Download App

DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Find the matrix X satisfying the matrix equation.
$\text{X}\begin{bmatrix}5 & 3 \\-1 & -2 \end{bmatrix}\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}$
✓

Answer

$\text{X}\begin{bmatrix}5 & 3 \\-1 & -2 \end{bmatrix}\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}=-10+3=-7\neq0$
Hence, A is invertible.
If $C_{ij}$ is a cofactor of $a_{ij}$ in A, then $C_{11} = -2, C_{12} = 1, C_{21} = -3$ and $C_{22} = 5$.
Now, $\text{adj A}\begin{bmatrix}-2 & 1 \\-3 & 5 \end{bmatrix}^\text{T}=\begin{bmatrix}-2 & -3 \\1 & 5 \end{bmatrix}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\frac{-1}{7}\begin{bmatrix}-2 & -3 \\1 & 5 \end{bmatrix}$
Let: $\text{B}=\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}$
$\Rightarrow\ \text{B}=\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}=98-49=49\neq0$
Hence, B is invertible.
The given matrix equation becomes XA = B.
$\Rightarrow (XA)A^{-1} = BA^{-1}$
$\Rightarrow\ \text{X(AA}^{-1})=\text{B}=\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}\times\frac{-1}{7}\times\begin{bmatrix}-2 & -3 \\1 & 5 \end{bmatrix}$
$\Rightarrow\ \text{X}=\frac{-1}{7}\begin{bmatrix}-28+7 & -42+35 \\ -14+7 & -21+35 \end{bmatrix}$
$\Rightarrow\ \text{X}=\frac{-1}{7}\begin{bmatrix}-21 & -7 \\ -7 & 14 \end{bmatrix}$
$\Rightarrow\ \text{X}=\begin{bmatrix} 3 & 1 \\1 & -2 \end{bmatrix}$

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 DETERMINANTS→DETERMINANTS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Differentiate $\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$ with respect to $\sec^{-1}\text{x}$
→
If sin y = x sin (a + y), prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^{2}\text{(a + y)}}{\sin\text{a}}.$
→
Evaluate the following:
$\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
→
Show that $\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=0$ where a, b, c are in A.P.
→
Write the following in the simplest form:
$\tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}},-\text{a}<\text{x}<\text{a}$
→
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{(\sin^3\text{x}\cos^3\text{x})}\text{ dx}$
→
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{y}=0,\text{y}(0)=2,\text{y}'(0)=0$Function $\text{y}=\text{e}^\text{x}+\text{e}^{-\text{x}}$
→
Show that the areas under the curves $\text{y}=\sin\text{x}\text{ and }\text{y}=\sin2\text{x}$ between $x = 0$ and $\text{x}=\frac{\pi}{3}$ are in the ratio $2 : 3$.
→
If $\sqrt{\text{y}+\text{x}}+\sqrt{\text{y}-\text{x}}=\text{c},$ show that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$
→
Find a unit vector perpendicular to each of the vector $\vec a + \vec b\;$ and $\vec a - \vec b$, where $\vec a = 3\hat i + 2\hat j + 2\hat k\;$ and $\;\vec b = \hat i + 2\hat j - 2\hat k$.
→