Determinants

A system of two linear equations in two unknowns has a unique solution exactly when a single number built from its coefficients is nonzero. That number, $a_1b_2 - a_2b_1$ , is the seed of an entire branch of algebra. Attach it to a square matrix $A$ and call it $\det A$ or $|A|$ , and you get a function that takes any square array of numbers and returns one number that packs an astonishing amount of structure: whether a matrix can be inverted, whether a system of equations has a unique solution, and even the area of a triangle drawn from three coordinate points. This chapter builds that function from the ground up for matrices of order one, two and three, and then puts it to work. You will learn to expand a $3\times 3$ determinant along any row or column and get the same answer every time, a fact that looks like a coincidence until you see why it must be true. You will meet minors and cofactors, the local building blocks that assemble into the determinant. You will define the adjoint of a matrix — the transpose of its cofactor matrix — and discover the single identity $A(\operatorname{adj} A) = |A|\,I$ that makes the formula for a matrix inverse fall out almost for free. Along the way, determinants will tell you the area of a triangle, detect three collinear points, and decide instantly whether a system of linear equations has one solution, none, or infinitely many, all without solving the system by elimination. Every idea here rests on arithmetic and pattern you already know from Class 11 matrices: nothing more exotic than multiplication, addition and careful bookkeeping of signs. The payoff is a genuinely powerful tool — the matrix method for solving simultaneous equations, a cornerstone technique for both board examinations and competitive tests such as JEE and NEET, where determinant identities, adjoint properties and consistency conditions appear constantly, often disguised inside larger problems on matrices, vectors and coordinate geometry.

Key takeaways

A determinant is a number associated with a square matrix; only square matrices have determinants, and $|A|$ denotes "determinant of $A$ ," never the modulus of $A$ .
For order 2, $\begin{vmatrix}a & b\\ c& d\end{vmatrix} = ad-bc$ ; for order 3, expand along any row or column using the sign pattern $(-1)^{i+j}$ — every choice of row or column gives the same value.
The minor $M_{ij}$ of an entry $a_{ij}$ is the determinant left after deleting row $i$ and column $j$ ; the cofactor is $A_{ij} = (-1)^{i+j}M_{ij}$ .
The area of a triangle with vertices $(x_1,y_1)$ , $(x_2,y_2)$ , $(x_3,y_3)$ is $\dfrac{1}{2}\left|\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}\right|$ , and this determinant is exactly zero when the three points are collinear.
The adjoint of $A$ is the transpose of its cofactor matrix, and it satisfies the master identity $A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|\,I$ .
A square matrix is invertible if and only if it is non-singular ( $|A| \neq 0$ ), and then $A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A$ .
For order $n$ , $|kA| = k^n|A|$ and $|\operatorname{adj} A| = |A|^{n-1}$ ; the determinant of a product equals the product of the determinants, $|AB| = |A||B|$ .
A linear system $AX = B$ with a square coefficient matrix has a unique solution $X = A^{-1}B$ when $|A| \neq 0$ (the matrix method); when $|A| = 0$ , the system is inconsistent if $(\operatorname{adj} A)B \neq O$ , and may be consistent or inconsistent if $(\operatorname{adj} A)B = O$ .
Key idea: the determinant is a single number built from a square matrix by cofactor expansion that instantly tells you whether the matrix is invertible, giving the formula A⁻¹ = adj(A)/|A| and deciding whether a linear system has a unique, no, or infinitely many solutions.

The core insight

A determinant is the one number, computed by cofactor expansion along any row or column, that decides whether a square matrix is invertible ( $|A| \neq 0 \Rightarrow A^{-1} = \operatorname{adj} A / |A|$ ) and whether a linear system built from that matrix has a unique solution.

What this chapter covers

The chapter opens by defining the determinant for orders one, two and three, and by showing the practical rule that expanding along a row or column full of zeros saves work without changing the answer. It then turns determinants toward coordinate geometry, using them to write the area of a triangle in a single compact formula and, as a special case, to test three points for collinearity. The middle of the chapter introduces minors and cofactors, the small determinants obtained by deleting a row and a column, and shows how the full determinant is simply the sum of any row's (or column's) entries times their cofactors. Those cofactors are then assembled, transposed, into the adjoint of a matrix, which leads directly to the formula for the inverse of a matrix and to the sharp dividing line between singular and non-singular matrices. The chapter closes by applying all of this machinery to solve systems of linear equations by the matrix method, including how to read off consistency or inconsistency purely from $|A|$ and $(\operatorname{adj} A)B$ without ever eliminating a variable by hand.

