Matrices and Determinants

In ordinary numbers, $ab=ac$ with $a\ne 0$ forces $b=c$. Cancel and move on.

Try it with matrices.

$A=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right),B=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right),C=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$

$AB=\left(\begin{array}{cc}2& 1\\ 2& 1\end{array}\right),AC=\left(\begin{array}{cc}2& 1\\ 2& 1\end{array}\right)$

So $AB=AC$, and $A$ is very far from the zero matrix. Cancel $A$ and you conclude $B=C$, which is plainly false.

One number explains it.

$\det A=(1)(1)-(1)(1)=0$

A matrix is an action, and its determinant says whether that action can be undone. This $A$ sends every point $(x,y)$ to $(x+y,x+y)$, collapsing the whole plane onto a single line. Once two different points have been sent to the same place, no amount of cleverness recovers which one you started from, so there is nothing to cancel with.

That single collapse wears four costumes, and the syllabus puts them in one unit for exactly that reason.

Roughly two of the twenty-five Mathematics questions come from here, and the arithmetic is heavy enough that method choice decides whether you finish in one minute or four.

1. Matrices and Their Types

Matrix. A rectangular array of numbers, of order $m\times n$ meaning $m$ rows and $n$ columns. It is not a number and has no value; it is an arrangement.

Trap. Skew-symmetry forces $a_{ii}=-a_{ii}$, so every diagonal entry of a skew-symmetric matrix is zero. Questions are built on that single consequence.

Transpose $A^T$: rows become columns. Two facts about it are worth committing.

$(AB{)}^T=B^T{A}^T,A=\underset{\text{symmetric}}{\underbrace{\frac{1}{2}(A+A^T)}}+\underset{\text{skew}}{\underbrace{\frac{1}{2}(A-A^T)}}$

The order reversal in the first is not a typo; it is forced by the shapes.

Illustration 1

Show that $A+A^T$ is symmetric and that $A{A}^T$ is symmetric, for any matrix $A$, square or not.

For the sum, transpose it and use the rule that transposing twice returns the original.

$(A+A^T{)}^T=A^T+(A^T{)}^T=A^T+A=A+A^T$

For the product, use the order-reversing rule.

$(A{A}^T{)}^T=(A^T{)}^T{A}^T=A{A}^T$

The second result holds even when $A$ is $3\times 5$: then $A{A}^T$ is $3\times 3$ and $A^{T}A$ is $5\times 5$, and both are symmetric while being different sizes.

This is why every decomposition question works. Any square matrix splits into a symmetric part and a skew part in exactly one way, and the halves are forced on you by adding and subtracting the two expressions.

Illustration 2

Identify the type of each matrix, and show that an orthogonal matrix always has determinant $\pm 1$.

$P=\left(\begin{array}{cc}2& -2\\ 1& -1\end{array}\right),Q=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),R=\left(\begin{array}{cc}0& 3\\ 0& 0\end{array}\right)$

$P^2=\left(\begin{array}{cc}4-2& -4+2\\ 2-1& -2+1\end{array}\right)=P$, so $P$ is idempotent.

$Q^2=I$, so $Q$ is involutory; it is also orthogonal, since $Q^T=Q$.

$R^2=O$, so $R$ is nilpotent, and it is a second witness that $AB=O$ does not force $A$ or $B$ to be zero.

For orthogonality, take determinants of $A{A}^T=I$ and use the two rules that a product multiplies and a transpose preserves.

$\det A\cdot \det A^T=1\Rightarrow (\det A{)}^2=1\Rightarrow \det A=\pm 1$

Determinant $\pm 1$ means an orthogonal matrix scales no area at all, which fits: it is a rotation or a reflection, and neither stretches anything.

2. Algebra of Matrices

Addition, subtraction and scalar multiplication act entrywise and require identical orders.

The row-by-column rule

$(AB{)}_{ij}={\sum }_k{a}_{ik}{b}_{kj}$

The entry in row $i$, column $j$ of the product comes from row $i$ of $A$ against column $j$ of $B$.

$\underset{A}{\underbrace{m\times n}}\cdot \underset{B}{\underbrace{n\times p}}=\underset{AB}{\underbrace{m\times p}}$

The inner numbers must match; the outer numbers survive. That one line settles every "is this product defined" question.

Matrix multiplication is associative and distributive, but three familiar properties fail.

