Determinants

1. Check this before you revise anything

Cramer's Rule does not appear anywhere in the current book — not once, in any section, example, or exercise. The only method taught for solving a system of linear equations is the matrix method: write the system as $AX=B$, and if $A$ is non-singular, $X=A^{-1}B$.

The old stub taught Cramer's Rule ($x_i=|A_i|/|A|$) as one of the chapter's core techniques — it has been removed from this rebuild entirely, matching exactly what the syllabus line itself asks for: "solving system of linear equations... using inverse of a matrix," nothing about determinant ratios.

The book's own Miscellaneous Exercise still points to questions that no longer exist. Its instructions read "Using properties of determinants in Exercises 11 to 15, prove that" and "Choose the correct answer in Exercise 17 to 19" — but the current exercise only runs from Q1 to Q9.

Several proof-based questions were dropped during the syllabus rationalisation, and the cross-references to their old numbers were never updated. Don't go looking for a "Question 12" or "Question 18" here; they were deleted, and the connecting sentences are leftover text, the same kind of rationalisation residue as Chapter 1's stray "binary operations" mention.

The old stub also collapsed all six real exercises into one invented 35-question group with no solutions file behind it. The book actually has Exercise 4.1 (8 questions, computing determinants), Exercise 4.2 (5 questions, area of a triangle), Exercise 4.3 (5 questions, minors and cofactors), Exercise 4.4 (18 questions, adjoint and inverse), Exercise 4.5 (14 questions, consistency and solving systems), and a Miscellaneous Exercise (9 questions).

59 questions in total — the largest exercise set in Class 12 Mathematics so far.

2. What this chapter covers

3. Computing a determinant

For a $3\times 3$ matrix, expanding along the first row:

$|A|=a_1\left|\begin{array}{cc}{b}_2& c_2\\ b_3& c_3\end{array}\right|-b_1\left|\begin{array}{cc}{a}_2& c_2\\ a_3& c_3\end{array}\right|+c_1\left|\begin{array}{cc}{a}_2& b_2\\ a_3& b_3\end{array}\right|$

You can expand along any row or column — the result is always the same, and picking a row or column with more zeros makes the arithmetic shorter. This is what "using cofactors" means formally: each entry $a_{ij}$ pairs with its cofactor $A_{ij}=(-1{)}^{i+j}{M}_{ij}$, where $M_{ij}$ (the minor) is the determinant left after deleting row $i$ and column $j$.

The smaller orders (4.2.1 to 4.2.2). A $1\times 1$ matrix $[a]$ has determinant $a$. A $2\times 2$ matrix has:

$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$

which is the building block every $3\times 3$ expansion reduces to.

The sign pattern. The factor $(-1{)}^{i+j}$ produces the checkerboard that must be applied before summing:

$\left[\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right]$

Forgetting the minus on the middle term of a first-row expansion is the single commonest arithmetic error in the chapter.

Properties of determinants. These are what make large determinants tractable, and several exercise questions are unreasonable without them:

• If any row or column is entirely zero, $|A|=0$.
• Interchanging two rows (or two columns) reverses the sign of the determinant.
• If two rows (or columns) are identical or proportional, $|A|=0$.
• Multiplying one row or column by $k$ multiplies the determinant by $k$ — hence $|kA|=k^n|A|$ for an $n\times n$ matrix, not $k|A|$.
• Adding a multiple of one row to another leaves the determinant unchanged. This is the workhorse for creating zeros before expanding.
• $|A^{\prime }|=|A|$: transposing does not change the determinant, which is why row and column operations are equally valid.
• $|AB|=|A||B|$, so the determinant of a product is the product of the determinants.

Why $|kA|=k^n|A|$. Scaling the whole matrix scales every one of the $n$ rows, and each row contributes one factor of $k$. Treating it as $k|A|$ is a frequent and heavily penalised error.

4. Area of a triangle, and the adjoint

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is:

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Since area can't be negative, take the absolute value of the determinant — and since three collinear points make a degenerate triangle, $\Delta =0$ is exactly the collinearity test.

The adjoint of $A$ is the transpose of its cofactor matrix: $\text{adj}(A)=[A_{ij}{]}^T$. It satisfies $A\cdot \text{adj}(A)=\text{adj}(A)\cdot A=|A|I$, which is exactly what makes the inverse formula work:

$A^{-1}=\frac{1}{|A|}\text{adj}(A),\text{provided }|A|\ne 0$

A matrix is called singular if $|A|=0$ and non-singular otherwise — and $A$ is invertible if and only if it is non-singular.