The core ideas

The determinant of orders one, two and three

To every square matrix $A = [a_{ij}]$ of order $n$ we associate a single real (or complex) number called its determinant, written $|A|$ , $\det A$ , or $\Delta$ . Only square matrices have determinants — a non-square array simply has no such number defined for it. For order one, the rule is trivial: $A = [a]$ gives $\det A = a$ . For order two, the determinant is the familiar cross-difference:

\det A = |A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{21}a_{12}. \qquad (1)

where: $a_{11}, a_{22}$ are the leading-diagonal entries and $a_{21}, a_{12}$ are the off-diagonal entries. A quick check: $\begin{vmatrix}2 & 4\\ -1 & 2\end{vmatrix} = 2(2) - 4(-1) = 4+4 = 8$ . For order three, the determinant is defined by reducing it to three order-two determinants — an expansion along a chosen row or column. Expanding along the first row of $A = [a_{ij}]_{3\times 3}$ ,

|A| = a_{11}\begin{vmatrix}a_{22} & a_{23}\\ a_{32} & a_{33}\end{vmatrix} - a_{12}\begin{vmatrix}a_{21} & a_{23}\\ a_{31} & a_{33}\end{vmatrix} + a_{13}\begin{vmatrix}a_{21} & a_{22}\\ a_{31} & a_{32}\end{vmatrix}. \qquad (2)

where: each $2\times 2$ determinant is obtained by striking out the row and column of the multiplying entry, and the signs alternate $+,-,+$ across the row. There are six such expansions in all — one for each of the three rows and three columns — and the chapter's first big theorem, proved below, is that every one of them gives the same number. Because of this freedom, the practical rule is to expand along whichever row or column has the most zeros, since a zero entry contributes nothing and can simply be skipped. A useful companion fact, used constantly in exercises, is the scalar rule: if every entry of an $n\times n$ matrix $A$ is multiplied by a constant $k$ , the determinant scales as $|kA| = k^n|A|$ , not $k|A|$ — the exponent $n$ , the order of the matrix, is the detail students most often forget.

Figure

A three by three grid showing the checkerboard sign pattern plus minus plus in the first row minus plus minus in the second row and plus minus plus in the third row used for cofactor expansion

Minors, cofactors, and the expansion theorem

The pieces used above have names. The minor $M_{ij}$ of an element $a_{ij}$ is the determinant of the smaller matrix left after deleting row $i$ and column $j$ — for an order- $n$ determinant, every minor is an order- $(n-1)$ determinant. The cofactor $A_{ij}$ attaches the checkerboard sign to the minor:

A_{ij} = (-1)^{i+j}M_{ij}. \qquad (3)

where: the sign is $+1$ when $i+j$ is even and $-1$ when $i+j$ is odd, matching the grid above. With this notation, expansion along row 1, equation (2), reads compactly as $|A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$ : the determinant equals the sum of any row's entries times their own cofactors. This is proved to hold for every row and every column in the first derivation below. A companion identity, equally important for later work on the adjoint, is that mixing a row's entries with a different row's cofactors always gives zero: $a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0$ , because that sum is secretly the determinant of a matrix with two identical rows.

Area of a triangle and the collinearity test

The area of a triangle with vertices $(x_1,y_1)$ , $(x_2,y_2)$ , $(x_3,y_3)$ , ordinarily written as $\tfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ , packs neatly into a determinant:

\Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix}. \qquad (4)

where: the three rows list each vertex's coordinates followed by a $1$ . Because area is always positive, the absolute value of the determinant is taken; if the raw signed value comes out negative, the points were listed in clockwise rather than counter-clockwise order, but the magnitude of the area is unaffected. If the area is given as a fixed number, both the positive and negative values of the determinant should be tried when solving for an unknown coordinate. The most useful special case is when $\Delta = 0$ : three points give a triangle with zero area precisely when they are collinear — this turns the determinant into a one-line collinearity test, and the same determinant, set to zero with $(x,y)$ left as unknowns, gives the equation of the line through two fixed points.