All three failures have the same root, and it is the one from the opening: a singular matrix destroys information.

Illustration 3

Show that $A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ and $B=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ do not commute, even though both are invertible.

$AB=\left(\begin{array}{cc}1+1& 1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right),BA=\left(\begin{array}{cc}1& 1\\ 1& 1+1\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$

Both have determinant $1$, so neither destroys anything, and yet the products differ.

Non-commutativity is therefore a separate phenomenon from singularity. $A$ shears horizontally and $B$ shears vertically, and doing them in the other order lands somewhere else, exactly as turning then walking differs from walking then turning.

Trap. $(A+B{)}^2=A^2+2AB+B^2$ is false in general. The correct expansion is $A^2+AB+BA+B^2$, and the middle terms combine only if $A$ and $B$ commute.

3. Determinants of Order Two and Three

A determinant is a number extracted from a square matrix. A matrix is an array; its determinant is a value.

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

For order three, expand along any row or column using cofactors, where the cofactor $C_{ij}$ is the minor $M_{ij}$ with the sign $(-1{)}^{i+j}$ attached.

$\det A=a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}$

The sign pattern alternates from the top left, and expanding along the row or column containing the most zeros saves the most work.

Illustration 4

Evaluate $\left|\begin{array}{ccc}2& 1& 4\\ 0& 5& 0\\ 3& 7& 1\end{array}\right|$.

Expanding along the first row costs three $2\times 2$ determinants. The second row already carries two zeros, so expand along it instead.

The middle entry sits at position $(2,2)$, whose sign board symbol is $+$.

$=+5\left|\begin{array}{cc}2& 4\\ 3& 1\end{array}\right|=5(2-12)=-50$

One product replaced six. Expansion along any row or column gives the same value, so the only question is which line costs least, and a row of zeros with one survivor costs almost nothing.

4. Properties That Make Evaluation Fast

Full expansion of a $3\times 3$ determinant is six products. Properties usually replace all six with one.

The third row of that table is the workhorse: it lets you manufacture zeros for free before expanding.

$\det (kA)=k^n\det A\text{ for order }n,\det (AB)=\det A\cdot \det B$

Trap. $\det (kA)$ is not $k\det A$. Every one of the $n$ rows gets multiplied, so the factor comes out $n$ times.

Illustration 5

Evaluate $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ without full expansion.

Every column sums to $a+b+c$, which is the signal to add all rows into one.

$R_1\to R_1+R_2+R_3\Rightarrow \left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ b& c& a\\ c& a& b\end{array}\right|$

Take the common factor out of the new first row.

$=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|$

Now clear the first row with column operations, which change nothing.

$C_2\to C_2-C_1,C_3\to C_3-C_1\Rightarrow (a+b+c)\left|\begin{array}{ccc}1& 0& 0\\ b& c-b& a-b\\ c& a-c& b-c\end{array}\right|$

Expanding along the first row now costs one $2\times 2$ determinant.

$=(a+b+c)\left[(c-b)(b-c)-(a-b)(a-c)\right]=-\left(a^3+b^3+c^3-3abc\right)$

The factor $(a+b+c)$ appeared before any expansion, which is the whole point. It also predicts the answer's structure: $a^3+b^3+c^3-3abc$ is known to have $(a+b+c)$ as a factor.

Illustration 6

Given $\det A=5$ for a $3\times 3$ matrix, use $\det (AB)=\det A\det B$ to explain why a singular matrix can never have an inverse.

Suppose $M$ is singular, so $\det M=0$, and suppose an inverse $N$ existed with $MN=I$.

$\det M\cdot \det N=\det I=1\Rightarrow 0\cdot \det N=1$

No number satisfies that, so no such $N$ exists. The multiplicative property turned a matrix question into an arithmetic impossibility.

The same property gives the scaling result instantly: $\det (3A)=3^3\det A=27\times 5=135$, because tripling a $3\times 3$ matrix triples all three rows.

5. The Determinant as a Scale Factor

Geometrically, a $2\times 2$ matrix maps the unit square to a parallelogram, and $|\det A|$ is that parallelogram's area. In three dimensions it is a volume.

The sign records orientation: negative means the plane was flipped over.

