Scalar multiplication: Exercises — Maths deep dive

This is the practice ground for Scalar Multiplication of Matrices. Every problem below has a complete, revealed solution hidden inside a collapsible callout — try it first with a pen, then open. The exercises climb five levels: Recognition → Application → Analysis → Synthesis → Mastery. Do not skip levels; each rung assumes you own the one below it.

Reminders you'll lean on:

Addition/subtraction happen entry-by-entry and only between matrices of the same shape — see matrix addition.
The properties $k (A + B) = k A + k B$ , $(k_{1} + k_{2}) A = k_{1} A + k_{2} A$ , $(k_{1} k_{2}) A = k_{1} (k_{2} A)$ come from properties of matrix operations.

Level 1 — Recognition

Can you spot the rule and apply it once?

L1.1 Let $A = [20 - 1 4]$ . Compute $3 A$ .

L1.2 Which of these equals $(- 2) [3 - 5]$ ? (a) $[- 6 10]$ (b) $[- 6 - 10]$ (c) $[610]$

L1.3 State the dimensions of $7 B$ if $B$ is a $3 \times 5$ matrix.

Recall Solution L1.1

Multiply each entry by $3$ : $3 A = [3 \cdot 2 3 \cdot 0 3 \cdot (- 1) 3 \cdot 4] = [60 - 3 12]$ Notice $3 \cdot 0 = 0$ — zeros stay zero, but you still "touch" them.

Recall Solution L1.2

Multiply each entry by $- 2$ : $(- 2) (3) = - 6$ and $(- 2) (- 5) = + 10$ (negative times negative is positive). Answer: (a) $[- 6 10]$ .

Recall Solution L1.3

Scalar multiplication never changes shape. So $7 B$ is still $3 \times 5$ .

Level 2 — Application

Chain the rule with addition and fractions.

L2.1 Let $A = [1324]$ , $B = [02 - 1 5]$ . Compute $2 A + 3 B$ .

L2.2 Let $E = [84 - 12 20]$ . Compute $\frac{1}{4} E$ .

L2.3 Compute $5 A - 5 B$ using $A, B$ from L2.1, then check it equals $5 (A - B)$ .

Recall Solution L2.1

First scale each matrix, then add corresponding entries (both are $2 \times 2$ , so addition is legal). $2 A = [2648], 3 B = [06 - 3 15]$ $2 A + 3 B = [2 + 0 6 + 6 4 + (- 3) 8 + 15] = [212123]$

Recall Solution L2.2

Multiplying by $\frac{1}{4}$ is the same as dividing every entry by $4$ : $\frac{1}{4} E = [8/4 4/4 - 12/4 20/4] = [21 - 3 5]$

Recall Solution L2.3

Direct route: $5 A = [5151020], 5 B = [010 - 5 25]$ $5 A - 5 B = [5 - 0 15 - 10 10 - (- 5) 20 - 25] = [55 15 - 5]$ Check via distributivity: $A - B = [11 3 - 1]$ , and $5 (A - B) = [55 15 - 5]$ . ✓ They match — this is $k (A - B) = k A - k B$ in action.

Level 3 — Analysis

Now you must reason backwards or test a property.

L3.1 Find the scalar $k$ such that $k [39 - 6 12] = [13 - 2 4]$ .

L3.2 Verify $(k_{1} k_{2}) A = k_{1} (k_{2} A)$ for $k_{1} = 2$ , $k_{2} = - 3$ , $A = [1 - 2 05]$ , by computing both sides.

L3.3 For what value(s) of $k$ is $k A$ equal to the zero matrix, where $A = [4 - 1 72]$ ? Justify.

Recall Solution L3.1

Every entry on the right is the left entry times $k$ . Compare one pair: $k \cdot 3 = 1 \Rightarrow k = \frac{1}{3}$ . Check it works everywhere: $\frac{1}{3} \cdot (- 6) = - 2$ ✓, $\frac{1}{3} \cdot 9 = 3$ ✓, $\frac{1}{3} \cdot 12 = 4$ ✓. $k = \frac{1}{3}$ (If even one entry disagreed, no single scalar would exist.)

Recall Solution L3.2

Left side: $k_{1} k_{2} = 2 \cdot (- 3) = - 6$ , so $(- 6) A = [- 6 12 0 - 30]$ Right side: first $k_{2} A = - 3 A = [- 3 6 0 - 15]$ , then $2 \cdot (- 3 A) = [- 6 12 0 - 30]$ Both sides identical ✓ — this is just associativity of ordinary number multiplication happening inside each cell.