Figure

A coordinate plane showing a triangle with vertices at three, eight and minus four, two and five, one with the signed determinant giving its area, next to a degenerate case of three collinear points lying on a straight line where the same determinant is zero

Quick-evaluation properties every JEE aspirant should know

Beyond the definitions above, a handful of standard determinant properties — heavily used in competitive problems though not restated as a separate list in this chapter — speed up evaluation. If two rows (or columns) are identical, the value is zero (exactly why $a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}=0$ above). Interchanging any two rows or columns flips the sign without changing the magnitude. A common factor of one row or column can be pulled outside as a multiplier — the row-by-row version of $|kA|=k^n|A|$ . Adding a multiple of one row to another leaves the determinant unchanged — the row-operation trick that creates zeros before expanding.

Adjoint of a matrix and the master identity

Collect every cofactor of $A = [a_{ij}]_{n\times n}$ into a matrix $[A_{ij}]$ , then transpose it: this transpose is the adjoint of $A$ , written $\operatorname{adj} A$ . For a $3\times 3$ matrix,

\operatorname{adj} A = \begin{bmatrix} A_{11} & A_{21} & A_{31}\\ A_{12} & A_{22} & A_{32}\\ A_{13} & A_{23} & A_{33} \end{bmatrix}. \qquad (5)

where: notice the transpose — row $i$ of the cofactor matrix becomes column $i$ of the adjoint; forgetting this transpose is the single most common slip in this chapter. For order $2$ , there is a shortcut: swap the two diagonal entries and change the sign of the two off-diagonal entries. The adjoint exists purely to satisfy one identity, proved in the first derivation below, that makes the inverse formula almost automatic:

A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|\,I. \qquad (6)

where: $I$ is the identity matrix of the same order as $A$ . Taking determinants of both sides of (6) and using $|AB| = |A||B|$ gives the useful order-independent fact $|\operatorname{adj} A| = |A|^{n-1}$ for an $n\times n$ matrix.

Singular matrices, invertibility, and the inverse formula

A square matrix $A$ is singular if $|A| = 0$ and non-singular if $|A| \neq 0$ . The central theorem of this chapter, proved in the second derivation below, is that these two conditions exactly separate the matrices that can be inverted from those that cannot:

A \text{ is invertible} \iff A \text{ is non-singular}, \qquad A^{-1} = \frac{1}{|A|}\operatorname{adj} A. \qquad (7)

where: the formula on the right only makes sense when $|A|\neq 0$ , matching the left-hand condition exactly — there is no partial or approximate inverse when $|A| = 0$ . Two related facts round out the picture: if $A$ and $B$ are both non-singular matrices of the same order, then $AB$ and $BA$ are also non-singular, and $|AB| = |A|\,|B|$ for any square matrices of matching order.

Applications: solving linear systems by the matrix method

A system of three linear equations in three unknowns, $a_1x+b_1y+c_1z=d_1$ and so on, can be written as a single matrix equation $AX = B$ , where $A$ is the $3\times 3$ coefficient matrix, $X$ is the column of unknowns, and $B$ is the column of constants. If $A$ is non-singular, premultiplying both sides by $A^{-1}$ gives the unique solution directly:

X = A^{-1}B. \qquad (8)

where: this is possible because $A^{-1}(AX) = (A^{-1}A)X = IX = X$ by associativity, so the whole system collapses to one matrix multiplication once $A^{-1}$ is known. This is called the matrix method, and it is the chapter's payoff: no elimination, no back-substitution, just one inverse and one multiplication. When $A$ is singular ( $|A|=0$ ), the matrix method breaks down and a separate consistency check is needed: compute $(\operatorname{adj} A)B$ . If $(\operatorname{adj} A)B \neq O$ (the zero matrix), the system has no solution and is called inconsistent. If $(\operatorname{adj} A)B = O$ , the system may have infinitely many solutions or none, and the two singular cases cannot be told apart from $|A|$ alone — a genuinely useful shortcut for the objective-type questions that dominate competitive papers, since it avoids solving the full system just to classify it.