$\det A=0⟺\text{area collapses to }0⟺\text{everything lands on a line}$

Area of a triangle

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

Setting that determinant to zero is the collinearity condition: zero area means the three points lie on a line. This is the third costume, and it is the same collapse as before.

Illustration 7

Find the area of the triangle with vertices $(1,1)$, $(4,5)$ and $(6,2)$, then shift all three vertices by $(+10,-7)$ and recompute.

$2\text{Area}=\left|\begin{array}{ccc}1& 1& 1\\ 4& 5& 1\\ 6& 2& 1\end{array}\right|$

Subtract the first row from the other two, which changes nothing and kills the last column.

$=\left|\begin{array}{ccc}1& 1& 1\\ 3& 4& 0\\ 5& 1& 0\end{array}\right|=1\times (3\times 1-4\times 5)=-17$

$\text{Area}=\frac{17}{2}=8.5$

After the shift the vertices are $(11,-6)$, $(14,-2)$ and $(16,-5)$. Subtracting the first row from the others gives exactly the same two rows $(3,4,0)$ and $(5,1,0)$ as before.

The translation cancelled in the row differences, so the area is unchanged, as geometry demands. That is also why the row-subtraction step is worth doing first: it strips out the position of the triangle and leaves only its shape.

6. Adjoint and Inverse

Adjoint. $\mathrm{adj}A$ is the transpose of the matrix of cofactors. Compute every cofactor, arrange them, then transpose.

$A(\mathrm{adj}A)=(\mathrm{adj}A)A=(\det A)I$

$A^{-1}=\frac{1}{\det A}\mathrm{adj}A,\det A\ne 0$

Singular means $\det A=0$ and no inverse exists. Non-singular means $\det A\ne 0$ and the inverse exists and is unique.

Illustration 8

For a $2\times 2$ matrix the inverse has a shortcut: swap the diagonal entries, negate the other two, divide by the determinant. Show where it comes from, and why it does not extend to $3\times 3$.

The cofactors of $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ are $d$, $-c$, $-b$, $a$, arranged as $\left(\begin{array}{cc}d& -c\\ -b& a\end{array}\right)$. Transposing gives the adjoint.

$\mathrm{adj}\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)\Rightarrow A^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$

The shortcut works because each cofactor of a $2\times 2$ matrix is a single entry, so no determinants are involved at all.

For a $3\times 3$ matrix each of the nine cofactors is itself a $2\times 2$ determinant, so no rearrangement of entries can produce them. There is no shortcut, and the adjoint has to be computed properly.

7. Consistency and Solving Linear Systems

Write $AX=B$ with $A$ the coefficient matrix. Cramer's rule uses determinants alone.

$x=\frac{{\Delta }_1}{\Delta },y=\frac{{\Delta }_2}{\Delta },z=\frac{{\Delta }_3}{\Delta }$

where $\Delta =\det A$ and ${\Delta }_i$ replaces the $i$-th column of $A$ by the constants.

The same three cases are reached by the matrix inversion method, which is what a JEE question usually names.

$AX=B\Rightarrow A^{-1}(AX)=A^{-1}B\Rightarrow X=A^{-1}B$

That derivation quietly assumes $A^{-1}$ exists, which is why the method answers only the first row of the table. When $\det A=0$ there is nothing to multiply by, and the question becomes which of the other two rows applies.

Trap. The last row is usually infinitely many, not always. All determinants can vanish while the equations still contradict each other, so a system that reaches that row must be checked directly by substitution.

What the three cases look like

Each equation in three unknowns is a plane, and the solution set is where all three meet.

Illustration 9

Two systems share the same coefficients. Explain why one has no solution and the other has infinitely many.

$\text{(i)}\{\begin{array}{l}x+2y=3\\ 2x+4y=7\end{array}\text{(ii)}\{\begin{array}{l}x+2y=3\\ 2x+4y=6\end{array}$

Both have $\Delta =(1)(4)-(2)(2)=0$, so neither has a unique solution. The constants decide the rest.

In (i), doubling the first equation gives $2x+4y=6$, contradicting $2x+4y=7$. The two lines are parallel and distinct, and ${\Delta }_1=\left|\begin{array}{cc}3& 2\\ 7& 4\end{array}\right|=-2\ne 0$ confirms it.

In (ii), the second equation is twice the first, so it adds nothing. Here ${\Delta }_1=\left|\begin{array}{cc}3& 2\\ 6& 4\end{array}\right|=0$ and ${\Delta }_2=0$ as well, and every point of the line $x+2y=3$ is a solution.

