Solution: $\left|\begin{array}{cc}\sqrt{3} & \sqrt{5} \\ -\sqrt{5} & 3 \sqrt{3}\end{array}\right| \cdot=3 \sqrt{3} \times \sqrt{3}-(-\sqrt{5} \times \sqrt{5})$ $=14$
Evaluate
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}14 & 9 \\ -8 & -7\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5d897e9c6160fa4dcad39ded81500421_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{cc} 14 & 9 \\ -8 & -7 \end{array}\right|=14 \times(-7)-9 \times(-8) \\ =-26 \end{array}$
For what value of
, the given matrix
is a singular matrix?
Solution: For $A$ to be singular matrix its determinant should be equal to 0 . $\begin{array}{l} 0=(3-2 x) \times 4-(x+1) \times 2 \\ 0=12-8 x-2 x-2 \\ 0=10-10 x \\ x=1 \end{array}$
Without expanding the determinant, prove that
. SINGULAR MATRIX A square matrix
is said to be singular if
. Also,
is called non singular if
.
Solution: We know that $C_{1} \Rightarrow C_{1}-C_{2}$, would not change anything for the determinant. Applying the same in above determinant, we get $\left[\begin{array}{lll}40 & 1 & 5 \\...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{lll}0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-32f51a6de1804af917fa79d89813b6ed_l3.png)
Solution: We know that expansion of determinant with respect to first row is $a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}$. $0(3 \times 6-5 \times 4)-2(2 \times 6-4 \times 4)+0(2 \times 5-4 \times 3)$...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\ \sin 75^{\circ} & \cos 75^{\circ}\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3d8be1e9a2eae4e88c55cbf66d16d191_l3.png)
Solution: $\begin{array}{l} \cos 15^{\circ} \cos 75^{\circ}-\sin 75^{\circ} \sin 15^{\circ} \\ =\cos \left(15^{\circ}+75^{\circ}\right) \because \cos A \cos B-\sin A \sin B=\cos (A+B) \\ =\cos...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}\cos 65^{\circ} & \sin 65^{\circ} \\ \sin 25^{\circ} & \cos 25^{\circ}\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-31eca4dd143debf57f6eb5cd3161f7aa_l3.png)
Solution: By directly opening this determinant $\begin{array}{l} \cos 65^{\circ} \times \cos 25^{\circ}-\sin 25^{\circ} \times \sin 65^{\circ} \\ =\cos \left(65^{\circ}+25^{\circ}\right) \because...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{ll}\sin 60^{\circ} & \cos 60^{\circ} \\ -\sin 30^{\circ} & \cos 30^{\circ}\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-88d0e92a3541f71a649886452cbc75a8_l3.png)
Solution: After finding determinant we will get, $\begin{array}{l} \operatorname{Sin} 60^{\circ}=\frac{\sqrt{3}}{2}=\cos 30^{\circ} \\ \operatorname{Cos} 60^{\circ}=\frac{1}{2}=\sin 30^{\circ} \\...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-59e09dfe6790101aa2e50ee530018d65_l3.png)
Solution: After finding determinant we will get a trigonometric identity. $\begin{array}{l} \cos ^{2} \alpha+\sin ^{2} \alpha \\ =1 \end{array}$ $\because \sin ^{2} \theta+\cos ^{2} \theta=1$
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}2 \cos \theta & -2 \sin \theta \\ \sin \theta & \cos \theta\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fb446379f26715435ce3931fc5938c4f_l3.png)
Solution: After finding determinant we will get a trigonometric identity. $\begin{array}{l} 2 \cos ^{2} \theta+2 \sin ^{2} \theta \\ =2 \\ \because \sin ^{2} \theta+\cos ^{2} \theta=1...
Evaluate
.
Solution: Find determinant $\begin{array}{l} \sqrt{6} \times \sqrt{24-\sqrt{2}} 20 \times \sqrt{5} \\ \sqrt{1} 144-\sqrt{1} 100 \\ =12-10 \\ =2 \end{array}$
Evaluate
.
Solution: It is determinant multiplied by a scalar number 2 , just find determinant of matrix and multiply it by 2 . $\begin{array}{l} 2 \times(35-20) \\ 2 \times 15=30 \end{array}$
If
, find the value of
.
Solution: Find the determinant of $A$ and then multiply it by 3 $\begin{array}{l} |A|=2 \\ 3|A|=3 \times 2 \\ =6 \end{array}$
If
, write the value of
.
Solution: Simply by equating both sides we can get the value of $x$. $\begin{array}{l} 2 x^{2}+2 x-2\left(x^{2}+4 x+3\right)=-12 \\ \Rightarrow-6 x-6=-12 \\ \Rightarrow-6 x=-6 \\ \Rightarrow x=1...
If
, write the value of
.
Solution: This question is having the same logic as above. $\begin{array}{l} 2 x^{2}-40=18+14 \\ \Rightarrow 2 x^{2}=72 \\ \Rightarrow x^{2}=36 \\ \Rightarrow x=\pm 6 \end{array}$
If
, write the value of
.
Solution: Here the determinant is compared so we need to take determinant both sides then find $\mathrm{x}$. $\begin{array}{l} 12 x+14=32-42 \\ \Rightarrow 12 x=-10-14 \\ \Rightarrow 12 x=-24 \\...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}\mathrm{a}+\mathrm{ib} & \mathrm{c}+\mathrm{id} \\ -\mathrm{c}+\mathrm{id} & \mathrm{a}-\mathrm{ib}\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-afe300cc28b1c8fe93675d27d2938b53_l3.png)
Solution: This we can very simply go through directly. $\begin{array}{l} ((a+i b)(a-i b))-((-c+i d)(c+i d)) \\ \Rightarrow\left(a^{2}+b^{2}\right)-\left(-c^{2}-d^{2}\right) \\ \Rightarrow...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}\mathrm{x}^{2}-\mathrm{x}+1 & \mathrm{x}-1 \\ \mathrm{x}+1 & \mathrm{x}+1\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a0399aa0c317584f6b5333ed893b4eff_l3.png)
Solution: Theorem: This evaluation can be done in two different ways either by taking out the common things anc then calculating the determinants or simply take determinant. I will prefer first...
If
is the cofactor of the element
of
then write the value of
.
Solution: Theorem: $A_{i j}$ is found by deleting $j^{t h}$ rowand $j^{t h}$ column, the determinant of left matrix is called cofactor with multiplied by $(-1)^{(i+j)}$ Given: $\mathrm{j}=3$ and...
If
is a
matrix such that
and
then write the value of
.
Solution: Theorem: If Let $A$ be $k \times k$ matrix then $|p A|=p^{k}|A|$. Given: $\mathrm{k}=3$ and $\mathrm{p}=3$. $\begin{array}{l} |3 \mathrm{~A}|=3^{3} \times|\mathrm{A}| \\ =27|\mathrm{~A}|...
If
is a
matrix such that
and
, write the value of
.
Solution: Theorem: If $A$ be $k \times k$ matrix then $|p A|=p^{k}|A|$. Given, $\mathrm{p}=4, \mathrm{k}=2$ and $|\mathrm{A}|=5$. $\begin{array}{l} |4 \mathrm{~A}|=4^{2} \times 5 \\ =16 \times 5 \\...