Three key derivations

Two results carry this chapter's applications: cofactor expansion along any row or column gives the same value, and the identity $A(\operatorname{adj} A) = |A|\,I$ that hands us the inverse formula. A third result, the consistency test for linear systems, shows how far a single number can go in classifying a whole system without solving it. The three derivations below prove each in turn.

Derivation AExpanding a determinant along any row or column gives the same value

The goal is to confirm the claim used silently throughout this chapter — that the six possible cofactor expansions of a $3\times3$ determinant (three rows, three columns) all produce the identical number — so that "the determinant of $A$ " is an unambiguous, well-defined quantity rather than six potentially different answers in disguise. We do this by expanding along the first row and along the second row independently, in raw coordinates, and comparing the fully multiplied-out expressions term by term.

Step 1. Expand along the first row $R_1$ using the definition (2). Multiplying out each $2\times2$ determinant gives $|A| = a_{11}(a_{22}a_{33}-a_{32}a_{23}) - a_{12}(a_{21}a_{33}-a_{31}a_{23}) + a_{13}(a_{21}a_{32}-a_{31}a_{22})$ , which expands to the six-term sum $a_{11}a_{22}a_{33} - a_{11}a_{32}a_{23} - a_{12}a_{21}a_{33} + a_{12}a_{31}a_{23} + a_{13}a_{21}a_{32} - a_{13}a_{31}a_{22}$ .

Step 2. Independently expand along the second row $R_2$ . By the same definition applied to row 2, with signs $(-1)^{2+1}, (-1)^{2+2}, (-1)^{2+3}$ , we get $|A| = -a_{21}(a_{12}a_{33}-a_{32}a_{13}) + a_{22}(a_{11}a_{33}-a_{31}a_{13}) - a_{23}(a_{11}a_{32}-a_{31}a_{12})$ .

Step 3. Multiply out the $R_2$ expansion fully. This gives $-a_{21}a_{12}a_{33} + a_{21}a_{32}a_{13} + a_{22}a_{11}a_{33} - a_{22}a_{31}a_{13} - a_{23}a_{11}a_{32} + a_{23}a_{31}a_{12}$ .

Step 4. Reorder the six terms from Step 3 alphabetically by subscript pattern: $a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{31}a_{22}$ . This is term-for-term identical to the Step 1 result. The same reordering-and-matching argument works for expansion along $R_3$ , $C_1$ , $C_2$ , or $C_3$ : each produces a permutation of the identical six signed products, so all six expansions of a $3\times3$ determinant agree, and $|A| = a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}$ using any row's cofactors is a safe, unambiguous definition. A useful byproduct: replacing a row's own cofactors with a different row's cofactors, such as $a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}$ , is secretly the expansion of a determinant with row 1 repeated in place of row 2, forced to equal zero because two identical rows always shrink a determinant to zero.

Go deeper · Derivation B · 2 minThe identity A adjoint A equals determinant of A times I

Derivation BThe identity A adjoint A equals determinant of A times I

The goal is to establish the single identity that makes the whole inverse-matrix formula possible: for a square matrix $A$ of order $n$ , the product $A$ times its own adjoint always equals $|A|$ times the identity matrix, on both sides. This is not obvious by inspection — it says every off-diagonal entry of the product $A(\operatorname{adj} A)$ vanishes while every diagonal entry equals the same number $|A|$ — so the proof has to track a general diagonal entry and a general off-diagonal entry of the product separately, using the two expansion facts proved above.

Step 1. Write $A$ and $\operatorname{adj} A$ in full for order $3$ , as in equation (5), and set up the matrix product $A(\operatorname{adj} A)$ using ordinary row-times-column multiplication. The $(1,1)$ entry of the product is row 1 of $A$ dotted with column 1 of $\operatorname{adj} A$ , which is $a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}$ .

Step 2. Recognise the $(1,1)$ entry from Step 1 as exactly the cofactor expansion of $|A|$ along its first row, proved in the previous derivation. So the $(1,1)$ entry of $A(\operatorname{adj} A)$ equals $|A|$ . By the identical argument applied to row 2 dotted with column 2, and row 3 dotted with column 3, every diagonal entry of $A(\operatorname{adj} A)$ equals $|A|$ .

