Class 12 Matrices: Exercise 3.2 Solved Questions

This lesson works through several questions from Exercise 3.2 on matrices. It covers adding matrices whose entries form perfect squares, simplifying matrices built from trigonometric functions, and solving matrix equations for unknown matrices and unknown variables.

Adding matrices with square identities

When two matrices are added, you add the corresponding entries. In this question each resulting entry matches a perfect square identity, $a^{2}+b^{2}+2ab=(a+b)^{2}$ and $a^{2}+b^{2}-2ab=(a-b)^{2}$.

$\begin{bmatrix} a^{2}+b^{2}+2ab & b^{2}+c^{2}+2bc \\ a^{2}+c^{2}-2ac & a^{2}+b^{2}-2ab \end{bmatrix} = \begin{bmatrix} (a+b)^{2} & (b+c)^{2} \\ (a-c)^{2} & (a-b)^{2} \end{bmatrix}$

A matrix of trigonometric squares

Given a matrix whose entries are $\cos^{2}x$ and $\sin^{2}x$, adding it to its partner so that each entry becomes $\sin^{2}x+\cos^{2}x$ uses the identity $\sin^{2}x+\cos^{2}x=1$.

$\begin{bmatrix} \cos^{2}x & \sin^{2}x \\ \sin^{2}x & \cos^{2}x \end{bmatrix} + \begin{bmatrix} \sin^{2}x & \cos^{2}x \\ \cos^{2}x & \sin^{2}x \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$

Simplifying a sum of two trigonometric products

Multiplying out and adding two matrices of $\sin\theta$ and $\cos\theta$ terms gives entries that simplify with $\cos^{2}\theta+\sin^{2}\theta=1$ and with the cross terms cancelling.

Diagonal entries: $\cos^{2}\theta+\sin^{2}\theta=1$.

Off-diagonal entries: $\sin\theta\cos\theta+(-\sin\theta\cos\theta)=0$.

$\begin{bmatrix} \cos^{2}\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^{2}\theta \end{bmatrix} + \begin{bmatrix} \sin^{2}\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^{2}\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Finding X and Y from their sum and difference

Given $X+Y=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$ and $X-Y=\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$, treat these like two linear equations and use elimination.

Add the two equations to remove $Y$:

$2X = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}, \qquad X = \tfrac{1}{2}\begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$

Substitute $X$ back into $X+Y$ to find $Y$:

$Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$

Solving for an unknown matrix

Given $Y=\begin{bmatrix} 1 & 2 \\ 1 & 4 \end{bmatrix}$ and $2X+Y=\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$, substitute the value of $Y$ and isolate $2X$.

$2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ -4 & -2 \end{bmatrix}$

Divide every entry by $2$:

$X = \tfrac{1}{2}\begin{bmatrix} 0 & -2 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -2 & -1 \end{bmatrix}$

Comparing entries to find x, y, z and t

Given $2\begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$, carry out the scalar multiplication and addition.

$\begin{bmatrix} 2x+3 & 2z-3 \\ 2y & 2t+6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$

Compare corresponding entries to get four equations:

$2x+3=9 \Rightarrow x=3, \qquad 2z-3=15 \Rightarrow z=9$ $2y=12 \Rightarrow y=6, \qquad 2t+6=18 \Rightarrow t=6$

Finding x, y, z and w from a matrix equation

Given $3\begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix}$, compare corresponding entries.

Top left: $3x=x+4 \Rightarrow 2x=4 \Rightarrow x=2$.

Top right: $3y=6+x+y \Rightarrow 2y=6+2=8 \Rightarrow y=4$.

Bottom right: $3w=2w+3 \Rightarrow w=3$.

Bottom left: $3z=-1+z+w \Rightarrow 2z=-1+w=-1+3=2 \Rightarrow z=1$.

So $x=2$, $y=4$, $z=1$, $w=3$.

Key takeaways

• Matrix addition works entry by entry, so look for algebraic and trigonometric identities hiding in the entries.
• A sum and a difference of two matrices behave like two linear equations: add to eliminate one matrix, then substitute back.
• Two matrices are equal only when every corresponding entry is equal, which turns one matrix equation into a set of ordinary equations to solve.