MCQ
Solve system of linear equations, using matrix method. $4 x-3 y=3$ ; $3 x-5 y=7$
- A$x=\frac{6}{11},y=\frac{-19}{11}$
- B$x=\frac{-6}{11},y=\frac{19}{11}$
- C$x=\frac{6}{11},y=\frac{19}{11}$
- ✓$x=\frac{-6}{11},y=\frac{-19}{11}$
$A=\left[\begin{array}{ll}4 & -3 \\ 3 & -5\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 7\end{array}\right]$
Now,
$|A|=-20+9=-11 \neq 0$
Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$A^{-1}=\frac{1}{|A|}(a d j A)=-\frac{1}{11}\left[\begin{array}{ll}
-5 & 3 \\
-3 & 4
\end{array}\right]=\frac{1}{11}\left[\begin{array}{ll}
5 & -3 \\
3 & -4
\end{array}\right]$
$\therefore X=A^{-1} B=\frac{1}{11}\left[\begin{array}{ll}5 & -3 \\ 3 & -4\end{array}\right]\left[\begin{array}{l}3 \\ 7\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{ll}5 & -3 \\ 3 & -4\end{array}\right]\left[\begin{array}{l}3 \\ 7\end{array}\right]$
$=\frac{1}{11}\left[\begin{array}{c}15-21 \\ 9-28\end{array}\right]$
$=\frac{1}{11}\left[\begin{array}{l}-6 \\ -19\end{array}\right]$
$=\left[\begin{array}{r}-\frac{6}{11} \\ -\frac{19}{11}\end{array}\right]$
Hence, $x=\frac{-6}{11}$ and $y=\frac{-19}{11}$
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If
$\text{P}(\text{A})=\frac{2}{5},\text{P}(\text{B})=\frac{3}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5},$ then $\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)\cdot\text{P}\Big(\frac{\text{B}'}{\text{A}'}\Big)$ is equas:Match each entry in $List-I$ to the correct entry in $List-II$.
| $List-I$ | $List-II$ |
| ($P$) The number of matrices $M=\left(a_{i j}\right)_3 \times 3$ with all entries in $T$ such that $R_i=C_j=0$ for all $i, j$ is | ($1$) ($1$) |
| ($Q$) The number of symmetric matrices $M=\left(a_{i j}\right) 3 \times 3$ with all entries in $T$ such that $C_j=0$ for all $j$ is | ($2$) ($2$) |
| ($R$) Let $M=\left(a_{i j}\right) 3 \times 3$ be a skew symmetric matrix such that $a_{i j} \in T$ for $i>j$. Then the number of elements in the set $\left\{\left(\begin{array}{l}x \\ y \\ z\end{array}\right): x, y \cdot z \in R, M\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}a_{12} \\ 0 \\ -a_{23}\end{array}\right)\right\}$ is is | ($3$) Infinite |
| ($S$) Let $M=\left(a_{i j}\right)_3 \times 3$ be a matrix with all entries in $T$ such that $R_i=0$ for all $i$. Then the absolute value of the determinant of $M$ is | ($4$) ($6$) |
| ($5$) ($0$) |
The correct option is
$(S1)$: $f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2$
$( S 2): \int_{-2}^{2} f ( x ) dx =12$Then,