Step 3. Examine an off-diagonal entry, say $(1,2)$ : row 1 of $A$ dotted with column 2 of $\operatorname{adj} A$ , which is $a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}$ — the "mixed" sum flagged at the end of the previous derivation, equal to the determinant of a matrix with row 1 repeated, hence zero. The same reasoning kills every other off-diagonal entry.

Step 4. Assemble the result: every diagonal entry of $A(\operatorname{adj} A)$ is $|A|$ and every off-diagonal entry is $0$ , so $A(\operatorname{adj} A) = |A|I$ . An identical row-versus-column argument, dotting rows of $\operatorname{adj} A$ against columns of $A$ , gives $(\operatorname{adj} A)A = |A|I$ as well, so both orders of multiplication agree.

Step 5. Extract the inverse formula. When $|A|\neq 0$ , divide the identity by the scalar $|A|$ : $A\left(\dfrac{1}{|A|}\operatorname{adj} A\right) = I$ . This exhibits a genuine two-sided inverse for $A$ , namely $\dfrac{1}{|A|}\operatorname{adj} A$ , proving the "if non-singular then invertible" direction. Conversely, if $A$ is invertible with $AB = BA = I$ for some $B$ , taking determinants gives $|A||B| = |I| = 1$ , which forces $|A|\neq 0$ — proving the "if invertible then non-singular" direction. Together these give the full equivalence $A$ is invertible if and only if $A$ is non-singular. Verified: for $A=\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}$ , $|A|=1(16-9)-3(4-3)+3(3-4)=7-3-3=1$ , and direct multiplication of $A$ with $\operatorname{adj}A=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}$ gives exactly the identity matrix, matching $|A|I$ with $|A|=1$ .

Go deeper · Derivation C · 1 minClassifying a linear system from determinant and adjoint alone

Derivation CClassifying a linear system from determinant and adjoint alone

The goal is to justify the classification rule stated in equation (8) and its surrounding text — that for a system $AX=B$ with square coefficient matrix $A$ , the sign of $|A|$ and, when $|A|=0$ , the value of $(\operatorname{adj}A)B$ , are enough to decide whether the system has a unique solution, no solution, or possibly infinitely many, all without eliminating a single variable. The proof splits naturally into the two cases $|A|\neq 0$ and $|A|=0$ , using the master identity from the previous derivation as the main tool.

Step 1. Suppose $|A|\neq 0$ . Premultiply $AX=B$ by $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A$ , which exists by the previous derivation. This gives $A^{-1}(AX)=A^{-1}B$ , and by associativity of matrix multiplication the left side becomes $(A^{-1}A)X = IX = X$ . Hence $X=A^{-1}B$ is forced — there is exactly one possible solution, and substituting it back into $AX=B$ confirms it works, since $A(A^{-1}B) = (AA^{-1})B = IB = B$ . This proves existence and uniqueness together whenever $A$ is non-singular.

Step 2. Now suppose $|A|=0$ . The matrix $A$ has no inverse, so Step 1's argument is simply unavailable — the system might still have solutions, but they cannot be produced by inverting $A$ . Multiply both sides of $AX=B$ on the left by $\operatorname{adj}A$ instead, which always exists regardless of whether $A$ is singular: $(\operatorname{adj}A)(AX) = (\operatorname{adj}A)B$ .

Step 3. Simplify the left side of Step 2 using the master identity $(\operatorname{adj}A)A = |A|I$ from the previous derivation. Since $|A|=0$ here, $(\operatorname{adj}A)A = 0\cdot I = O$ , the zero matrix. So the left side of Step 2 becomes $O\cdot X = O$ , meaning any genuine solution $X$ of the original system must satisfy $(\operatorname{adj}A)B = O$ as a necessary condition.

Step 4. Read Step 3 as a test. If $(\operatorname{adj}A)B\neq O$ , no vector $X$ can satisfy the original system, so it has no solution and is inconsistent. If $(\operatorname{adj}A)B = O$ , the necessary condition is satisfied but this does not manufacture an actual solution — it only fails to rule one out — so the system may have infinitely many solutions or, in some configurations, still none; distinguishing those two singular sub-cases requires examining the equations directly.

