Sure Questions on Matrices (Class 12)

This lesson works through several frequently asked Class 12 matrix questions: the commutative and associative laws of matrix addition, solving a matrix equation for two unknowns, and finding an unknown constant from a matrix product.

Commutative law of matrix addition

With $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 4 & -2 \\ 3 & 4 & 6 \end{bmatrix}$, we check that $A + B = B + A$.

Left side. Adding entry by entry:

$A + B = \begin{bmatrix} 1+1 & 2+4 & 3+(-2) \\ 2+3 & 4+4 & 5+6 \end{bmatrix} = \begin{bmatrix} 2 & 6 & 1 \\ 5 & 8 & 11 \end{bmatrix}$

Right side. Adding in the other order gives the same result:

$B + A = \begin{bmatrix} 1+1 & 4+2 & -2+3 \\ 3+2 & 4+4 & 6+5 \end{bmatrix} = \begin{bmatrix} 2 & 6 & 1 \\ 5 & 8 & 11 \end{bmatrix}$

Since both sides match, $A + B = B + A$. This is the commutative property of matrix addition.

Associative law of matrix addition

If $A$, $B$, $C$ are matrices of the same order, then $(A + B) + C = A + (B + C)$. Take

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & 5 \\ 6 & 7 \end{bmatrix}, \quad C = \begin{bmatrix} 0 & 1 \\ 2 & 3 \end{bmatrix}$

Left side. First $A + B = \begin{bmatrix} 5 & 7 \\ 9 & 11 \end{bmatrix}$, then add $C$:

$(A + B) + C = \begin{bmatrix} 5+0 & 7+1 \\ 9+2 & 11+3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 11 & 14 \end{bmatrix}$

Right side. First $B + C = \begin{bmatrix} 4 & 6 \\ 8 & 10 \end{bmatrix}$, then add $A$:

$A + (B + C) = \begin{bmatrix} 1+4 & 2+6 \\ 3+8 & 4+10 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 11 & 14 \end{bmatrix}$

Both sides give the same matrix, so $(A + B) + C = A + (B + C)$. This is the associative law.

Solving a matrix equation for x and y

Given

$x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$

the left side becomes $\begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix}$. Comparing entries gives two equations:

$2x - y = 10 \quad (1) \qquad 3x + y = 5 \quad (2)$

Adding $(1)$ and $(2)$ removes $y$:

$5x = 15 \implies x = 3$

Substituting $x = 3$ into $(1)$: $6 - y = 10$, so $-y = 4$, giving $y = -4$.

So $x = 3$ and $y = -4$. Check in $(2)$: $3(3) + (-4) = 5$. ✓

Finding k from a matrix squared

Given $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, find $k$ such that $A^2 = kA - 2I$.

Compute $A^2 = A \cdot A$ by multiplying rows into columns:

$A^2 = \begin{bmatrix} 3 \cdot 3 + (-2)(4) & 3(-2) + (-2)(-2) \\ 4 \cdot 3 + (-2)(4) & 4(-2) + (-2)(-2) \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$

Set up the right side. With $kA - 2I = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix}$.

Compare entries. Equating $A^2$ with $kA - 2I$ entry by entry:

$1 = 3k - 2 \implies 3k = 3 \implies k = 1$

The other entries agree too: $-2 = -2k$, $4 = 4k$, and $-4 = -2k - 2$ all give $k = 1$.

Key takeaways

• Matrix addition is commutative ($A + B = B + A$) and associative ($(A + B) + C = A + (B + C)$).
• A matrix equation in unknowns can be split into ordinary equations by comparing corresponding entries, then solved like simultaneous equations.
• To compare two equal matrices, match their corresponding entries; any one matching pair can be enough to solve for an unknown constant.