Recall Solution L3.3

$k A = O$ means every entry $k \cdot a_{ij} = 0$ . Since $A$ has non-zero entries (e.g. $a_{11} = 4$ ), we need $k \cdot 4 = 0$ , forcing $k = 0$ . $k = 0 only$ Because $A \neq = O$ , no non-zero scalar can annihilate it. (Contrast: if $A$ were already $O$ , any $k$ would work.)

Level 4 — Synthesis

Combine scalar multiplication with geometry and structure.

The unit square below has corners collected as columns of a matrix. Watch what scaling does to it.

L4.1 The points $P = (1, 0)$ and $Q = (0, 1)$ are stored as columns of $M = [1001]$ . Compute $2 M$ and describe geometrically what happened to the two points.

L4.2 Using the same $M$ , compute $(- 1) M$ and describe the geometric effect on the points.

L4.3 Solve for the matrix $X$ : $2 X + [4862] = [102012]$ .

Recall Solution L4.1

$2 M = [2002]$ The columns are now $(2, 0)$ and $(0, 2)$ . Both points moved twice as far from the origin along their own directions — the whole picture doubles in size (an expansion by factor $2$ ). See the outer teal square in the figure. This connection to stretching is exactly what linear transformations make precise.

Recall Solution L4.2

$(- 1) M = [- 1 0 0 - 1]$ Columns become $(- 1, 0)$ and $(0, - 1)$ : each point is flung to the opposite side of the origin — a reflection through the origin (a $180°$ turn). See the plum points in the figure.

Recall Solution L4.3

Isolate $X$ like an ordinary equation. Subtract the known matrix from both sides: $2 X = [102012] - [4862] = [6 - 6 - 6 10]$ Now divide every entry by $2$ (multiply by $\frac{1}{2}$ ): $X = \frac{1}{2} [6 - 6 - 6 10] = [3 - 3 - 3 5]$

Level 5 — Mastery

General reasoning, edge cases, and proof-flavoured problems.

L5.1 Let $A$ be any $2 \times 2$ matrix. Prove that $3 A - A = 2 A$ using the entry-wise definition.

L5.2 A matrix $A$ satisfies $k A = A$ for a scalar $k \neq = 1$ . What must be true about $A$ ? Prove it.

L5.3 Find all scalars $k$ such that $k^{2} A = 9 A$ , where $A = [13 2 - 1]$ (note $A \neq = O$ ).

Recall Solution L5.1

Write $A = [a_{ij}]$ . Look at a single entry of the left side: $(3 A - A)_{ij} = 3 a_{ij} - a_{ij} = (3 - 1) a_{ij} = 2 a_{ij} = (2 A)_{ij} .$ This used $(k_{1} - k_{2}) A = k_{1} A - k_{2} A$ (distributivity over scalar subtraction). Since every entry matches, $3 A - A = 2 A$ . ∎

Recall Solution L5.2

Entry-wise, $k A = A$ means $k a_{ij} = a_{ij}$ for every $i, j$ , i.e. $(k - 1) a_{ij} = 0$ . Since $k \neq = 1$ , the factor $(k - 1) \neq = 0$ . For a product to be zero with a non-zero factor, the other factor must be zero: $a_{ij} = 0$ for all $i, j$ . Therefore $A = O$ (the zero matrix). ∎

Recall Solution L5.3

$k^{2} A = 9 A$ means $(k^{2} - 9) A = O$ . Since $A \neq = O$ (it has non-zero entries), the scalar must vanish: $k^{2} - 9 = 0 \Rightarrow k^{2} = 9 \Rightarrow k = 3 or k = - 3.$ Both work: $3^{2} = 9$ and $(- 3)^{2} = 9$ . $k = 3 or k = - 3$

Wrap-up

Recall One-line summary of every level

L1 apply the rule ::: multiply each entry by $k$ L2 combine with add/fractions ::: scale each matrix fully, then add entry-wise L3 reason backwards ::: a single $k$ must satisfy every entry equation L4 geometry & equations ::: scaling = stretch/reflect; isolate then reciprocal-scale L5 proofs & edge cases ::: turn $(scalar) A = O$ into "scalar $= 0$ or $A = O$ "

Next stop after you've mastered scaling and adding: how matrices multiply each other in matrix multiplication — a genuinely different operation, and a classic source of the "L1 trap" grown up.