Worked examples

Board level - evaluating a 3 by 3 determinant by expanding along the column with the most zeros (Example 3 pattern)

Evaluate $\Delta = \begin{vmatrix} 1 & 2 & 4\\ -1 & 3 & 0\\ 4 & 1 & 0 \end{vmatrix}$ .

Step 1. Scan the three rows and three columns for the one with the most zeros. Column 3 has two zeros ( $0$ in row 2 and row 3), so expanding along $C_3$ will make two of the three terms vanish immediately.

Step 2. Write the expansion along $C_3$ using the sign pattern for column 3, which is $+,-,+$ down the column: $\Delta = 4\begin{vmatrix}-1 & 3\\ 4 & 1\end{vmatrix} - 0\begin{vmatrix}1 & 2\\ 4 & 1\end{vmatrix} + 0\begin{vmatrix}1 & 2\\ -1 & 3\end{vmatrix}$ .

Step 3. Only the first term survives. Compute the surviving $2\times2$ determinant: $\begin{vmatrix}-1 & 3\\ 4 & 1\end{vmatrix} = (-1)(1) - (3)(4) = -1-12 = -13$ .

Step 4. Multiply by the cofactor sign and the entry: $\Delta = 4(-13) = -52$ .

$\Delta = \mathbf{-52}$ .

Board level - finding the adjoint and inverse of a 3 by 3 matrix (Example 13 pattern)

If $A = \begin{bmatrix}1 & 3 & 3\\ 1 & 4 & 3\\ 1 & 3 & 4\end{bmatrix}$ , verify $A(\operatorname{adj}A) = |A|I$ and find $A^{-1}$ .

Step 1. Compute $|A|$ by expanding along row 1: $|A| = 1(16-9) - 3(4-3) + 3(3-4) = 7 - 3 - 3 = 1$ . Since $|A|\neq 0$ , $A$ is non-singular and $A^{-1}$ exists.

Step 2. Compute all nine cofactors. $A_{11}=(4)(4)-(3)(3)=7$ , $A_{12}=-[(1)(4)-(3)(1)]=-1$ , $A_{13}=(1)(3)-(4)(1)=-1$ , $A_{21}=-[(3)(4)-(3)(3)]=-3$ , $A_{22}=(1)(4)-(3)(1)=1$ , $A_{23}=-[(1)(3)-(3)(1)]=0$ , $A_{31}=(3)(3)-(3)(4)=-3$ , $A_{32}=-[(1)(3)-(3)(1)]=0$ , $A_{33}=(1)(4)-(3)(1)=1$ .

Step 3. Assemble $\operatorname{adj}A$ by transposing the cofactor matrix — row $i$ of cofactors becomes column $i$ of the adjoint: $\operatorname{adj}A = \begin{bmatrix}7 & -3 & -3\\ -1 & 1 & 0\\ -1 & 0 & 1\end{bmatrix}$ .

Step 4. Multiply $A(\operatorname{adj}A)$ directly and confirm it equals $|A|I = I$ (since $|A|=1$ ); the product comes out to the $3\times3$ identity matrix, confirming the master identity numerically.

Step 5. Divide the adjoint by $|A|=1$ to get the inverse: $A^{-1} = \dfrac{1}{|A|}\operatorname{adj}A = \operatorname{adj}A$ .

$A^{-1} = \begin{bmatrix}7 & -3 & -3\\ -1 & 1 & 0\\ -1 & 0 & 1\end{bmatrix}$ .

Competitive level (JEE / NEET) - using the determinant test to find the value that makes a system inconsistent

For what value of $\lambda$ does the system $x+y+z=6$ , $x+2y+3z=10$ , $x+2y+\lambda z=12$ fail to have a solution?

Step 1. Write the coefficient matrix $A = \begin{bmatrix}1&1&1\\ 1&2&3\\ 1&2&\lambda\end{bmatrix}$ and compute $|A|$ by expanding along row 1: $|A| = 1(2\lambda-6) - 1(\lambda-3) + 1(2-2) = (2\lambda-6)-(\lambda-3)+0 = \lambda-3$ .