Minors and cofactors are not the same thing (4.4). The minor $M_{ij}$ is a determinant; the cofactor $A_{ij}=(-1{)}^{i+j}{M}_{ij}$ is that determinant with a sign attached. Questions ask for one or the other specifically, and answering with the unsigned minor where a cofactor was wanted loses the mark.

A useful consequence. Expanding along a row using the cofactors of a different row always gives zero:

$a_{i1}{A}_{j1}+a_{i2}{A}_{j2}+a_{i3}{A}_{j3}=0\text{when }i\ne j$

This is precisely what makes the off-diagonal entries of $A\cdot \text{adj}(A)$ vanish, leaving $|A|I$.

Properties of the adjoint. For an $n\times n$ matrix, $|\text{adj}(A)|=|A{|}^{n-1}$, and $\text{adj}(AB)=\text{adj}(B)\text{adj}(A)$ — the order reverses, as it does for transposes and inverses.

Checking an inverse cheaply. Since $|A^{-1}|=1/|A|$, a quick determinant check catches most arithmetic slips before you commit to the full adjoint computation.

A worked line of the derivation. The equation $A\cdot \text{adj}(A)=|A|I$ rearranges directly: dividing both sides by the scalar $|A|$ gives $A\cdot \frac{\text{adj}(A)}{|A|}=I$, so the matrix multiplying $A$ to give $I$ is $\frac{1}{|A|}\text{adj}(A)$, which is the definition of $A^{-1}$. The requirement $|A|\ne 0$ is visible here as the prohibition on dividing by zero.

5. Solving a linear system and checking consistency

Write the system as $AX=B$. If $A$ is non-singular, the system has a unique solution $X=A^{-1}B$, and it's automatically consistent. If $A$ is singular ($|A|=0$), check $(\text{adj}A)B$: if it's non-zero, the system is inconsistent (no solution); if it's the zero vector, the system may or may not be consistent, depending on further detail the book doesn't ask you to resolve.

The three outcomes, summarised. For $AX=B$:

Homogeneous systems. When $B=O$, the system $AX=O$ always has the trivial solution $X=O$. It has non-trivial solutions precisely when $|A|=0$, which is how "find $k$ such that the system has a non-zero solution" questions are answered — set the determinant to zero and solve for the parameter.

Worked, mirroring the textbook's own final example. Solve $x-y+2z=7$, $3x+4y-5z=-5$, $2x-y+3z=12$. Here $A=\left[\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right]$ is non-singular, so $X=A^{-1}B$ gives the unique solution directly — no need to fall back on any determinant-ratio shortcut.

Summary

• $|A|$ expands along any row or column using cofactors $A_{ij}=(-1{)}^{i+j}{M}_{ij}$; the result is the same regardless of which row/column you pick.
• Area of a triangle from its vertices is $\frac{1}{2}|\det [\dots ]|$; a zero determinant means the three points are collinear.
• $\text{adj}(A)$ is the transpose of the cofactor matrix; $A\cdot \text{adj}(A)=|A|I$.
• $A^{-1}=\text{adj}(A)/|A|$, valid only when $|A|\ne 0$ (i.e. $A$ is non-singular).
• Solve $AX=B$ via $X=A^{-1}B$ when $A$ is non-singular; when $A$ is singular, check $(\text{adj}A)B$ to distinguish "no solution" from "possibly consistent."
• A $2\times 2$ determinant is $ad-bc$; the $(-1{)}^{i+j}$ checkerboard must be applied before summing a $3\times 3$ expansion.
• Determinant properties: a zero row gives $0$; swapping rows flips the sign; identical or proportional rows give $0$; adding a multiple of one row to another changes nothing.
• $|kA|=k^n|A|$ for an $n\times n$ matrix, because every one of the $n$ rows is scaled — not $k|A|$.
• $|A^{\prime }|=|A|$ and $|AB|=|A||B|$.
• A minor is unsigned; a cofactor carries $(-1{)}^{i+j}$ — answer the one the question actually asks for.
• $|\text{adj}(A)|=|A{|}^{n-1}$, $\text{adj}(AB)=\text{adj}(B)\text{adj}(A)$, and $|A^{-1}|=1/|A|$.
• $AX=O$ has non-trivial solutions exactly when $|A|=0$ — the basis of every "find $k$ for a non-zero solution" question.
• Cramer's Rule is not part of the current book — the matrix method above is the only technique taught or examined here.
• The Miscellaneous Exercise's own instruction lines reference questions 11-15 and 17-19 that don't exist in the current 9-question version — a rationalisation leftover, not something missing from your copy of the book.