9.1 Matrices & Determinants

Matrices & Determinants

The NTS Algebra block reads: "Fundamental of Geometry, Practical Geometry, Areas and Volumes, Matrices and Determinants, Basic Statistics, Arithmetic and Geometric Sequences, Linear Graphs, Sets and Functions" — 33% of the quantitative segment. Matrices are standard Pakistani intermediate mathematics, which is exactly why they appear here and never in GRE-derived material. If your preparation came from an international source, this topic is a blind spot.

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Order, Equality and Special Types

A matrix of $m$ rows and $n$ columns has order $m \times n$. Rows first, always.

Two matrices are equal only if they have the same order and every corresponding entry matches.

The transpose $A^{T}$ turns rows into columns. Note the reversal rule: $(AB)^{T} = B^{T}A^{T}$.

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Addition and Scalar Multiplication

Addition and subtraction are entry-by-entry and require identical order:

$$\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} + \begin{bmatrix} 4 & -1 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 6 & 4 \\ 1 & 9 \end{bmatrix}$$

A scalar multiplies every entry: $3\begin{bmatrix} 2 & -1 \\ 0 & 4\end{bmatrix} = \begin{bmatrix} 6 & -3 \\ 0 & 12\end{bmatrix}$. Multiplying only the first row or the diagonal is a common slip.

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Matrix Multiplication

Conformability: an $m \times n$ matrix can multiply an $n \times p$ matrix, giving an $m \times p$ result. The inner dimensions must match and they vanish; the outer dimensions survive.

$$(3 \times 2)(2 \times 4) = 3 \times 4 \qquad\qquad (3 \times 2)(3 \times 2) = \text{undefined}$$

Each entry of the product is a row-by-column dot product:

$$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 1(5)+2(7) & 1(6)+2(8) \\ 3(5)+4(7) & 3(6)+4(8) \end{bmatrix} = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}$$

Reversing the order gives a different answer:

$$\begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 23 & 34 \\ 31 & 46 \end{bmatrix}$$

Matrix multiplication is not commutative. It is associative, $(AB)C = A(BC)$, and distributive over addition. NTS items frequently ask which algebraic property fails, and the answer is commutativity.

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Determinants

For a $2 \times 2$ matrix:

$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \qquad |A| = ad - bc$$

For a $3 \times 3$ matrix, expand along the first row using minors and the alternating sign pattern $+\,-\,+$:

$$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)$$

Useful Determinant Facts

• Only square matrices have determinants.
• If any row or column is entirely zero, $|A| = 0$.
• If two rows or two columns are identical, $|A| = 0$.
• $|AB| = |A| \times |B|$.
• $|A^{T}| = |A|$.
• Multiplying a single row by $k$ multiplies the determinant by $k$; multiplying the whole $n \times n$ matrix by $k$ multiplies the determinant by $k^{n}$.

A matrix with $|A| = 0$ is called singular and has no inverse.

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The Inverse of a 2 by 2 Matrix

$$A^{-1} = \frac{1}{|A|}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

In words: swap the leading diagonal, negate the other diagonal, divide by the determinant.

For $A = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}$, the determinant is $4(5) - 3(2) = 14$, so

$$A^{-1} = \frac{1}{14}\begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}$$

The defining property is $AA^{-1} = A^{-1}A = I$, and the reversal rule applies here too: $(AB)^{-1} = B^{-1}A^{-1}$.

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Solving a Linear System by Matrices

Solve $\;3x + 2y = 16\;$ and $\;x + 4y = 12$.

Write it as $AX = B$:

$$\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 16 \\ 12 \end{bmatrix}, \qquad |A| = 12 - 2 = 10$$ $$X = A^{-1}B = \frac{1}{10}\begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} 16 \\ 12 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 64 - 24 \\ -16 + 36 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 40 \\ 20 \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix}$$

So $x = 4$, $y = 2$. Substituting back: $3(4) + 2(2) = 16$ ✓ and $4 + 4(2) = 12$ ✓.

Cramer's rule reaches the same place by replacing one column at a time:

$$x = \frac{\begin{vmatrix} 16 & 2 \\ 12 & 4\end{vmatrix}}{10} = \frac{64 - 24}{10} = 4, \qquad y = \frac{\begin{vmatrix} 3 & 16 \\ 1 & 12\end{vmatrix}}{10} = \frac{36 - 16}{10} = 2$$

Interpretation worth remembering: if $|A| = 0$ the system has no unique solution — either no solution at all or infinitely many, because the two lines are parallel or coincident. NTS asks this as a conceptual item without any computation.