Step 2. Set $|A|=0$ to find the singular value: $\lambda-3=0 \Rightarrow \lambda=3$ . For every $\lambda\neq3$ , $|A|\neq0$ , the matrix method applies directly, and the system has a unique solution — so $\lambda=3$ is the only candidate for a breakdown.

Step 3. Check the equations directly at $\lambda=3$ rather than computing the full $(\operatorname{adj}A)B$ product, since the structure is transparent here: the second equation is $x+2y+3z=10$ and the third, at $\lambda=3$ , becomes $x+2y+3z=12$ . Both equations have the identical left-hand side $x+2y+3z$ but demand two different values, $10$ and $12$ , for it simultaneously — a direct contradiction that no $(x,y,z)$ can satisfy.

Step 4. Confirm this matches the theory: since the left-hand sides of equations 2 and 3 are proportional (in fact identical) while the right-hand sides are not, $(\operatorname{adj}A)B\neq O$ at $\lambda=3$ , so by the classification derivation the system is inconsistent — no solution exists, for every value taken by the free left-over parameter, not merely on average.

$\lambda = \mathbf{3}$ makes the system inconsistent; for every $\lambda \neq 3$ , the unique solution $X = A^{-1}B$ exists. Timing note: budget under two minutes. The determinant computation to locate the singular value takes seconds once you expand along the row with the simplest entries; the real time-saver is noticing, before computing any adjoint, that two equations share the same left-hand side — this instantly proves inconsistency without needing $(\operatorname{adj}A)B$ at all. Reach for this direct-contradiction check first in MCQs; fall back to computing $(\operatorname{adj}A)B$ only when the equations do not simplify so transparently.

Common traps

The most damaging error is forgetting to transpose the cofactor matrix when forming the adjoint — writing $[A_{ij}]$ itself instead of its transpose gives a matrix correct only when $A$ is symmetric, silently wrong otherwise, poisoning every inverse computed from it. A second slip is a sign error in the cofactor: the minor is computed correctly but the factor $(-1)^{i+j}$ is forgotten or misjudged, especially at the $(1,2)$ , $(2,1)$ , $(2,3)$ and $(3,2)$ positions where the sign is negative. Third, the scalar rule is routinely misapplied: $|kA| = k^n|A|$ , not $k|A|$ — a factor of $2$ pulled from every entry of a $3\times3$ matrix multiplies the determinant by $2^3=8$ , not $2$ . Fourth, in the area formula, students drop the outer $\tfrac{1}{2}$ , forget the absolute value, or misread a zero-area answer as "no triangle" rather than "collinear points." Finally, when $|A|=0$ , many stop and declare "no solution" outright without checking $(\operatorname{adj}A)B$ — a singular coefficient matrix does not automatically mean an inconsistent system.

Common trapForgetting to transpose the cofactor matrix when building the adjoint is the single biggest misconception in this chapter

The adjoint is defined as the transpose of the cofactor matrix, not the cofactor matrix itself — row $i$ of cofactors must become column $i$ of the adjoint. Students who skip this step get every diagonal entry right (since $A_{ii}$ sits on the diagonal either way) but silently swap every off-diagonal pair, such as putting $A_{12}$ where $A_{21}$ should be. For a symmetric matrix this error is invisible because the cofactor matrix already equals its own transpose, which is precisely why the mistake survives unnoticed through years of practice on symmetric examples until it produces a wrong inverse, a wrong solution to a linear system, or a wrong answer on a competitive exam where the matrix is deliberately not symmetric.

