Download App

Algebra of Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices5 Marks
Question
Let $\text{A}=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix},$ compute $A^2 - B^2.$
✓

Answer

Given: $\text{A}=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+3-5&-1-3-5&1+3-5\\-3-9+15&3+9+15&-3-9+15\\-5+15+25&5-15+25&-5+15+25\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&-9&-1\\3&27&3\\35&15&35\end{bmatrix}$
$\text{B}^2=\text{BB}$
$\Rightarrow\text{B}^2=\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix}\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix}$
$ \Rightarrow\text{B}^2=\begin{bmatrix}0+4-3&0-12+12&0-12+12\\0-3+3&4+9-12&3+9-12\\0+4-4&-4-12+16&-3-12+16\end{bmatrix}$
$\Rightarrow\text{B}^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\text{A}^2-\text{B}^2$
$\Rightarrow\text{A}^2-\text{B}^2=\begin{bmatrix}-1&-9&-1\\3&27&3\\35&15&35\end{bmatrix}-\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{B}^2=\begin{bmatrix}-1-1&-9-0&-1-0\\3-0&27-1&3-0\\35-0&15-0&35-1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{B}^2=\begin{bmatrix}-2&-1&-9\\3&26&3\\35&15&34\end{bmatrix}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Algebra of Matrices→Algebra of Matrices — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Find the equation of the plane passing through the intersection of the planes.
$ \vec{r}.(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k})}=7,\text{ }\vec{r}.(2\hat{\text{i}}+5\hat{\text{j}}+3\hat{\text{k}})=9$and the point (2, 1, 3).
→
Evaluate the following intregals:
$\int\frac{1}{4\sin^2\text{x}+5\cos^2\text{x}}\text{ dx}$
→
Using integeration, find the area of the region bounded by the y - 1 = x, the x-axis and the ordinates x = -2 and x = 3.
→
Find the corrdinates of the points P where the line throught A(3, -4,-5) and B(2, -3, 1) crosses the plane passing throught three points L(2, 2, 1), M(3, 0, 1) and N(4, -1, 0). Also, find the ratio in which P diveides the line segment AB.
→
Express the matrix $B=\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right]$ as the sum of a symmetric and a skew $-$ symmetric matrix.
→
Find the equation of a normal to the curve $y = x \log_e x$ which is parallel to the line $2x − 2y + 3 = 0.$
→
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin2\text{x},\text{ y}=2,\text{ when x}=\frac{\pi}{2}$
→
Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=(1+\text{s}-\text{t})\hat{\text{t}}+(2-\text{s})\hat{\text{j}}+(3-2\text{s}+2\text{t})\hat{\text{k}}$
→
Find the angle between the following pairs of lines:$\frac{-\text{x}+2}{-2}=\frac{\text{y}-1}{7}=\frac{\text{z}+3}{-3}$ and $\frac{\text{x}+2}{-1}=\frac{2\text{y}-8}{4}=\frac{\text{z}-5}{4}$
→
A trust invested some money in two type of bonds. The first bond pays 10% interest and bond pays 12% interest. The trust received 2,800 as interest. However, if trust had interchanged money in bonds, they would have got 100 less as interest. Using matrix method, find the amount invested by the trust. Which value is reflected in this question?
→