$\Delta =0\text{ says the lines are parallel};\text{the }{\Delta }_i\text{ say whether they are the same line}$

Homogeneous systems

$AX=O$, with all constants zero.

$X=O$ always works, so a homogeneous system is never inconsistent. The only question is whether anything else works.

$\Delta \ne 0\Rightarrow \text{only the trivial solution};\Delta =0\Rightarrow \text{infinitely many non-trivial ones}$

Illustration 10

Solve $2x+3y=0$ and $4x+6y=0$, and say what the answer looks like geometrically.

$\Delta =(2)(6)-(3)(4)=0$

so non-trivial solutions exist. The second equation is twice the first, so there is really only one condition.

$2x+3y=0\Rightarrow (x,y)=t(3,-2)\text{ for any real }t$

Both "planes" are the same line through the origin, and the solution set is that entire line rather than a point.

Note what could not have happened. Two parallel but distinct lines would need different constants, and homogeneous systems have none: every equation is forced through the origin, which is why they always have at least the trivial solution.

Illustration 11

Find $A^{-1}$ for $A=\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)$ without computing an adjoint, using $A^2=(\mathrm{tr}A)A-(\det A)I$.

Here $\mathrm{tr}A=2+2=4$ and $\det A=4-3=1$, so the identity reads $A^2=4A-I$.

Verify it once: $A^2=\left(\begin{array}{cc}7& 12\\ 4& 7\end{array}\right)$, and $4A-I=\left(\begin{array}{cc}8& 12\\ 4& 8\end{array}\right)-\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}7& 12\\ 4& 7\end{array}\right)$.

Now rearrange to isolate $I$ and read off the inverse.

$A^2-4A=-I\Rightarrow A(4I-A)=I\Rightarrow A^{-1}=4I-A=\left(\begin{array}{cc}2& -3\\ -1& 2\end{array}\right)$

Check: $\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)\left(\begin{array}{cc}2& -3\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}4-3& -6+6\\ 2-2& -3+4\end{array}\right)=I$.

The identity holds for every $2\times 2$ matrix, so it also gives $A^2$, $A^3$ and any power in terms of $A$ and $I$, which is far faster than repeated multiplication.

Summary

A matrix is an action; its determinant is the number saying whether the action can be undone. Zero means information was destroyed.

That single collapse explains all four failures: no cancellation, no inverse, zero area, no unique solution.

Only square matrices have determinants. Skew-symmetric matrices have zero diagonals, and any square matrix splits uniquely into symmetric and skew parts.

For products, inner dimensions must match and outer ones survive. $(AB{)}^T=B^T{A}^T$ and $(AB{)}^{-1}=B^{-1}{A}^{-1}$ both reverse the order.

$AB\ne BA$ in general even for invertible matrices, so $(A+B{)}^2=A^2+AB+BA+B^2$.

Idempotent, involutory, nilpotent and orthogonal are defined by one equation each, and an orthogonal matrix has determinant plus or minus one because it scales no area.

Expand along the row or column already carrying the most zeros; a cofactor is a minor with the sign board's symbol attached.

Row and column operations beat expansion. Adding a multiple of one row to another leaves the value unchanged, and manufacturing zeros first turns six products into one.

$\det (kA)=k^n\det A$, not $k\det A$. And $\det (AB)=\det A\det B$, which is why a singular matrix can never have an inverse.

The determinant is an area scale factor, so the triangle formula and the collinearity condition are the same statement read two ways.

$A(\mathrm{adj}A)=(\det A)I$ gives the inverse. The $2\times 2$ swap-and-negate shortcut works only because its cofactors are single entries.

The matrix inversion method gives $X=A^{-1}B$, but its derivation assumes the inverse exists, so it answers the unique-solution case only.

Non-zero $\Delta$ gives a unique solution. When $\Delta =0$, the ${\Delta }_i$ decide between no solution and infinitely many, and the all-zero case still deserves a substitution check.

Homogeneous systems are never inconsistent; $\Delta =0$ is exactly the condition for a non-trivial solution.

The identity $A^2=(\mathrm{tr}A)A-(\det A)I$ holds for every $2\times 2$ matrix and turns inverses and powers into linear combinations of $A$ and $I$.