You should be able to…

Define the determinant of a square matrix of order one, two, and three, and evaluate each using the appropriate expansion.
Choose the row or column with the most zeros before expanding a $3\times3$ determinant, to minimise arithmetic.
Apply the scalar rule $|kA| = k^n|A|$ correctly for a matrix of order $n$ , distinguishing it from the incorrect $k|A|$ .
Write the area-of-a-triangle formula as a determinant, use it to compute area given three vertices, and use it to test three points for collinearity.
Define the minor and cofactor of an entry, compute both for any position in a determinant, and state the sign rule $(-1)^{i+j}$ .
State and use the expansion theorem $|A| = a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3}$ along any row or column, and explain why mixing a row's entries with another row's cofactors gives zero.
Construct the adjoint of a matrix as the transpose of its cofactor matrix, and state and verify the identity $A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|I$ .
Distinguish singular from non-singular matrices, and state precisely the theorem that a square matrix is invertible if and only if it is non-singular.
Compute $A^{-1} = \dfrac{1}{|A|}\operatorname{adj}A$ for a $2\times2$ or $3\times3$ non-singular matrix.
Solve a system of linear equations by the matrix method $X=A^{-1}B$ when $|A|\neq0$ , and classify a singular system as inconsistent or possibly-consistent using $(\operatorname{adj}A)B$ .

Ask the tutor: If you are stuck deciding whether a singular system is inconsistent or has infinitely many solutions, ask the tutor to walk through the $(\operatorname{adj}A)B$ test on your specific system, step by step.

Why is the determinant of A written with vertical bars and not confused with the modulus of A even though the notation looks the same?

The vertical-bar notation $|A|$ is a historical convention shared with the absolute-value symbol, but the two mean entirely different things. The modulus of a real number strips its sign and is always non-negative; the determinant of a matrix can be positive, negative, or zero, and it carries essential information (its sign, in area problems, tells you the orientation of the vertices). When you see $|A|$ next to a matrix, always read it as "determinant of $A$ "; the modulus interpretation only applies to numbers, never to a matrix as a whole.

What exactly is the difference between a minor and a cofactor, and why do we need both?

The minor $M_{ij}$ is the smaller determinant left after deleting row $i$ and column $j$ — a raw number with no sign adjustment. The cofactor $A_{ij} = (-1)^{i+j}M_{ij}$ takes that same minor and multiplies it by $+1$ or $-1$ according to the checkerboard pattern at position $(i,j)$ . Both are needed because the minor alone cannot correctly reconstruct the determinant by expansion — without the sign correction, "odd" positions would be added when they should be subtracted. The cofactor is what appears in expansion formulas and the adjoint; the minor is only an intermediate step toward it.

If the determinant of the coefficient matrix is zero, does that always mean the system of equations has no solution?

No, and this is one of the most commonly misunderstood points in the chapter. A zero determinant only rules out the matrix method's shortcut $X=A^{-1}B$ , because $A^{-1}$ simply does not exist when $|A|=0$ . Whether the system actually has no solution or infinitely many solutions depends on a further check: compute $(\operatorname{adj}A)B$ . If this product is not the zero matrix, the system genuinely has no solution (inconsistent). If it is the zero matrix, the system could have infinitely many solutions, or it could still be inconsistent in a way this test cannot distinguish, and you would need to examine the original equations directly, for instance by checking whether one equation is a scalar multiple of another with a mismatched constant.

Is there a quicker way to solve a two-variable or three-variable linear system using determinants directly, without computing the full inverse matrix?

Yes — this is Cramer's rule, which replaces one column of the coefficient matrix at a time with the constants column and takes ratios of determinants; for example in two variables, $x = \Delta_x/\Delta$ and $y = \Delta_y/\Delta$ , where $\Delta$ is the coefficient determinant and $\Delta_x, \Delta_y$ come from substituting the constants column into the $x$ - and $y$ -columns. This rule is a direct consequence of the same $A^{-1}=\operatorname{adj}A/|A|$ machinery built here, just packaged differently, and many students find it faster for a $2\times2$ or $3\times3$ system than writing out the full adjoint matrix.

Why does the area-of-a-triangle determinant use a column of all ones rather than just the two coordinate columns?

The column of ones is what turns a simple algebraic expression into a determinant in the first place, and it is also exactly what makes the collinearity test work as "determinant equals zero." Without that third column, you would have only a $3\times2$ array, which is not square and has no determinant at all. With the ones column included, three points being collinear (lying on one straight line) makes the third row a linear combination of the pattern set by the first two rows in exactly the way that forces a $3\times3$ determinant to vanish — so the same formula that computes area for a genuine triangle automatically returns zero for a degenerate one, without needing a separate collinearity formula.