse — a useful way to remember why the condition is unavoidable, not a technicality to check off.

Figure

A number line reciprocal beside a matrix inverse diagram showing the same one over structure

Practical recipe

Step 1. Compute $|A|$ ; if it is $0$ , stop — no inverse exists. Step 2. Compute $\operatorname{adj}(A)$ (via cofactors and transpose, or the $2\times2$ shortcut of swapping the diagonal and negating the off-diagonal). Step 3. Divide every entry of $\operatorname{adj}(A)$ by $|A|$ to get $A^{-1}$ .

Derivation ADeriving the adjoint formula for a matrix inverse

The goal is to show, starting only from the previously established identity $A \cdot \operatorname{adj}(A) = |A|\,I$ and the definition of a matrix inverse ( $A^{-1}$ is whatever matrix satisfies $A \cdot A^{-1} = I$ ), that $A^{-1} = \frac{1}{|A|}\operatorname{adj}(A)$ is forced upon us whenever $|A| \neq 0$ , and that this is exactly why the condition $|A| \neq 0$ is non-negotiable rather than a side remark.

Step 1. Start from the identity connecting a matrix to its own adjoint: $A \cdot \operatorname{adj}(A) = |A|\,I$ . This was proved in the adjoint topic using cofactor expansion, and it holds for every square matrix $A$ , singular or not.

Step 2. Suppose $|A| \neq 0$ . Since $|A|$ is then a non-zero scalar, we are free to divide both sides of the identity by it: $\dfrac{A \cdot \operatorname{adj}(A)}{|A|} = \dfrac{|A|\,I}{|A|} = I$ .

Step 3. Pull the scalar $\frac{1}{|A|}$ inside, next to $\operatorname{adj}(A)$ , using the rule that scalar multiplication can be regrouped with matrix multiplication: $A \cdot \left(\frac{1}{|A|}\operatorname{adj}(A)\right) = I$ .

Step 4. Compare this to the defining property of an inverse, $A \cdot A^{-1} = I$ . The matrix $\frac{1}{|A|}\operatorname{adj}(A)$ satisfies exactly that equation, so by definition it is $A^{-1}$ : $A^{-1} = \frac{1}{|A|}\operatorname{adj}(A)$ .

Step 5. Now see why $|A| \neq 0$ is unavoidable, not optional. If $|A| = 0$ , Step 2's division is undefined — there is no legal way to isolate $A^{-1}$ from the identity. Trying to force it, $A \cdot \operatorname{adj}(A) = 0 \cdot I = 0$ would be the only true statement left, which says nothing about any inverse. No matrix can then satisfy $A \cdot A^{-1} = I$ , confirming that a singular matrix has no inverse at all.

Worked example

Find $A^{-1}$ for $A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$ , extending the calculation from the adjoint topic.

Step 1. Compute the determinant: $|A| = (2)(3) - (5)(1) = 6 - 5 = 1$ .

Step 2. Since $|A| = 1 \neq 0$ , $A^{-1}$ exists — no need to stop.

Step 3. Compute $\operatorname{adj}(A)$ using the $2\times2$ shortcut (swap the diagonal entries, negate the off-diagonal entries):

\operatorname{adj}(A) = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}

Step 4. Divide every entry by $|A| = 1$ :

A^{-1} = \frac{1}{1}\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}

Step 5. Check by verifying $A \cdot A^{-1} = I$ : $\begin{bmatrix}2&5\\1&3\end{bmatrix}\begin{bmatrix}3&-5\\-1&2\end{bmatrix} = \begin{bmatrix}2(3)+5(-1) & 2(-5)+5(2)\\1(3)+3(-1)&1(-5)+3(2)\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$ . This confirms $A^{-1} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$ .

Competitive stretch

For $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix}$ , the adjoint topic already found $\operatorname{adj}(A) = \begin{bmatrix} 1 & 0 & -2 \\ 0 & -5 & 0 \\ -3 & 0 & 1 \end{bmatrix}$ and $|A| = -5$ . Dividing every entry by $-5$ gives $A^{-1} = \begin{bmatrix} -\tfrac15 & 0 & \tfrac25 \\ 0 & 1 & 0 \\ \tfrac35 & 0 & -\tfrac15 \end{bmatrix}$ . A quick partial check: row $1$ of $A^{-1}$ , $\left(-\tfrac15, 0, \tfrac25\right)$ , dotted with column $1$ of $A$ , $(1,0,3)$ , gives $-\tfrac15(1) + 0(0) + \tfrac25(3) = -\tfrac15 + \tfrac65 = 1$ , exactly as $A \cdot A^{-1} = I$ demands.

Common traps

Trying to find $A^{-1}$ without first checking $|A| \neq 0$ — if $|A| = 0$ , no amount of further cofactor work produces an inverse; stop at Step 1.
Forgetting to divide every entry of $\operatorname{adj}(A)$ by $|A|$ — a partial division leaves an incorrect matrix.
Confusing $\operatorname{adj}(A)$ with $A^{-1}$ itself — the adjoint is only a stepping stone; $A^{-1}$ needs the extra division by $|A|$ .
Sign slips when $|A|$ is negative, as in the competitive stretch above — track the resulting negative fractions carefully.

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Why do we divide by the determinant specifically, and not some other number?

The division falls out of the identity $A \cdot \operatorname{adj}(A) = |A|\,I$ , which is proved directly from cofactor expansion — it is not a choice, it is a consequence. Dividing both sides by the scalar $|A|$ is the only way to turn that identity into $A \cdot (\text{something}) = I$ , and whatever matches that "something" is, by definition, $A^{-1}$ . If $|A|$ were replaced by any other number, the right-hand side would no longer equal $I$ , and the resulting matrix would fail the defining test of an inverse.

Can a non-square matrix have an inverse found this way?

No. The adjoint method, and the very concept of $A^{-1}$ satisfying $A \cdot A^{-1} = I$ , is defined only for square matrices, because only square matrices have a determinant and a matching-size identity matrix to multiply into. Non-square matrices are outside this topic's scope entirely — CBSE Class 12 restricts the inverse-via-adjoint method to square, non-singular matrices, almost always $2\times2$ or $3\times3$ in examination questions.

If I compute A inverse and then multiply it back by A, what should I always get, and why check it?

You should always get the identity matrix $I$ , because that equation, $A \cdot A^{-1} = I$ , is the very definition an inverse must satisfy — it is not an optional extra property but the test of correctness. Checking it catches nearly every mistake: an arithmetic slip in a cofactor, a forgotten sign, or a division applied to only some entries of the adjoint. In an exam, this verification step also demonstrates you understand why the formula works, not just that you can apply it.